What are the detailed steps of the integral $int_-infty^+inftyrm e^(-frac(x-(a+bc^2))^2 2c^2),rm dx$? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Calculus Question: Improper integral $int_0^inftyfraccos(2x+1)sqrt[3]xdx$Finding the integral of $int_-infty^inftye^$.Computing a double integral $int_-infty^inftyint_-infty^inftyfracf(t)1+(x+g(t))^2dt dx$How do I solve this improper integral: $int_-infty^infty e^-x^2-xdx$?Evaluate the improper integral $int_-infty^0 e^xtextsin, xdx$Evaluate the $int_0^infty xe^-xdx$.Evaluate $int_-infty^inftycos(x)operatornamesech(x)dx$$int_-infty^infty e^-2ax^2dx=fracsqrtpisqrt2a$Improper integral $int_1^inftyfracdxxsqrtx^2 - 1$Improper integral with a parameter $int_0^infty e^-cx^2sin(tx)dx$
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What are the detailed steps of the integral $int_-infty^+inftyrm e^(-frac(x-(a+bc^2))^2 2c^2),rm dx$?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Calculus Question: Improper integral $int_0^inftyfraccos(2x+1)sqrt[3]xdx$Finding the integral of $int_-infty^inftye^$.Computing a double integral $int_-infty^inftyint_-infty^inftyfracf(t)1+(x+g(t))^2dt dx$How do I solve this improper integral: $int_-infty^infty e^-x^2-xdx$?Evaluate the improper integral $int_-infty^0 e^xtextsin, xdx$Evaluate the $int_0^infty xe^-xdx$.Evaluate $int_-infty^inftycos(x)operatornamesech(x)dx$$int_-infty^infty e^-2ax^2dx=fracsqrtpisqrt2a$Improper integral $int_1^inftyfracdxxsqrtx^2 - 1$Improper integral with a parameter $int_0^infty e^-cx^2sin(tx)dx$
$begingroup$
What are the detailed steps of calculating the integral $int_-infty^+inftyrm e^(-frac(x-(a+bc^2))^2 2c^2),rm dx$, I got an answer of $sqrtpi b$ from WolframAlpha but couldn't get a detailed answer of how it is derived.
Also, the indefinite integral is $-frac12sqrtpi brm erf(fraca-xsqrtb)+C$, I wonder how it is derived too.
Thanks.
calculus integration definite-integrals improper-integrals
edited Feb 25 at 11:45
Harry Peter
5,53411439
asked Feb 22 at 5:53
user598604user598604
62
$endgroup$
add a comment |
$begingroup$
What are the detailed steps of calculating the integral $int_-infty^+inftyrm e^(-frac(x-(a+bc^2))^2 2c^2),rm dx$, I got an answer of $sqrtpi b$ from WolframAlpha but couldn't get a detailed answer of how it is derived.
Also, the indefinite integral is $-frac12sqrtpi brm erf(fraca-xsqrtb)+C$, I wonder how it is derived too.
Thanks.
calculus integration definite-integrals improper-integrals
edited Feb 25 at 11:45
Harry Peter
5,53411439
asked Feb 22 at 5:53
user598604user598604
62
$endgroup$
$begingroup$
after some suitable substitutions problem will boil down to just finding fourier transform of gaussian function
$endgroup$
– Faraday Pathak
Feb 22 at 6:31
2
$begingroup$
The answer is wrong, it should be $sqrtpi c$ instead. This is just the Gaussian integral en.wikipedia.org/wiki/Gaussian_integral
$endgroup$
– Andrei
Feb 22 at 6:35
add a comment |
$begingroup$
What are the detailed steps of calculating the integral $int_-infty^+inftyrm e^(-frac(x-(a+bc^2))^2 2c^2),rm dx$, I got an answer of $sqrtpi b$ from WolframAlpha but couldn't get a detailed answer of how it is derived.
Also, the indefinite integral is $-frac12sqrtpi brm erf(fraca-xsqrtb)+C$, I wonder how it is derived too.
Thanks.
calculus integration definite-integrals improper-integrals
edited Feb 25 at 11:45
Harry Peter
5,53411439
asked Feb 22 at 5:53
user598604user598604
62
$endgroup$
What are the detailed steps of calculating the integral $int_-infty^+inftyrm e^(-frac(x-(a+bc^2))^2 2c^2),rm dx$, I got an answer of $sqrtpi b$ from WolframAlpha but couldn't get a detailed answer of how it is derived.
Also, the indefinite integral is $-frac12sqrtpi brm erf(fraca-xsqrtb)+C$, I wonder how it is derived too.
Thanks.
