The integral $int_{-infty}^{infty}frac{1}{R^2+z^2}e^{-alpha sqrt{R^2+z^2}}mathrm{d}z$
$begingroup$
I need to evaluate the following integral (for a physical application):
$$I=int_{-infty}^{infty}frac{1}{R^2+z^2}e^{-alpha sqrt{R^2+z^2}}mathrm{d}z, $$
where $alpha>0$ and $R>0$.
I tried putting it in Mathematica, but it failed. I thought about using the residue theorem. I thought that the only pole inside the upper half of the complex plane should be at $z=iR$ (because $R^2+z^2=(z-iR)(z+iR)$). But when evaluating the residue at the simple pole $z=iR$, I am left with an answer independent of $alpha$, which does not make sense physically.
integration complex-analysis
$endgroup$
add a comment |
$begingroup$
I need to evaluate the following integral (for a physical application):
$$I=int_{-infty}^{infty}frac{1}{R^2+z^2}e^{-alpha sqrt{R^2+z^2}}mathrm{d}z, $$
where $alpha>0$ and $R>0$.
I tried putting it in Mathematica, but it failed. I thought about using the residue theorem. I thought that the only pole inside the upper half of the complex plane should be at $z=iR$ (because $R^2+z^2=(z-iR)(z+iR)$). But when evaluating the residue at the simple pole $z=iR$, I am left with an answer independent of $alpha$, which does not make sense physically.
integration complex-analysis
$endgroup$
$begingroup$
You could try differentiating $I$ with respect to $alpha$ twice and see where that takes you.
$endgroup$
– Zachary
Dec 10 '18 at 20:25
$begingroup$
I got a result using modified Struve functions and Modified Bessel functions: $$-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+frac{pi}{R}$$ I will write an answer, shortly.
$endgroup$
– projectilemotion
Dec 10 '18 at 20:57
add a comment |
$begingroup$
I need to evaluate the following integral (for a physical application):
$$I=int_{-infty}^{infty}frac{1}{R^2+z^2}e^{-alpha sqrt{R^2+z^2}}mathrm{d}z, $$
where $alpha>0$ and $R>0$.
I tried putting it in Mathematica, but it failed. I thought about using the residue theorem. I thought that the only pole inside the upper half of the complex plane should be at $z=iR$ (because $R^2+z^2=(z-iR)(z+iR)$). But when evaluating the residue at the simple pole $z=iR$, I am left with an answer independent of $alpha$, which does not make sense physically.
integration complex-analysis
$endgroup$
I need to evaluate the following integral (for a physical application):
$$I=int_{-infty}^{infty}frac{1}{R^2+z^2}e^{-alpha sqrt{R^2+z^2}}mathrm{d}z, $$
where $alpha>0$ and $R>0$.
I tried putting it in Mathematica, but it failed. I thought about using the residue theorem. I thought that the only pole inside the upper half of the complex plane should be at $z=iR$ (because $R^2+z^2=(z-iR)(z+iR)$). But when evaluating the residue at the simple pole $z=iR$, I am left with an answer independent of $alpha$, which does not make sense physically.
integration complex-analysis
integration complex-analysis
asked Dec 10 '18 at 20:06
EwoudEwoud
22517
22517
$begingroup$
You could try differentiating $I$ with respect to $alpha$ twice and see where that takes you.
$endgroup$
– Zachary
Dec 10 '18 at 20:25
$begingroup$
I got a result using modified Struve functions and Modified Bessel functions: $$-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+frac{pi}{R}$$ I will write an answer, shortly.
$endgroup$
– projectilemotion
Dec 10 '18 at 20:57
add a comment |
$begingroup$
You could try differentiating $I$ with respect to $alpha$ twice and see where that takes you.
$endgroup$
– Zachary
Dec 10 '18 at 20:25
$begingroup$
I got a result using modified Struve functions and Modified Bessel functions: $$-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+frac{pi}{R}$$ I will write an answer, shortly.
$endgroup$
– projectilemotion
Dec 10 '18 at 20:57
$begingroup$
You could try differentiating $I$ with respect to $alpha$ twice and see where that takes you.
$endgroup$
– Zachary
Dec 10 '18 at 20:25
$begingroup$
You could try differentiating $I$ with respect to $alpha$ twice and see where that takes you.
