An integral for which I don't trust WolframAlpha The Next CEO of Stack OverflowIntegral $int_0^4 int_sqrty^2 y^2 e^x^7 operatorname d!x operatorname d!y,$Holomorphic Functions and Cauchy TheoremShould it be obvious that this integral is zero?How to integrate the following? $int_0^+inftyfrac1-cos xx^alpha+1,dx$Complex integral with Residues TheoremEvaluation of integral related to Gamma functionGive a recursive formula for calculating the value of $ int_0^pi sin^n x , mathrm dx$Convergence of This integral?Function defined by Cauchy IntegralNeed verification for this integral
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An integral for which I don't trust WolframAlpha
The Next CEO of Stack OverflowIntegral $int_0^4 int_sqrty^2 y^2 e^x^7 operatorname d!x operatorname d!y,$Holomorphic Functions and Cauchy TheoremShould it be obvious that this integral is zero?How to integrate the following? $int_0^+inftyfrac1-cos xx^alpha+1,dx$Complex integral with Residues TheoremEvaluation of integral related to Gamma functionGive a recursive formula for calculating the value of $ int_0^pi sin^n x , mathrm dx$Convergence of This integral?Function defined by Cauchy IntegralNeed verification for this integral
$begingroup$
WolframAlpha tells me that $$int_-infty^infty e^-(t-i)^4 , mathrmdt = 0.$$ I am suspicious of this, because of the idea that for nice enough $f$, since $$int_-infty^infty f(t-a) , mathrmdt$$ is holomorphic in $a$ and takes the same value for all $a in mathbbR$, it should be independent of $a$ by the identity principle; and then $$int_-infty^infty e^-(t-i)^4 , mathrmdt = int_-infty^infty e^-t^4 , mathrmdt = 2 cdot Gamma(5/4).$$ But WolframAlpha even tells me numerically that $int_-infty^infty e^-(t-i)^4 , mathrmdt$ is close to $0$. Am I doing something wrong?
integration complex-analysis proof-verification
asked Jun 26 '17 at 3:01
user457218user457218
17525
$endgroup$
|
show 6 more comments
$begingroup$
WolframAlpha tells me that $$int_-infty^infty e^-(t-i)^4 , mathrmdt = 0.$$ I am suspicious of this, because of the idea that for nice enough $f$, since $$int_-infty^infty f(t-a) , mathrmdt$$ is holomorphic in $a$ and takes the same value for all $a in mathbbR$, it should be independent of $a$ by the identity principle; and then $$int_-infty^infty e^-(t-i)^4 , mathrmdt = int_-infty^infty e^-t^4 , mathrmdt = 2 cdot Gamma(5/4).$$ But WolframAlpha even tells me numerically that $int_-infty^infty e^-(t-i)^4 , mathrmdt$ is close to $0$. Am I doing something wrong?
integration complex-analysis proof-verification
asked Jun 26 '17 at 3:01
user457218user457218
17525
$endgroup$
3
$begingroup$
The integrals are equal provided that $i$ is a real number, but it isn't.
$endgroup$
– WW1
Jun 26 '17 at 3:23
2
$begingroup$
@WW1 That shouldn't matter by the identity principle, or as User8128 suggests (essentially) the contour of integration can be shifted in the $y$ direction without changing the integral, since it only matters that it connects $mathrmRe[z] = -infty$ and $mathrmRe[z] = infty.$ I am pretty confident in either argument but I thought I would ask here just to double check. (I usually trust wolframalpha blindly so it is irritating when something strange like this turns up)
$endgroup$
– user457218
Jun 26 '17 at 4:06
3
$begingroup$
@WW1: The argument for $a=i$ isn't about making a change of variable: it's about analytic continuation.
$endgroup$
– Hurkyl
Jun 26 '17 at 10:04
1
$begingroup$
For the fun type $int_-infty^infty e^-(t-colorred1.0i)^4 ,dt $ and you will get the value !!!
$endgroup$
– Claude Leibovici
Jun 26 '17 at 10:07
1
$begingroup$
They seem to have already corrected this.
$endgroup$
– sophros
Jun 26 '17 at 17:11
|
show 6 more comments
$begingroup$
WolframAlpha tells me that $$int_-infty^infty e^-(t-i)^4 , mathrmdt = 0.$$ I am suspicious of this, because of the idea that for nice enough $f$, since $$int_-infty^infty f(t-a) , mathrmdt$$ is holomorphic in $a$ and takes the same value for all $a in mathbbR$, it should be independent of $a$ by the identity principle; and then $$int_-infty^infty e^-(t-i)^4 , mathrmdt = int_-infty^infty e^-t^4 , mathrmdt = 2 cdot Gamma(5/4).$$ But WolframAlpha even tells me numerically that $int_-infty^infty e^-(t-i)^4 , mathrmdt$ is close to $0$. Am I doing something wrong?
