Prove $int^infty_0 fraccos(ax)cosh(beta x)dx = fracpi2betaoperatornamesech(fracapi2beta)$ The 2019 Stack Overflow Developer Survey Results Are InFourier transform of 1/coshShow $int_0^infty fraccos a x-cos b xsinh beta xfracdxx=logbig( fraccosh fracbpi2 betacosh fracapi2betabig)$Integral of the product of squared exponential and two erf functionsIntegration of $int e^axcos bxcosh cx,dx$Derivation of Gradshteyn and Ryzhik integral 3.876.1 (in question)The entry-level PhD integral: $int_0^inftyfracsin 3xsin 4xsin5xcos6xxsin^2 xcosh x dx$Calculate using residues $int_0^inftyint_0^inftycosfracpi2Big(nx^2-fracy^2nBig)cospi xyovercoshpi xcoshpi ydxdy,ninmathbbN$Computing the Integral $inttanh[b(x-a)]cosbeta x,dx$Evaluate $int_-infty^inftycos(x)operatornamesech(x)dx$Integral $intlimits_0^inftye^-frac12left(y^2+fract^2y^2right),dy$Evaluating $int_-infty^inftyfrac1-acosh(alpha x)(cosh(alpha x)-a)^2cos(beta x),dx$
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Prove $int^infty_0 fraccos(ax)cosh(beta x)dx = fracpi2betaoperatornamesech(fracapi2beta)$
The 2019 Stack Overflow Developer Survey Results Are InFourier transform of 1/coshShow $int_0^infty fraccos a x-cos b xsinh beta xfracdxx=logbig( fraccosh fracbpi2 betacosh fracapi2betabig)$Integral of the product of squared exponential and two erf functionsIntegration of $int e^axcos bxcosh cx,dx$Derivation of Gradshteyn and Ryzhik integral 3.876.1 (in question)The entry-level PhD integral: $int_0^inftyfracsin 3xsin 4xsin5xcos6xxsin^2 xcosh x dx$Calculate using residues $int_0^inftyint_0^inftycosfracpi2Big(nx^2-fracy^2nBig)cospi xyovercoshpi xcoshpi ydxdy,ninmathbbN$Computing the Integral $inttanh[b(x-a)]cosbeta x,dx$Evaluate $int_-infty^inftycos(x)operatornamesech(x)dx$Integral $intlimits_0^inftye^-frac12left(y^2+fract^2y^2right),dy$Evaluating $int_-infty^inftyfrac1-acosh(alpha x)(cosh(alpha x)-a)^2cos(beta x),dx$
$begingroup$
Proof of 3.981.3 Gradshteyn ed.8.
$$int^infty_0 fraccos(ax)cosh(beta x)dx = fracpi2betaoperatornamesech(fracapi2beta)$$
I was interested in the derivation (not necessarily rigorously proved) of the above result. I've tried using derivation under integral sign using $$I(a) = int^infty_0 fraccos(ax)cosh(x)dx$$ but with no success.
A hint would be highly appreciated.
calculus integration definite-integrals trigonometric-integrals
edited Mar 24 at 4:39
Andrews
1,2812423
asked Mar 24 at 3:53
EliantherElianther
12
$endgroup$
add a comment |
$begingroup$
Proof of 3.981.3 Gradshteyn ed.8.
$$int^infty_0 fraccos(ax)cosh(beta x)dx = fracpi2betaoperatornamesech(fracapi2beta)$$
I was interested in the derivation (not necessarily rigorously proved) of the above result. I've tried using derivation under integral sign using $$I(a) = int^infty_0 fraccos(ax)cosh(x)dx$$ but with no success.
A hint would be highly appreciated.
calculus integration definite-integrals trigonometric-integrals
edited Mar 24 at 4:39
Andrews
1,2812423
asked Mar 24 at 3:53
EliantherElianther
12
$endgroup$
$begingroup$
You can look at this answer by Ron Gordon and use some basic substitutions in the integral to transform it to yours.
$endgroup$
– aleden
Mar 24 at 13:55
$begingroup$
@aleden Thanks, the cited procedure was a nice insight.
$endgroup$
– Elianther
Mar 25 at 0:17
add a comment |
$begingroup$
Proof of 3.981.3 Gradshteyn ed.8.
$$int^infty_0 fraccos(ax)cosh(beta x)dx = fracpi2betaoperatornamesech(fracapi2beta)$$
I was interested in the derivation (not necessarily rigorously proved) of the above result. I've tried using derivation under integral sign using $$I(a) = int^infty_0 fraccos(ax)cosh(x)dx$$ but with no success.