$endgroup$
– Zachary
Dec 10 '18 at 20:25
$begingroup$
I got a result using modified Struve functions and Modified Bessel functions: $$-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+frac{pi}{R}$$ I will write an answer, shortly.
$endgroup$
– projectilemotion
Dec 10 '18 at 20:57
$begingroup$
I got a result using modified Struve functions and Modified Bessel functions: $$-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+frac{pi}{R}$$ I will write an answer, shortly.
$endgroup$
– projectilemotion
Dec 10 '18 at 20:57
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
In this answer, we use the following well-known general results:
$$K_{n}(z)=frac{sqrt{pi}}{Gamma(n+frac{1}{2})}(z/2)^nint_1^{infty} frac{e^{-zx}}{(x^2-1)^{1/2-n}}~dx tag{1}$$
And:
$$int K_0(z)~dz=frac{pi z}{2}(K_0(z)L_{-1}(z)+K_1(z)L_0(z))+C tag{2}$$
Where $K_n(z)$ is the modified Bessel function of the second kind and $L_n(z)$ is the modified Struve function. These results have been found here on equation (7) and here respectively.
We use differentiation under the integral sign:
$$I(alpha)=int_{-infty}^{infty} frac{1}{R^2+z^2}e^{-alphasqrt{R^2+z^2}}~dz$$
$$I'(alpha)=-int_{-infty}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz$$
The second integral can be computed using the substitution $t=sqrt{R^2+z^2}$ and the substitution $u=t/R$:
$$begin{align}int_{-infty}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz&=2int_{0}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz\&=2int_R^{infty} frac{e^{-at}}{sqrt{t^2-R^2}}~dt\&=2int_1^{infty} frac{e^{-alpha Ru}}{sqrt{u^2-1}}~du end{align}$$
Using equation $(1)$ with $n=0$ gives us a very simple result:
$$I'(alpha)=-2K_0(alpha R)$$
We can now integrate with respect to $alpha$, using equation $(2)$. We obtain:
$$I(alpha)=-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+C$$
Using that $I(0)=int_{-infty}^{infty} frac{1}{R^2+z^2}~dz=pi/R$, we obtain that $C=pi/R$. Thus, the result is:
$$I=frac{pi}{R}-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))$$
I highly doubt that one can simplify this further.
$endgroup$
add a comment |
$begingroup$
- Differentiate w.r.t $alpha$
- Sub $z=R sinh(x)$
- do the integral over x $int_{mathbb{R}}dxexp(-ccosh(x))=K_0(c)$, $K_0(c)$ is a modified Bessel-function
- integrate back w.r.t. $alpha$
Use $tfrac{2}{pi z}int dz K_0(z)= L_0(z)
K_0(z)+L_{-1}(z)K_1(z)+C$, $L_b(z)$ is a so called Struve
function - Find Integration constant $C$ by noting that the inital integral is clearly $0$ for $alpha=infty$
PS: Your contour Approach fails since you didn't take the branchcut $[i,iinfty]$ into account :/
$endgroup$
add a comment |
$begingroup$
When you integrate using residue theorem, you often integrate over a semicircumference with radius that goes to infinity. Remember that you have to integrate over ALL the semicircumference. Often the integral on the semicircumference (curved part) is 0, in this case it probably is not. You have to sum both integrals (real line AND semicircumference) to use residue theorem. Actually you can use other curves, like rectangles, the important thing is that you have to consider all the pieces of the curve.
$endgroup$
$begingroup$
The semicircumference should go to 0 I thought, because my integrand has an upper bound by $1/z^2$, so it decays faster than than $1/z$, so this upper circumference integral should be zero.
$endgroup$
– Ewoud
Dec 10 '18 at 20:33
$begingroup$
Yeah I think I remember a result saying something like that. I’d suggest you to give it a try, though. Also, did you remember you had to multiply by the derivative of the parametrization?
$endgroup$
– tommy1996q
Dec 10 '18 at 20:36
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034403%2fthe-integral-int-infty-infty-frac1r2z2e-alpha-sqrtr2z2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In this answer, we use the following well-known general results:
$$K_{n}(z)=frac{sqrt{pi}}{Gamma(n+frac{1}{2})}(z/2)^nint_1^{infty} frac{e^{-zx}}{(x^2-1)^{1/2-n}}~dx tag{1}$$
And:
$$int K_0(z)~dz=frac{pi z}{2}(K_0(z)L_{-1}(z)+K_1(z)L_0(z))+C tag{2}$$
Where $K_n(z)$ is the modified Bessel function of the second kind and $L_n(z)$ is the modified Struve function. These results have been found here on equation (7) and here respectively.