integration complex-analysis proof-verification
asked Jun 26 '17 at 3:01
user457218user457218
17525
$endgroup$
WolframAlpha tells me that $$int_-infty^infty e^-(t-i)^4 , mathrmdt = 0.$$ I am suspicious of this, because of the idea that for nice enough $f$, since $$int_-infty^infty f(t-a) , mathrmdt$$ is holomorphic in $a$ and takes the same value for all $a in mathbbR$, it should be independent of $a$ by the identity principle; and then $$int_-infty^infty e^-(t-i)^4 , mathrmdt = int_-infty^infty e^-t^4 , mathrmdt = 2 cdot Gamma(5/4).$$ But WolframAlpha even tells me numerically that $int_-infty^infty e^-(t-i)^4 , mathrmdt$ is close to $0$. Am I doing something wrong?
integration complex-analysis proof-verification
integration complex-analysis proof-verification
asked Jun 26 '17 at 3:01
user457218user457218
17525
asked Jun 26 '17 at 3:01
user457218user457218
17525
asked Jun 26 '17 at 3:01
user457218user457218
17525
asked Jun 26 '17 at 3:01
user457218user457218
17525
asked Jun 26 '17 at 3:01
user457218user457218
17525
17525
3
$begingroup$
The integrals are equal provided that $i$ is a real number, but it isn't.
$endgroup$
– WW1
Jun 26 '17 at 3:23
2
$begingroup$
@WW1 That shouldn't matter by the identity principle, or as User8128 suggests (essentially) the contour of integration can be shifted in the $y$ direction without changing the integral, since it only matters that it connects $mathrmRe[z] = -infty$ and $mathrmRe[z] = infty.$ I am pretty confident in either argument but I thought I would ask here just to double check. (I usually trust wolframalpha blindly so it is irritating when something strange like this turns up)
$endgroup$
– user457218
Jun 26 '17 at 4:06
3
$begingroup$
@WW1: The argument for $a=i$ isn't about making a change of variable: it's about analytic continuation.
$endgroup$
– Hurkyl
Jun 26 '17 at 10:04
1
$begingroup$
For the fun type $int_-infty^infty e^-(t-colorred1.0i)^4 ,dt $ and you will get the value !!!
$endgroup$
– Claude Leibovici
Jun 26 '17 at 10:07
1
$begingroup$
They seem to have already corrected this.
$endgroup$
– sophros
Jun 26 '17 at 17:11
|
show 6 more comments
3
$begingroup$
The integrals are equal provided that $i$ is a real number, but it isn't.
$endgroup$
– WW1
Jun 26 '17 at 3:23
2
$begingroup$
@WW1 That shouldn't matter by the identity principle, or as User8128 suggests (essentially) the contour of integration can be shifted in the $y$ direction without changing the integral, since it only matters that it connects $mathrmRe[z] = -infty$ and $mathrmRe[z] = infty.$ I am pretty confident in either argument but I thought I would ask here just to double check. (I usually trust wolframalpha blindly so it is irritating when something strange like this turns up)
$endgroup$
– user457218
Jun 26 '17 at 4:06
3
$begingroup$
@WW1: The argument for $a=i$ isn't about making a change of variable: it's about analytic continuation.
$endgroup$
– Hurkyl
Jun 26 '17 at 10:04
1
$begingroup$
For the fun type $int_-infty^infty e^-(t-colorred1.0i)^4 ,dt $ and you will get the value !!!
$endgroup$
– Claude Leibovici
Jun 26 '17 at 10:07
1
$begingroup$
They seem to have already corrected this.
$endgroup$
– sophros
Jun 26 '17 at 17:11
3
3
$begingroup$
The integrals are equal provided that $i$ is a real number, but it isn't.
$endgroup$
– WW1
Jun 26 '17 at 3:23
$begingroup$
The integrals are equal provided that $i$ is a real number, but it isn't.
$endgroup$
– WW1
Jun 26 '17 at 3:23
2
2
$begingroup$
@WW1 That shouldn't matter by the identity principle, or as User8128 suggests (essentially) the contour of integration can be shifted in the $y$ direction without changing the integral, since it only matters that it connects $mathrmRe[z] = -infty$ and $mathrmRe[z] = infty.$ I am pretty confident in either argument but I thought I would ask here just to double check. (I usually trust wolframalpha blindly so it is irritating when something strange like this turns up)
$endgroup$
– user457218
Jun 26 '17 at 4:06
$begingroup$
@WW1 That shouldn't matter by the identity principle, or as User8128 suggests (essentially) the contour of integration can be shifted in the $y$ direction without changing the integral, since it only matters that it connects $mathrmRe[z] = -infty$ and $mathrmRe[z] = infty.$ I am pretty confident in either argument but I thought I would ask here just to double check. (I usually trust wolframalpha blindly so it is irritating when something strange like this turns up)
$endgroup$
– user457218
Jun 26 '17 at 4:06
3
3
$begingroup$
@WW1: The argument for $a=i$ isn't about making a change of variable: it's about analytic continuation.