A hint would be highly appreciated.
calculus integration definite-integrals trigonometric-integrals
edited Mar 24 at 4:39
Andrews
1,2812423
asked Mar 24 at 3:53
EliantherElianther
12
$endgroup$
Proof of 3.981.3 Gradshteyn ed.8.
$$int^infty_0 fraccos(ax)cosh(beta x)dx = fracpi2betaoperatornamesech(fracapi2beta)$$
I was interested in the derivation (not necessarily rigorously proved) of the above result. I've tried using derivation under integral sign using $$I(a) = int^infty_0 fraccos(ax)cosh(x)dx$$ but with no success.
A hint would be highly appreciated.
calculus integration definite-integrals trigonometric-integrals
calculus integration definite-integrals trigonometric-integrals
edited Mar 24 at 4:39
Andrews
1,2812423
asked Mar 24 at 3:53
EliantherElianther
12
edited Mar 24 at 4:39
Andrews
1,2812423
asked Mar 24 at 3:53
EliantherElianther
12
edited Mar 24 at 4:39
Andrews
1,2812423
edited Mar 24 at 4:39
Andrews
1,2812423
edited Mar 24 at 4:39
Andrews
1,2812423
1,2812423
asked Mar 24 at 3:53
EliantherElianther
12
asked Mar 24 at 3:53
EliantherElianther
12
asked Mar 24 at 3:53
EliantherElianther
12
12
$begingroup$
You can look at this answer by Ron Gordon and use some basic substitutions in the integral to transform it to yours.
$endgroup$
– aleden
Mar 24 at 13:55
$begingroup$
@aleden Thanks, the cited procedure was a nice insight.
$endgroup$
– Elianther
Mar 25 at 0:17
add a comment |
$begingroup$
You can look at this answer by Ron Gordon and use some basic substitutions in the integral to transform it to yours.
$endgroup$
– aleden
Mar 24 at 13:55
$begingroup$
@aleden Thanks, the cited procedure was a nice insight.
$endgroup$
– Elianther
Mar 25 at 0:17
$begingroup$
You can look at this answer by Ron Gordon and use some basic substitutions in the integral to transform it to yours.
$endgroup$
– aleden
Mar 24 at 13:55
$begingroup$
You can look at this answer by Ron Gordon and use some basic substitutions in the integral to transform it to yours.
$endgroup$
– aleden
Mar 24 at 13:55
$begingroup$
@aleden Thanks, the cited procedure was a nice insight.
$endgroup$
– Elianther
Mar 25 at 0:17
$begingroup$
@aleden Thanks, the cited procedure was a nice insight.
$endgroup$
– Elianther
Mar 25 at 0:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: The function is even, so it's easily turned into an integral over the real line. Try rewriting it as a contour integral to turn it into an infinite sum over the residues at the poles of $mathrmsech$.
answered Mar 24 at 5:12
eyeballfrogeyeballfrog
7,212633
$endgroup$
$begingroup$
Thanks, I was rather hesitant in using the residue theorem (since it just too powerful and might as well be considered "cheating"). Now, I think it is interesting (and might as well be another separate question) to note the procedure cited by @aleden where Gordon used a different criteria of determining the residue: $sum_kRe_z=z_kleft[pi cot(pi z)f(z)right] sim sum_kRe_z=z_kleft[pi csc(pi z)f(z)right]$ Is there any difference between both criteria? I have used the r.h.s. and the residue is zero however, using the l.h.s. the desired result is obtained.
$endgroup$
– Elianther
Mar 25 at 0:29
add a comment |
$begingroup$
Interesting problem, for sure.
Using a CAS, the antiderivative can be found. If as
$$I=int cos (a x), textsech(b x),dx$$ write $$(a^2+b^2),e^-bx,I=(b+i a)e^- i a x , _2F_1left(1,fracb-i a2 b;frac3b-ia2;-e^2 b
xright)+$$ $$(b-i a) e^ i a x , _2F_1left(1,fracb+i a2 b;frac3b+i a2
b;-e^2 b xright)$$ where appear the gaussian hypergeometric functions.