We use differentiation under the integral sign:
$$I(alpha)=int_{-infty}^{infty} frac{1}{R^2+z^2}e^{-alphasqrt{R^2+z^2}}~dz$$
$$I'(alpha)=-int_{-infty}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz$$
The second integral can be computed using the substitution $t=sqrt{R^2+z^2}$ and the substitution $u=t/R$:
$$begin{align}int_{-infty}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz&=2int_{0}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz\&=2int_R^{infty} frac{e^{-at}}{sqrt{t^2-R^2}}~dt\&=2int_1^{infty} frac{e^{-alpha Ru}}{sqrt{u^2-1}}~du end{align}$$
Using equation $(1)$ with $n=0$ gives us a very simple result:
$$I'(alpha)=-2K_0(alpha R)$$
We can now integrate with respect to $alpha$, using equation $(2)$. We obtain:
$$I(alpha)=-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+C$$
Using that $I(0)=int_{-infty}^{infty} frac{1}{R^2+z^2}~dz=pi/R$, we obtain that $C=pi/R$. Thus, the result is:
$$I=frac{pi}{R}-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))$$
I highly doubt that one can simplify this further.
$endgroup$
add a comment |
$begingroup$
In this answer, we use the following well-known general results:
$$K_{n}(z)=frac{sqrt{pi}}{Gamma(n+frac{1}{2})}(z/2)^nint_1^{infty} frac{e^{-zx}}{(x^2-1)^{1/2-n}}~dx tag{1}$$
And:
$$int K_0(z)~dz=frac{pi z}{2}(K_0(z)L_{-1}(z)+K_1(z)L_0(z))+C tag{2}$$
Where $K_n(z)$ is the modified Bessel function of the second kind and $L_n(z)$ is the modified Struve function. These results have been found here on equation (7) and here respectively.
We use differentiation under the integral sign:
$$I(alpha)=int_{-infty}^{infty} frac{1}{R^2+z^2}e^{-alphasqrt{R^2+z^2}}~dz$$
$$I'(alpha)=-int_{-infty}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz$$
The second integral can be computed using the substitution $t=sqrt{R^2+z^2}$ and the substitution $u=t/R$:
$$begin{align}int_{-infty}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz&=2int_{0}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz\&=2int_R^{infty} frac{e^{-at}}{sqrt{t^2-R^2}}~dt\&=2int_1^{infty} frac{e^{-alpha Ru}}{sqrt{u^2-1}}~du end{align}$$
Using equation $(1)$ with $n=0$ gives us a very simple result:
$$I'(alpha)=-2K_0(alpha R)$$
We can now integrate with respect to $alpha$, using equation $(2)$. We obtain:
$$I(alpha)=-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+C$$
Using that $I(0)=int_{-infty}^{infty} frac{1}{R^2+z^2}~dz=pi/R$, we obtain that $C=pi/R$. Thus, the result is:
$$I=frac{pi}{R}-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))$$
I highly doubt that one can simplify this further.
$endgroup$
add a comment |
$begingroup$
In this answer, we use the following well-known general results:
$$K_{n}(z)=frac{sqrt{pi}}{Gamma(n+frac{1}{2})}(z/2)^nint_1^{infty} frac{e^{-zx}}{(x^2-1)^{1/2-n}}~dx tag{1}$$
And:
$$int K_0(z)~dz=frac{pi z}{2}(K_0(z)L_{-1}(z)+K_1(z)L_0(z))+C tag{2}$$
Where $K_n(z)$ is the modified Bessel function of the second kind and $L_n(z)$ is the modified Struve function. These results have been found here on equation (7) and here respectively.