$endgroup$
– Hurkyl
Jun 26 '17 at 10:04
$begingroup$
@WW1: The argument for $a=i$ isn't about making a change of variable: it's about analytic continuation.
$endgroup$
– Hurkyl
Jun 26 '17 at 10:04
1
1
$begingroup$
For the fun type $int_-infty^infty e^-(t-colorred1.0i)^4 ,dt $ and you will get the value !!!
$endgroup$
– Claude Leibovici
Jun 26 '17 at 10:07
$begingroup$
For the fun type $int_-infty^infty e^-(t-colorred1.0i)^4 ,dt $ and you will get the value !!!
$endgroup$
– Claude Leibovici
Jun 26 '17 at 10:07
1
1
$begingroup$
They seem to have already corrected this.
$endgroup$
– sophros
Jun 26 '17 at 17:11
$begingroup$
They seem to have already corrected this.
$endgroup$
– sophros
Jun 26 '17 at 17:11
|
show 6 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Just as User8128 commented, I believe that you are perfectly correct.
Without changing variable and using another CAS
$$int mathrme^-(t-i)^4 , mathrmdt=colorred-frac(t-i), Gamma left(frac14,(t-i)^4right)4 sqrt[4](t-i)^4=-frac14 (t-i) E_frac34left((t-i)^4right)$$
$$int_-infty^infty mathrme^-(t-i)^4 , mathrmdt=i left(E_frac34(-4)-Releft(E_frac34(-4)right)right)$$ $$int_-infty^infty mathrme^-(t-i)^4 , mathrmdt=frac(-1)^1/4, Gamma left(frac14,-4right)sqrt2-i
Releft(-frac(-1)^3/4, Gamma left(frac14,-4right)sqrt2right)=2, Gamma left(frac54right) $$ Notice that, for the indefinite integral, Wolfram Alpha correctly reports the expression written in red.
What is interesting is that, if you ask Wolfram Alpha
$$int_0^infty mathrme^-(t-i)^4 , mathrmdt=-frac i4E_frac34(-1)$$
$$int^0_-infty mathrme^-(t-i)^4 , mathrmdt=+frac i4E_frac34(-1)$$ leading to the $0$ you obtained.
There is obviously a bug that I suggest you to report.
edited Mar 18 at 10:57
YuiTo Cheng
2,1512937
answered Jun 26 '17 at 3:37
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
3
$begingroup$
The bug is kind of interesting. Just below i it still works, e.g. for 0.99 i see here at 1 i it fails.
$endgroup$
– Jannick
Jun 26 '17 at 9:36
10
$begingroup$
@Jannick. Type $1.0i$ and it works !!!
$endgroup$
– Claude Leibovici
Jun 26 '17 at 10:09
$begingroup$
Yes true, that is really funny. For integer one it already fails but for float one it still works. For 1.01 it fails however.
$endgroup$
– Jannick
Jun 26 '17 at 10:13
3
$begingroup$
@Jannick.**Funny** may not be the most appropriate term here ! Cheers and thanks for your previous comment which made me trying $1.0i$.
$endgroup$
– Claude Leibovici
Jun 26 '17 at 10:15
1
$begingroup$
@MeniRosenfeld: The original bug does, at the least in Mathematica 9.Integrate[Exp[-(t - I)^4], t, -Infinity, Infinity]yields 0. NIntegrate with the same argument returns 1.8128 + very little i, just like what you see.
$endgroup$
– user22961
Jun 26 '17 at 18:27
|
show 6 more comments
$begingroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
int_-infty^inftyexpo-parst - ic^4dd t & =
int_0^inftybracksexpo-parst - ic^4 +
expo-pars-t - ic^4dd t =
2,Reint_0^inftyexpo-parst + ic^4dd t =
2,Reint_ic^infty + icexpo-t^4dd t
\[5mm] & =
-2,lim_R to inftyReint_1^0expo-parsR + ic y^4ic,dd y -
2,Reint_infty^0expo-t^4dd t -
2,Reint_0^1expo-parsic y^4ic,dd y
\[5mm] & =
-2,lim_R to inftyImint_0^1expo-parsR + ic y^4,dd y +
1 over 2int_0^inftyt^-3/4expo-t,dd t
\[5mm] & =
-2,lim_R to inftyImint_0^1exppars-R^4 - 4R^3yic + 6R^2y^2 + 4Ry^3ic - y^4,dd y +
1 over 2,Gammapars1 over 4
\[1cm] & =
1 over 2,Gammapars1 over 4
\[2mm] & -
2lim_R to inftybraces%
expo-R^4int_0^1expparsy^2bracks6R^2 - y^2
sinpars4Rybracksy^2 - R^2,dd y
endalign
- $dsy^2pars6R^2 - y^2$ has a maximum at $dsy_m = root3R$.
$dsy_m > 1$ when $dsR > root3 over 3$.
- It vanishes at $dsy = 0$ and at $dsy = root6R$.
- It $dsto -infty$ when $dsy to infty$.