Using the limits
$$K=int_0^infty cos (a x), textsech(b x),dx=fracpsi left(frac3b-ia4 bright)+psi left(frac3b+i
a4 bright)-psi left(fracb-i a4 bright)-psi
left(fracb+i a4 bright)4 b$$ and using the properties of the digamma function
$$K=fracpi 2 b textsechleft(fracpi a2 bright)$$
answered Mar 24 at 5:32
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
$begingroup$
I have found that using the following substitutions: $t = e^-x rightarrow u = t^2beta$ one gets the following integral: $frac12betaint_0^1fracu^ia/2beta-1/2-u^-ia/2beta-1/21-udu = frac12betaleft[psileft(frac1+beta+ia2right)-psileft(frac1+beta-ia2right)right]$. From here, the result may be near but not quite apparent; it seems it is somehow logically conected to the procedure you have posted.
$endgroup$
– Elianther
Mar 25 at 0:43
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: The function is even, so it's easily turned into an integral over the real line. Try rewriting it as a contour integral to turn it into an infinite sum over the residues at the poles of $mathrmsech$.
answered Mar 24 at 5:12
eyeballfrogeyeballfrog
7,212633
$endgroup$
$begingroup$
Thanks, I was rather hesitant in using the residue theorem (since it just too powerful and might as well be considered "cheating"). Now, I think it is interesting (and might as well be another separate question) to note the procedure cited by @aleden where Gordon used a different criteria of determining the residue: $sum_kRe_z=z_kleft[pi cot(pi z)f(z)right] sim sum_kRe_z=z_kleft[pi csc(pi z)f(z)right]$ Is there any difference between both criteria? I have used the r.h.s. and the residue is zero however, using the l.h.s. the desired result is obtained.
$endgroup$
– Elianther
Mar 25 at 0:29
add a comment |
$begingroup$
Hint: The function is even, so it's easily turned into an integral over the real line. Try rewriting it as a contour integral to turn it into an infinite sum over the residues at the poles of $mathrmsech$.
answered Mar 24 at 5:12
eyeballfrogeyeballfrog
7,212633
$endgroup$
$begingroup$
Thanks, I was rather hesitant in using the residue theorem (since it just too powerful and might as well be considered "cheating"). Now, I think it is interesting (and might as well be another separate question) to note the procedure cited by @aleden where Gordon used a different criteria of determining the residue: $sum_kRe_z=z_kleft[pi cot(pi z)f(z)right] sim sum_kRe_z=z_kleft[pi csc(pi z)f(z)right]$ Is there any difference between both criteria? I have used the r.h.s. and the residue is zero however, using the l.h.s. the desired result is obtained.
$endgroup$
– Elianther
Mar 25 at 0:29
add a comment |
$begingroup$
Hint: The function is even, so it's easily turned into an integral over the real line. Try rewriting it as a contour integral to turn it into an infinite sum over the residues at the poles of $mathrmsech$.
answered Mar 24 at 5:12
eyeballfrogeyeballfrog
7,212633
$endgroup$
Hint: The function is even, so it's easily turned into an integral over the real line. Try rewriting it as a contour integral to turn it into an infinite sum over the residues at the poles of $mathrmsech$.
answered Mar 24 at 5:12
eyeballfrogeyeballfrog
7,212633
answered Mar 24 at 5:12
eyeballfrogeyeballfrog
7,212633
answered Mar 24 at 5:12
eyeballfrogeyeballfrog
7,212633
answered Mar 24 at 5:12
eyeballfrogeyeballfrog
7,212633
7,212633
$begingroup$
Thanks, I was rather hesitant in using the residue theorem (since it just too powerful and might as well be considered "cheating"). Now, I think it is interesting (and might as well be another separate question) to note the procedure cited by @aleden where Gordon used a different criteria of determining the residue: $sum_kRe_z=z_kleft[pi cot(pi z)f(z)right] sim sum_kRe_z=z_kleft[pi csc(pi z)f(z)right]$ Is there any difference between both criteria? I have used the r.h.s. and the residue is zero however, using the l.h.s. the desired result is obtained.
$endgroup$
– Elianther
Mar 25 at 0:29
add a comment |
$begingroup$
Thanks, I was rather hesitant in using the residue theorem (since it just too powerful and might as well be considered "cheating"). Now, I think it is interesting (and might as well be another separate question) to note the procedure cited by @aleden where Gordon used a different criteria of determining the residue: $sum_kRe_z=z_kleft[pi cot(pi z)f(z)right] sim sum_kRe_z=z_kleft[pi csc(pi z)f(z)right]$ Is there any difference between both criteria? I have used the r.h.s. and the residue is zero however, using the l.h.s. the desired result is obtained.