calculus integration definite-integrals improper-integrals
calculus integration definite-integrals improper-integrals
edited Feb 25 at 11:45
Harry Peter
5,53411439
asked Feb 22 at 5:53
user598604user598604
62
edited Feb 25 at 11:45
Harry Peter
5,53411439
asked Feb 22 at 5:53
user598604user598604
62
edited Feb 25 at 11:45
Harry Peter
5,53411439
edited Feb 25 at 11:45
Harry Peter
5,53411439
edited Feb 25 at 11:45
Harry Peter
5,53411439
5,53411439
asked Feb 22 at 5:53
user598604user598604
62
asked Feb 22 at 5:53
user598604user598604
62
asked Feb 22 at 5:53
user598604user598604
62
62
$begingroup$
after some suitable substitutions problem will boil down to just finding fourier transform of gaussian function
$endgroup$
– Faraday Pathak
Feb 22 at 6:31
2
$begingroup$
The answer is wrong, it should be $sqrtpi c$ instead. This is just the Gaussian integral en.wikipedia.org/wiki/Gaussian_integral
$endgroup$
– Andrei
Feb 22 at 6:35
add a comment |
$begingroup$
after some suitable substitutions problem will boil down to just finding fourier transform of gaussian function
$endgroup$
– Faraday Pathak
Feb 22 at 6:31
2
$begingroup$
The answer is wrong, it should be $sqrtpi c$ instead. This is just the Gaussian integral en.wikipedia.org/wiki/Gaussian_integral
$endgroup$
– Andrei
Feb 22 at 6:35
$begingroup$
after some suitable substitutions problem will boil down to just finding fourier transform of gaussian function
$endgroup$
– Faraday Pathak
Feb 22 at 6:31
$begingroup$
after some suitable substitutions problem will boil down to just finding fourier transform of gaussian function
$endgroup$
– Faraday Pathak
Feb 22 at 6:31
2
2
$begingroup$
The answer is wrong, it should be $sqrtpi c$ instead. This is just the Gaussian integral en.wikipedia.org/wiki/Gaussian_integral
$endgroup$
– Andrei
Feb 22 at 6:35
$begingroup$
The answer is wrong, it should be $sqrtpi c$ instead. This is just the Gaussian integral en.wikipedia.org/wiki/Gaussian_integral
$endgroup$
– Andrei
Feb 22 at 6:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
put $x-(a+bc^2)=t$
then,
$I=displaystyleint_-infty^inftye^-t^2/2c^2 dt$
we know,
$mathcalF[e^-at^2]=displaystyleint_t=-infty^t=inftye^-at^2e^-jomega tdt=sqrtdfracpiaexpleft(dfrac-omega^24aright)$
for your case put $omega =0$ and $a=dfrac12c^2$
you'll get
$I=csqrt2pi$ notery along
note:
in general
$displaystyle int_-infty^inftyexp(-mx^n)dx=dfracdfrac1nGammaleft(dfrac1nright)n (m)^1/n$
answered Feb 22 at 6:46
Faraday PathakFaraday Pathak
1,155316
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
put $x-(a+bc^2)=t$
then,
$I=displaystyleint_-infty^inftye^-t^2/2c^2 dt$
we know,
$mathcalF[e^-at^2]=displaystyleint_t=-infty^t=inftye^-at^2e^-jomega tdt=sqrtdfracpiaexpleft(dfrac-omega^24aright)$
for your case put $omega =0$ and $a=dfrac12c^2$
you'll get
$I=csqrt2pi$ notery along
note:
in general
$displaystyle int_-infty^inftyexp(-mx^n)dx=dfracdfrac1nGammaleft(dfrac1nright)n (m)^1/n$
answered Feb 22 at 6:46
Faraday PathakFaraday Pathak
1,155316
$endgroup$
add a comment |
$begingroup$
put $x-(a+bc^2)=t$
then,
$I=displaystyleint_-infty^inftye^-t^2/2c^2 dt$
we know,
$mathcalF[e^-at^2]=displaystyleint_t=-infty^t=inftye^-at^2e^-jomega tdt=sqrtdfracpiaexpleft(dfrac-omega^24aright)$
for your case put $omega =0$ and $a=dfrac12c^2$
you'll get
$I=csqrt2pi$ notery along
note:
in general
$displaystyle int_-infty^inftyexp(-mx^n)dx=dfracdfrac1nGammaleft(dfrac1nright)n (m)^1/n$
answered Feb 22 at 6:46
Faraday PathakFaraday Pathak
1,155316
$endgroup$
add a comment |
$begingroup$
put $x-(a+bc^2)=t$
then,
$I=displaystyleint_-infty^inftye^-t^2/2c^2 dt$
we know,
$mathcalF[e^-at^2]=displaystyleint_t=-infty^t=inftye^-at^2e^-jomega tdt=sqrtdfracpiaexpleft(dfrac-omega^24aright)$
for your case put $omega =0$ and $a=dfrac12c^2$
you'll get
$I=csqrt2pi$ notery along
note:
in general
$displaystyle int_-infty^inftyexp(-mx^n)dx=dfracdfrac1nGammaleft(dfrac1nright)n (m)^1/n$
answered Feb 22 at 6:46
Faraday PathakFaraday Pathak
1,155316
$endgroup$
put $x-(a+bc^2)=t$
then,
$I=displaystyleint_-infty^inftye^-t^2/2c^2 dt$
we know,
$mathcalF[e^-at^2]=displaystyleint_t=-infty^t=inftye^-at^2e^-jomega tdt=sqrtdfracpiaexpleft(dfrac-omega^24aright)$
for your case put $omega =0$ and $a=dfrac12c^2$
you'll get
$I=csqrt2pi$ notery along
note:
in general
$displaystyle int_-infty^inftyexp(-mx^n)dx=dfracdfrac1nGammaleft(dfrac1nright)n (m)^1/n$
answered Feb 22 at 6:46
Faraday PathakFaraday Pathak
1,155316
edited Mar 26 at 8:07
answered Feb 22 at 6:46
Faraday PathakFaraday Pathak
1,155316
answered Feb 22 at 6:46
Faraday PathakFaraday Pathak
1,155316
answered Feb 22 at 6:46
Faraday PathakFaraday Pathak
1,155316
1,155316
add a comment |
add a comment |
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$begingroup$
after some suitable substitutions problem will boil down to just finding fourier transform of gaussian function
$endgroup$
– Faraday Pathak
Feb 22 at 6:31
2
$begingroup$
The answer is wrong, it should be $sqrtpi c$ instead. This is just the Gaussian integral en.wikipedia.org/wiki/Gaussian_integral
$endgroup$
– Andrei
Feb 22 at 6:35