- It's clear that $ds0 < y^2pars6R^2 - y^2 < 6R^2 - 1$ when $dsy in pars0,1$ with $dsR > root3 over 3$.
Then,
beginalign
0 & < vertsexpo-R^4int_0^1expparsy^2bracks6R^2 - y^2
sinpars4Rybracksy^2 - R^2,dd y_ R > !root3/3
\[5mm] & <
expo-R^4int_0^1exppars1^2bracks6R^2 - 1^2,dd y =
exppars-R^4 + 6R^2 - 1
,,,stackrelmrmas R to inftyto,,,
color#f00large 0
endalign
beginalign
mboxsuch thatquad
bbxint_-infty^inftyexpo-parst - ic^4dd t =
1 over 2,Gammapars1 over 4 approx 1.8128
endalign
answered Jun 26 '17 at 6:27
Felix MarinFelix Marin
68.8k7109146
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Just as User8128 commented, I believe that you are perfectly correct.
Without changing variable and using another CAS
$$int mathrme^-(t-i)^4 , mathrmdt=colorred-frac(t-i), Gamma left(frac14,(t-i)^4right)4 sqrt[4](t-i)^4=-frac14 (t-i) E_frac34left((t-i)^4right)$$
$$int_-infty^infty mathrme^-(t-i)^4 , mathrmdt=i left(E_frac34(-4)-Releft(E_frac34(-4)right)right)$$ $$int_-infty^infty mathrme^-(t-i)^4 , mathrmdt=frac(-1)^1/4, Gamma left(frac14,-4right)sqrt2-i
Releft(-frac(-1)^3/4, Gamma left(frac14,-4right)sqrt2right)=2, Gamma left(frac54right) $$ Notice that, for the indefinite integral, Wolfram Alpha correctly reports the expression written in red.
What is interesting is that, if you ask Wolfram Alpha
$$int_0^infty mathrme^-(t-i)^4 , mathrmdt=-frac i4E_frac34(-1)$$
$$int^0_-infty mathrme^-(t-i)^4 , mathrmdt=+frac i4E_frac34(-1)$$ leading to the $0$ you obtained.
There is obviously a bug that I suggest you to report.
edited Mar 18 at 10:57
YuiTo Cheng
2,1512937
answered Jun 26 '17 at 3:37
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
3
$begingroup$
The bug is kind of interesting. Just below i it still works, e.g. for 0.99 i see here at 1 i it fails.
$endgroup$
– Jannick
Jun 26 '17 at 9:36
10
$begingroup$
@Jannick. Type $1.0i$ and it works !!!
$endgroup$
– Claude Leibovici
Jun 26 '17 at 10:09
$begingroup$
Yes true, that is really funny. For integer one it already fails but for float one it still works. For 1.01 it fails however.
$endgroup$
– Jannick
Jun 26 '17 at 10:13
3
$begingroup$
@Jannick.**Funny** may not be the most appropriate term here ! Cheers and thanks for your previous comment which made me trying $1.0i$.
$endgroup$
– Claude Leibovici
Jun 26 '17 at 10:15
1
$begingroup$
@MeniRosenfeld: The original bug does, at the least in Mathematica 9.Integrate[Exp[-(t - I)^4], t, -Infinity, Infinity]yields 0. NIntegrate with the same argument returns 1.8128 + very little i, just like what you see.
$endgroup$
– user22961
Jun 26 '17 at 18:27
|
show 6 more comments
$begingroup$
Just as User8128 commented, I believe that you are perfectly correct.
Without changing variable and using another CAS
$$int mathrme^-(t-i)^4 , mathrmdt=colorred-frac(t-i), Gamma left(frac14,(t-i)^4right)4 sqrt[4](t-i)^4=-frac14 (t-i) E_frac34left((t-i)^4right)$$
$$int_-infty^infty mathrme^-(t-i)^4 , mathrmdt=i left(E_frac34(-4)-Releft(E_frac34(-4)right)right)$$ $$int_-infty^infty mathrme^-(t-i)^4 , mathrmdt=frac(-1)^1/4, Gamma left(frac14,-4right)sqrt2-i
Releft(-frac(-1)^3/4, Gamma left(frac14,-4right)sqrt2right)=2, Gamma left(frac54right) $$ Notice that, for the indefinite integral, Wolfram Alpha correctly reports the expression written in red.
What is interesting is that, if you ask Wolfram Alpha
$$int_0^infty mathrme^-(t-i)^4 , mathrmdt=-frac i4E_frac34(-1)$$
$$int^0_-infty mathrme^-(t-i)^4 , mathrmdt=+frac i4E_frac34(-1)$$ leading to the $0$ you obtained.
There is obviously a bug that I suggest you to report.
edited Mar 18 at 10:57
YuiTo Cheng
2,1512937
answered Jun 26 '17 at 3:37
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
3
$begingroup$
The bug is kind of interesting. Just below i it still works, e.g. for 0.99 i see here at 1 i it fails.
$endgroup$
– Jannick
Jun 26 '17 at 9:36
10
$begingroup$
@Jannick. Type $1.0i$ and it works !!!