$endgroup$
– Elianther
Mar 25 at 0:29
$begingroup$
Thanks, I was rather hesitant in using the residue theorem (since it just too powerful and might as well be considered "cheating"). Now, I think it is interesting (and might as well be another separate question) to note the procedure cited by @aleden where Gordon used a different criteria of determining the residue: $sum_kRe_z=z_kleft[pi cot(pi z)f(z)right] sim sum_kRe_z=z_kleft[pi csc(pi z)f(z)right]$ Is there any difference between both criteria? I have used the r.h.s. and the residue is zero however, using the l.h.s. the desired result is obtained.
$endgroup$
– Elianther
Mar 25 at 0:29
$begingroup$
Thanks, I was rather hesitant in using the residue theorem (since it just too powerful and might as well be considered "cheating"). Now, I think it is interesting (and might as well be another separate question) to note the procedure cited by @aleden where Gordon used a different criteria of determining the residue: $sum_kRe_z=z_kleft[pi cot(pi z)f(z)right] sim sum_kRe_z=z_kleft[pi csc(pi z)f(z)right]$ Is there any difference between both criteria? I have used the r.h.s. and the residue is zero however, using the l.h.s. the desired result is obtained.
$endgroup$
– Elianther
Mar 25 at 0:29
add a comment |
$begingroup$
Interesting problem, for sure.
Using a CAS, the antiderivative can be found. If as
$$I=int cos (a x), textsech(b x),dx$$ write $$(a^2+b^2),e^-bx,I=(b+i a)e^- i a x , _2F_1left(1,fracb-i a2 b;frac3b-ia2;-e^2 b
xright)+$$ $$(b-i a) e^ i a x , _2F_1left(1,fracb+i a2 b;frac3b+i a2
b;-e^2 b xright)$$ where appear the gaussian hypergeometric functions.
Using the limits
$$K=int_0^infty cos (a x), textsech(b x),dx=fracpsi left(frac3b-ia4 bright)+psi left(frac3b+i
a4 bright)-psi left(fracb-i a4 bright)-psi
left(fracb+i a4 bright)4 b$$ and using the properties of the digamma function
$$K=fracpi 2 b textsechleft(fracpi a2 bright)$$
answered Mar 24 at 5:32
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
$begingroup$
I have found that using the following substitutions: $t = e^-x rightarrow u = t^2beta$ one gets the following integral: $frac12betaint_0^1fracu^ia/2beta-1/2-u^-ia/2beta-1/21-udu = frac12betaleft[psileft(frac1+beta+ia2right)-psileft(frac1+beta-ia2right)right]$. From here, the result may be near but not quite apparent; it seems it is somehow logically conected to the procedure you have posted.
$endgroup$
– Elianther
Mar 25 at 0:43
add a comment |
$begingroup$
Interesting problem, for sure.
Using a CAS, the antiderivative can be found. If as
$$I=int cos (a x), textsech(b x),dx$$ write $$(a^2+b^2),e^-bx,I=(b+i a)e^- i a x , _2F_1left(1,fracb-i a2 b;frac3b-ia2;-e^2 b
xright)+$$ $$(b-i a) e^ i a x , _2F_1left(1,fracb+i a2 b;frac3b+i a2
b;-e^2 b xright)$$ where appear the gaussian hypergeometric functions.
Using the limits
$$K=int_0^infty cos (a x), textsech(b x),dx=fracpsi left(frac3b-ia4 bright)+psi left(frac3b+i
a4 bright)-psi left(fracb-i a4 bright)-psi
left(fracb+i a4 bright)4 b$$ and using the properties of the digamma function
$$K=fracpi 2 b textsechleft(fracpi a2 bright)$$
answered Mar 24 at 5:32
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
$begingroup$
I have found that using the following substitutions: $t = e^-x rightarrow u = t^2beta$ one gets the following integral: $frac12betaint_0^1fracu^ia/2beta-1/2-u^-ia/2beta-1/21-udu = frac12betaleft[psileft(frac1+beta+ia2right)-psileft(frac1+beta-ia2right)right]$. From here, the result may be near but not quite apparent; it seems it is somehow logically conected to the procedure you have posted.