$endgroup$
– Claude Leibovici
Jun 26 '17 at 10:09
$begingroup$
Yes true, that is really funny. For integer one it already fails but for float one it still works. For 1.01 it fails however.
$endgroup$
– Jannick
Jun 26 '17 at 10:13
3
$begingroup$
@Jannick.**Funny** may not be the most appropriate term here ! Cheers and thanks for your previous comment which made me trying $1.0i$.
$endgroup$
– Claude Leibovici
Jun 26 '17 at 10:15
1
$begingroup$
@MeniRosenfeld: The original bug does, at the least in Mathematica 9.Integrate[Exp[-(t - I)^4], t, -Infinity, Infinity]yields 0. NIntegrate with the same argument returns 1.8128 + very little i, just like what you see.
$endgroup$
– user22961
Jun 26 '17 at 18:27
|
show 6 more comments
$begingroup$
Just as User8128 commented, I believe that you are perfectly correct.
Without changing variable and using another CAS
$$int mathrme^-(t-i)^4 , mathrmdt=colorred-frac(t-i), Gamma left(frac14,(t-i)^4right)4 sqrt[4](t-i)^4=-frac14 (t-i) E_frac34left((t-i)^4right)$$
$$int_-infty^infty mathrme^-(t-i)^4 , mathrmdt=i left(E_frac34(-4)-Releft(E_frac34(-4)right)right)$$ $$int_-infty^infty mathrme^-(t-i)^4 , mathrmdt=frac(-1)^1/4, Gamma left(frac14,-4right)sqrt2-i
Releft(-frac(-1)^3/4, Gamma left(frac14,-4right)sqrt2right)=2, Gamma left(frac54right) $$ Notice that, for the indefinite integral, Wolfram Alpha correctly reports the expression written in red.
What is interesting is that, if you ask Wolfram Alpha
$$int_0^infty mathrme^-(t-i)^4 , mathrmdt=-frac i4E_frac34(-1)$$
$$int^0_-infty mathrme^-(t-i)^4 , mathrmdt=+frac i4E_frac34(-1)$$ leading to the $0$ you obtained.
There is obviously a bug that I suggest you to report.
edited Mar 18 at 10:57
YuiTo Cheng
2,1512937
answered Jun 26 '17 at 3:37
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
Just as User8128 commented, I believe that you are perfectly correct.
Without changing variable and using another CAS
$$int mathrme^-(t-i)^4 , mathrmdt=colorred-frac(t-i), Gamma left(frac14,(t-i)^4right)4 sqrt[4](t-i)^4=-frac14 (t-i) E_frac34left((t-i)^4right)$$
$$int_-infty^infty mathrme^-(t-i)^4 , mathrmdt=i left(E_frac34(-4)-Releft(E_frac34(-4)right)right)$$ $$int_-infty^infty mathrme^-(t-i)^4 , mathrmdt=frac(-1)^1/4, Gamma left(frac14,-4right)sqrt2-i
Releft(-frac(-1)^3/4, Gamma left(frac14,-4right)sqrt2right)=2, Gamma left(frac54right) $$ Notice that, for the indefinite integral, Wolfram Alpha correctly reports the expression written in red.
What is interesting is that, if you ask Wolfram Alpha
$$int_0^infty mathrme^-(t-i)^4 , mathrmdt=-frac i4E_frac34(-1)$$
$$int^0_-infty mathrme^-(t-i)^4 , mathrmdt=+frac i4E_frac34(-1)$$ leading to the $0$ you obtained.
There is obviously a bug that I suggest you to report.
edited Mar 18 at 10:57
YuiTo Cheng
2,1512937
answered Jun 26 '17 at 3:37
Claude LeiboviciClaude Leibovici
125k1158135
edited Mar 18 at 10:57
YuiTo Cheng
2,1512937
edited Mar 18 at 10:57
YuiTo Cheng
2,1512937
edited Mar 18 at 10:57
YuiTo Cheng
2,1512937
2,1512937
answered Jun 26 '17 at 3:37
Claude LeiboviciClaude Leibovici
125k1158135
answered Jun 26 '17 at 3:37
Claude LeiboviciClaude Leibovici
125k1158135
answered Jun 26 '17 at 3:37
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
3
$begingroup$
The bug is kind of interesting. Just below i it still works, e.g. for 0.99 i see here at 1 i it fails.
$endgroup$
– Jannick
Jun 26 '17 at 9:36
10
$begingroup$
@Jannick. Type $1.0i$ and it works !!!
$endgroup$
– Claude Leibovici
Jun 26 '17 at 10:09
$begingroup$
Yes true, that is really funny. For integer one it already fails but for float one it still works. For 1.01 it fails however.
$endgroup$
– Jannick
Jun 26 '17 at 10:13
3
$begingroup$
@Jannick.**Funny** may not be the most appropriate term here ! Cheers and thanks for your previous comment which made me trying $1.0i$.
$endgroup$
– Claude Leibovici
Jun 26 '17 at 10:15
1
$begingroup$
@MeniRosenfeld: The original bug does, at the least in Mathematica 9.Integrate[Exp[-(t - I)^4], t, -Infinity, Infinity]yields 0. NIntegrate with the same argument returns 1.8128 + very little i, just like what you see.
$endgroup$
– user22961
Jun 26 '17 at 18:27
|
show 6 more comments
3
$begingroup$
The bug is kind of interesting. Just below i it still works, e.g. for 0.99 i see here at 1 i it fails.
$endgroup$
– Jannick
Jun 26 '17 at 9:36
10
$begingroup$
@Jannick. Type $1.0i$ and it works !!!
$endgroup$
– Claude Leibovici
Jun 26 '17 at 10:09
$begingroup$
Yes true, that is really funny. For integer one it already fails but for float one it still works. For 1.01 it fails however.
$endgroup$
– Jannick
Jun 26 '17 at 10:13
3
$begingroup$
@Jannick.**Funny** may not be the most appropriate term here ! Cheers and thanks for your previous comment which made me trying $1.0i$.
$endgroup$
– Claude Leibovici
Jun 26 '17 at 10:15
1
$begingroup$
@MeniRosenfeld: The original bug does, at the least in Mathematica 9.Integrate[Exp[-(t - I)^4], t, -Infinity, Infinity]yields 0. NIntegrate with the same argument returns 1.8128 + very little i, just like what you see.
$endgroup$
– user22961
Jun 26 '17 at 18:27
3
3
$begingroup$
The bug is kind of interesting. Just below i it still works, e.g. for 0.99 i see here at 1 i it fails.
$endgroup$
– Jannick
Jun 26 '17 at 9:36
$begingroup$
The bug is kind of interesting. Just below i it still works, e.g. for 0.99 i see here at 1 i it fails.
$endgroup$
– Jannick
Jun 26 '17 at 9:36
10
10
$begingroup$
@Jannick. Type $1.0i$ and it works !!!
$endgroup$
– Claude Leibovici
Jun 26 '17 at 10:09
$begingroup$
@Jannick. Type $1.0i$ and it works !!!
$endgroup$
– Claude Leibovici
Jun 26 '17 at 10:09
$begingroup$
Yes true, that is really funny. For integer one it already fails but for float one it still works. For 1.01 it fails however.
$endgroup$
– Jannick
Jun 26 '17 at 10:13
$begingroup$
Yes true, that is really funny. For integer one it already fails but for float one it still works. For 1.01 it fails however.
$endgroup$
– Jannick
Jun 26 '17 at 10:13
3
3
$begingroup$
@Jannick.**Funny** may not be the most appropriate term here ! Cheers and thanks for your previous comment which made me trying $1.0i$.
$endgroup$
– Claude Leibovici
Jun 26 '17 at 10:15
$begingroup$
@Jannick.**Funny** may not be the most appropriate term here ! Cheers and thanks for your previous comment which made me trying $1.0i$.
$endgroup$
– Claude Leibovici
Jun 26 '17 at 10:15
1
1
$begingroup$
@MeniRosenfeld: The original bug does, at the least in Mathematica 9.
Integrate[Exp[-(t - I)^4], t, -Infinity, Infinity] yields 0. NIntegrate with the same argument returns 1.8128 + very little i, just like what you see.$endgroup$
– user22961
Jun 26 '17 at 18:27
$begingroup$
@MeniRosenfeld: The original bug does, at the least in Mathematica 9.
Integrate[Exp[-(t - I)^4], t, -Infinity, Infinity] yields 0. NIntegrate with the same argument returns 1.8128 + very little i, just like what you see.$endgroup$
– user22961
Jun 26 '17 at 18:27
|
show 6 more comments
$begingroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
int_-infty^inftyexpo-parst - ic^4dd t & =
int_0^inftybracksexpo-parst - ic^4 +
expo-pars-t - ic^4dd t =
2,Reint_0^inftyexpo-parst + ic^4dd t =
2,Reint_ic^infty + icexpo-t^4dd t
\[5mm] & =
-2,lim_R to inftyReint_1^0expo-parsR + ic y^4ic,dd y -
2,Reint_infty^0expo-t^4dd t -
2,Reint_0^1expo-parsic y^4ic,dd y
\[5mm] & =
-2,lim_R to inftyImint_0^1expo-parsR + ic y^4,dd y +
1 over 2int_0^inftyt^-3/4expo-t,dd t
\[5mm] & =
-2,lim_R to inftyImint_0^1exppars-R^4 - 4R^3yic + 6R^2y^2 + 4Ry^3ic - y^4,dd y +
1 over 2,Gammapars1 over 4
\[1cm] & =
1 over 2,Gammapars1 over 4
\[2mm] & -
2lim_R to inftybraces%
expo-R^4int_0^1expparsy^2bracks6R^2 - y^2
sinpars4Rybracksy^2 - R^2,dd y
endalign
- $dsy^2pars6R^2 - y^2$ has a maximum at $dsy_m = root3R$.