$endgroup$
– Elianther
Mar 25 at 0:43
add a comment |
$begingroup$
Interesting problem, for sure.
Using a CAS, the antiderivative can be found. If as
$$I=int cos (a x), textsech(b x),dx$$ write $$(a^2+b^2),e^-bx,I=(b+i a)e^- i a x , _2F_1left(1,fracb-i a2 b;frac3b-ia2;-e^2 b
xright)+$$ $$(b-i a) e^ i a x , _2F_1left(1,fracb+i a2 b;frac3b+i a2
b;-e^2 b xright)$$ where appear the gaussian hypergeometric functions.
Using the limits
$$K=int_0^infty cos (a x), textsech(b x),dx=fracpsi left(frac3b-ia4 bright)+psi left(frac3b+i
a4 bright)-psi left(fracb-i a4 bright)-psi
left(fracb+i a4 bright)4 b$$ and using the properties of the digamma function
$$K=fracpi 2 b textsechleft(fracpi a2 bright)$$
answered Mar 24 at 5:32
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
Interesting problem, for sure.
Using a CAS, the antiderivative can be found. If as
$$I=int cos (a x), textsech(b x),dx$$ write $$(a^2+b^2),e^-bx,I=(b+i a)e^- i a x , _2F_1left(1,fracb-i a2 b;frac3b-ia2;-e^2 b
xright)+$$ $$(b-i a) e^ i a x , _2F_1left(1,fracb+i a2 b;frac3b+i a2
b;-e^2 b xright)$$ where appear the gaussian hypergeometric functions.
Using the limits
$$K=int_0^infty cos (a x), textsech(b x),dx=fracpsi left(frac3b-ia4 bright)+psi left(frac3b+i
a4 bright)-psi left(fracb-i a4 bright)-psi
left(fracb+i a4 bright)4 b$$ and using the properties of the digamma function
$$K=fracpi 2 b textsechleft(fracpi a2 bright)$$
answered Mar 24 at 5:32
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 24 at 5:32
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 24 at 5:32
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 24 at 5:32
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
$begingroup$
I have found that using the following substitutions: $t = e^-x rightarrow u = t^2beta$ one gets the following integral: $frac12betaint_0^1fracu^ia/2beta-1/2-u^-ia/2beta-1/21-udu = frac12betaleft[psileft(frac1+beta+ia2right)-psileft(frac1+beta-ia2right)right]$. From here, the result may be near but not quite apparent; it seems it is somehow logically conected to the procedure you have posted.
$endgroup$
– Elianther
Mar 25 at 0:43
add a comment |
$begingroup$
I have found that using the following substitutions: $t = e^-x rightarrow u = t^2beta$ one gets the following integral: $frac12betaint_0^1fracu^ia/2beta-1/2-u^-ia/2beta-1/21-udu = frac12betaleft[psileft(frac1+beta+ia2right)-psileft(frac1+beta-ia2right)right]$. From here, the result may be near but not quite apparent; it seems it is somehow logically conected to the procedure you have posted.
$endgroup$
– Elianther
Mar 25 at 0:43
$begingroup$
I have found that using the following substitutions: $t = e^-x rightarrow u = t^2beta$ one gets the following integral: $frac12betaint_0^1fracu^ia/2beta-1/2-u^-ia/2beta-1/21-udu = frac12betaleft[psileft(frac1+beta+ia2right)-psileft(frac1+beta-ia2right)right]$. From here, the result may be near but not quite apparent; it seems it is somehow logically conected to the procedure you have posted.
$endgroup$
– Elianther
Mar 25 at 0:43
$begingroup$
I have found that using the following substitutions: $t = e^-x rightarrow u = t^2beta$ one gets the following integral: $frac12betaint_0^1fracu^ia/2beta-1/2-u^-ia/2beta-1/21-udu = frac12betaleft[psileft(frac1+beta+ia2right)-psileft(frac1+beta-ia2right)right]$. From here, the result may be near but not quite apparent; it seems it is somehow logically conected to the procedure you have posted.
$endgroup$
– Elianther
Mar 25 at 0:43
add a comment |
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$begingroup$
You can look at this answer by Ron Gordon and use some basic substitutions in the integral to transform it to yours.
$endgroup$
– aleden
Mar 24 at 13:55
$begingroup$
@aleden Thanks, the cited procedure was a nice insight.
$endgroup$
– Elianther
Mar 25 at 0:17