$dsy_m > 1$ when $dsR > root3 over 3$.
- It vanishes at $dsy = 0$ and at $dsy = root6R$.
- It $dsto -infty$ when $dsy to infty$.
- It's clear that $ds0 < y^2pars6R^2 - y^2 < 6R^2 - 1$ when $dsy in pars0,1$ with $dsR > root3 over 3$.
Then,
beginalign
0 & < vertsexpo-R^4int_0^1expparsy^2bracks6R^2 - y^2
sinpars4Rybracksy^2 - R^2,dd y_ R > !root3/3
\[5mm] & <
expo-R^4int_0^1exppars1^2bracks6R^2 - 1^2,dd y =
exppars-R^4 + 6R^2 - 1
,,,stackrelmrmas R to inftyto,,,
color#f00large 0
endalign
beginalign
mboxsuch thatquad
bbxint_-infty^inftyexpo-parst - ic^4dd t =
1 over 2,Gammapars1 over 4 approx 1.8128
endalign
answered Jun 26 '17 at 6:27
Felix MarinFelix Marin
68.8k7109146
$endgroup$
add a comment |
$begingroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
int_-infty^inftyexpo-parst - ic^4dd t & =
int_0^inftybracksexpo-parst - ic^4 +
expo-pars-t - ic^4dd t =
2,Reint_0^inftyexpo-parst + ic^4dd t =
2,Reint_ic^infty + icexpo-t^4dd t
\[5mm] & =
-2,lim_R to inftyReint_1^0expo-parsR + ic y^4ic,dd y -
2,Reint_infty^0expo-t^4dd t -
2,Reint_0^1expo-parsic y^4ic,dd y
\[5mm] & =
-2,lim_R to inftyImint_0^1expo-parsR + ic y^4,dd y +
1 over 2int_0^inftyt^-3/4expo-t,dd t
\[5mm] & =
-2,lim_R to inftyImint_0^1exppars-R^4 - 4R^3yic + 6R^2y^2 + 4Ry^3ic - y^4,dd y +
1 over 2,Gammapars1 over 4
\[1cm] & =
1 over 2,Gammapars1 over 4
\[2mm] & -
2lim_R to inftybraces%
expo-R^4int_0^1expparsy^2bracks6R^2 - y^2
sinpars4Rybracksy^2 - R^2,dd y
endalign
- $dsy^2pars6R^2 - y^2$ has a maximum at $dsy_m = root3R$.
$dsy_m > 1$ when $dsR > root3 over 3$.
- It vanishes at $dsy = 0$ and at $dsy = root6R$.
- It $dsto -infty$ when $dsy to infty$.
- It's clear that $ds0 < y^2pars6R^2 - y^2 < 6R^2 - 1$ when $dsy in pars0,1$ with $dsR > root3 over 3$.
Then,
beginalign
0 & < vertsexpo-R^4int_0^1expparsy^2bracks6R^2 - y^2
sinpars4Rybracksy^2 - R^2,dd y_ R > !root3/3
\[5mm] & <
expo-R^4int_0^1exppars1^2bracks6R^2 - 1^2,dd y =
exppars-R^4 + 6R^2 - 1
,,,stackrelmrmas R to inftyto,,,
color#f00large 0
endalign
beginalign
mboxsuch thatquad
bbxint_-infty^inftyexpo-parst - ic^4dd t =
1 over 2,Gammapars1 over 4 approx 1.8128
endalign
answered Jun 26 '17 at 6:27
Felix MarinFelix Marin
68.8k7109146
$endgroup$
add a comment |
$begingroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
int_-infty^inftyexpo-parst - ic^4dd t & =
int_0^inftybracksexpo-parst - ic^4 +
expo-pars-t - ic^4dd t =
2,Reint_0^inftyexpo-parst + ic^4dd t =
2,Reint_ic^infty + icexpo-t^4dd t
\[5mm] & =
-2,lim_R to inftyReint_1^0expo-parsR + ic y^4ic,dd y -
2,Reint_infty^0expo-t^4dd t -
2,Reint_0^1expo-parsic y^4ic,dd y
\[5mm] & =
-2,lim_R to inftyImint_0^1expo-parsR + ic y^4,dd y +
1 over 2int_0^inftyt^-3/4expo-t,dd t
\[5mm] & =
-2,lim_R to inftyImint_0^1exppars-R^4 - 4R^3yic + 6R^2y^2 + 4Ry^3ic - y^4,dd y +
1 over 2,Gammapars1 over 4
\[1cm] & =
1 over 2,Gammapars1 over 4
\[2mm] & -
2lim_R to inftybraces%
expo-R^4int_0^1expparsy^2bracks6R^2 - y^2
sinpars4Rybracksy^2 - R^2,dd y
endalign
- $dsy^2pars6R^2 - y^2$ has a maximum at $dsy_m = root3R$.
$dsy_m > 1$ when $dsR > root3 over 3$.
- It vanishes at $dsy = 0$ and at $dsy = root6R$.
- It $dsto -infty$ when $dsy to infty$.
- It's clear that $ds0 < y^2pars6R^2 - y^2 < 6R^2 - 1$ when $dsy in pars0,1$ with $dsR > root3 over 3$.
Then,
beginalign
0 & < vertsexpo-R^4int_0^1expparsy^2bracks6R^2 - y^2
sinpars4Rybracksy^2 - R^2,dd y_ R > !root3/3
\[5mm] & <
expo-R^4int_0^1exppars1^2bracks6R^2 - 1^2,dd y =
exppars-R^4 + 6R^2 - 1
,,,stackrelmrmas R to inftyto,,,
color#f00large 0
endalign
beginalign
mboxsuch thatquad
bbxint_-infty^inftyexpo-parst - ic^4dd t =
1 over 2,Gammapars1 over 4 approx 1.8128
endalign
answered Jun 26 '17 at 6:27
Felix MarinFelix Marin
68.8k7109146
$endgroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
int_-infty^inftyexpo-parst - ic^4dd t & =
int_0^inftybracksexpo-parst - ic^4 +
expo-pars-t - ic^4dd t =
2,Reint_0^inftyexpo-parst + ic^4dd t =
2,Reint_ic^infty + icexpo-t^4dd t
\[5mm] & =
-2,lim_R to inftyReint_1^0expo-parsR + ic y^4ic,dd y -
2,Reint_infty^0expo-t^4dd t -
2,Reint_0^1expo-parsic y^4ic,dd y
\[5mm] & =
-2,lim_R to inftyImint_0^1expo-parsR + ic y^4,dd y +
1 over 2int_0^inftyt^-3/4expo-t,dd t
\[5mm] & =
-2,lim_R to inftyImint_0^1exppars-R^4 - 4R^3yic + 6R^2y^2 + 4Ry^3ic - y^4,dd y +
1 over 2,Gammapars1 over 4
\[1cm] & =
1 over 2,Gammapars1 over 4
\[2mm] & -
2lim_R to inftybraces%
expo-R^4int_0^1expparsy^2bracks6R^2 - y^2
sinpars4Rybracksy^2 - R^2,dd y
endalign
- $dsy^2pars6R^2 - y^2$ has a maximum at $dsy_m = root3R$.
$dsy_m > 1$ when $dsR > root3 over 3$.
- It vanishes at $dsy = 0$ and at $dsy = root6R$.
- It $dsto -infty$ when $dsy to infty$.
- It's clear that $ds0 < y^2pars6R^2 - y^2 < 6R^2 - 1$ when $dsy in pars0,1$ with $dsR > root3 over 3$.
Then,
beginalign
0 & < vertsexpo-R^4int_0^1expparsy^2bracks6R^2 - y^2
sinpars4Rybracksy^2 - R^2,dd y_ R > !root3/3
\[5mm] & <
expo-R^4int_0^1exppars1^2bracks6R^2 - 1^2,dd y =
exppars-R^4 + 6R^2 - 1
,,,stackrelmrmas R to inftyto,,,
color#f00large 0
endalign
beginalign
mboxsuch thatquad
bbxint_-infty^inftyexpo-parst - ic^4dd t =
1 over 2,Gammapars1 over 4 approx 1.8128
endalign
answered Jun 26 '17 at 6:27
Felix MarinFelix Marin
68.8k7109146
edited Jun 26 '17 at 6:33
answered Jun 26 '17 at 6:27
Felix MarinFelix Marin
68.8k7109146
answered Jun 26 '17 at 6:27
Felix MarinFelix Marin
68.8k7109146
answered Jun 26 '17 at 6:27
Felix MarinFelix Marin
68.8k7109146
68.8k7109146
add a comment |
add a comment |
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3
$begingroup$
The integrals are equal provided that $i$ is a real number, but it isn't.
$endgroup$
– WW1
Jun 26 '17 at 3:23
2
$begingroup$
@WW1 That shouldn't matter by the identity principle, or as User8128 suggests (essentially) the contour of integration can be shifted in the $y$ direction without changing the integral, since it only matters that it connects $mathrmRe[z] = -infty$ and $mathrmRe[z] = infty.$ I am pretty confident in either argument but I thought I would ask here just to double check. (I usually trust wolframalpha blindly so it is irritating when something strange like this turns up)
$endgroup$
– user457218
Jun 26 '17 at 4:06
3
$begingroup$
@WW1: The argument for $a=i$ isn't about making a change of variable: it's about analytic continuation.
$endgroup$
– Hurkyl
Jun 26 '17 at 10:04
1
$begingroup$
For the fun type $int_-infty^infty e^-(t-colorred1.0i)^4 ,dt $ and you will get the value !!!
$endgroup$
– Claude Leibovici
Jun 26 '17 at 10:07
1
$begingroup$
They seem to have already corrected this.
$endgroup$
– sophros
Jun 26 '17 at 17:11