We use differentiation under the integral sign:
$$I(alpha)=int_{-infty}^{infty} frac{1}{R^2+z^2}e^{-alphasqrt{R^2+z^2}}~dz$$
$$I'(alpha)=-int_{-infty}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz$$
The second integral can be computed using the substitution $t=sqrt{R^2+z^2}$ and the substitution $u=t/R$:
$$begin{align}int_{-infty}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz&=2int_{0}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz\&=2int_R^{infty} frac{e^{-at}}{sqrt{t^2-R^2}}~dt\&=2int_1^{infty} frac{e^{-alpha Ru}}{sqrt{u^2-1}}~du end{align}$$
Using equation $(1)$ with $n=0$ gives us a very simple result:
$$I'(alpha)=-2K_0(alpha R)$$
We can now integrate with respect to $alpha$, using equation $(2)$. We obtain:
$$I(alpha)=-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+C$$
Using that $I(0)=int_{-infty}^{infty} frac{1}{R^2+z^2}~dz=pi/R$, we obtain that $C=pi/R$. Thus, the result is:
$$I=frac{pi}{R}-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))$$
I highly doubt that one can simplify this further.
$endgroup$
In this answer, we use the following well-known general results:
$$K_{n}(z)=frac{sqrt{pi}}{Gamma(n+frac{1}{2})}(z/2)^nint_1^{infty} frac{e^{-zx}}{(x^2-1)^{1/2-n}}~dx tag{1}$$
And:
$$int K_0(z)~dz=frac{pi z}{2}(K_0(z)L_{-1}(z)+K_1(z)L_0(z))+C tag{2}$$
Where $K_n(z)$ is the modified Bessel function of the second kind and $L_n(z)$ is the modified Struve function. These results have been found here on equation (7) and here respectively.
We use differentiation under the integral sign:
$$I(alpha)=int_{-infty}^{infty} frac{1}{R^2+z^2}e^{-alphasqrt{R^2+z^2}}~dz$$
$$I'(alpha)=-int_{-infty}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz$$
The second integral can be computed using the substitution $t=sqrt{R^2+z^2}$ and the substitution $u=t/R$:
$$begin{align}int_{-infty}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz&=2int_{0}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz\&=2int_R^{infty} frac{e^{-at}}{sqrt{t^2-R^2}}~dt\&=2int_1^{infty} frac{e^{-alpha Ru}}{sqrt{u^2-1}}~du end{align}$$
Using equation $(1)$ with $n=0$ gives us a very simple result:
$$I'(alpha)=-2K_0(alpha R)$$
We can now integrate with respect to $alpha$, using equation $(2)$. We obtain:
$$I(alpha)=-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+C$$
Using that $I(0)=int_{-infty}^{infty} frac{1}{R^2+z^2}~dz=pi/R$, we obtain that $C=pi/R$. Thus, the result is:
$$I=frac{pi}{R}-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))$$
I highly doubt that one can simplify this further.
edited Dec 10 '18 at 21:57
answered Dec 10 '18 at 21:09
projectilemotionprojectilemotion
11.4k62141
11.4k62141
add a comment |
add a comment |
$begingroup$
- Differentiate w.r.t $alpha$
- Sub $z=R sinh(x)$
- do the integral over x $int_{mathbb{R}}dxexp(-ccosh(x))=K_0(c)$, $K_0(c)$ is a modified Bessel-function
- integrate back w.r.t. $alpha$
Use $tfrac{2}{pi z}int dz K_0(z)= L_0(z)
K_0(z)+L_{-1}(z)K_1(z)+C$, $L_b(z)$ is a so called Struve
function - Find Integration constant $C$ by noting that the inital integral is clearly $0$ for $alpha=infty$
PS: Your contour Approach fails since you didn't take the branchcut $[i,iinfty]$ into account :/
$endgroup$
add a comment |
$begingroup$
- Differentiate w.r.t $alpha$
- Sub $z=R sinh(x)$
- do the integral over x $int_{mathbb{R}}dxexp(-ccosh(x))=K_0(c)$, $K_0(c)$ is a modified Bessel-function
- integrate back w.r.t. $alpha$
Use $tfrac{2}{pi z}int dz K_0(z)= L_0(z)
K_0(z)+L_{-1}(z)K_1(z)+C$, $L_b(z)$ is a so called Struve
function - Find Integration constant $C$ by noting that the inital integral is clearly $0$ for $alpha=infty$
PS: Your contour Approach fails since you didn't take the branchcut $[i,iinfty]$ into account :/
$endgroup$
add a comment |
$begingroup$
- Differentiate w.r.t $alpha$
- Sub $z=R sinh(x)$
- do the integral over x $int_{mathbb{R}}dxexp(-ccosh(x))=K_0(c)$, $K_0(c)$ is a modified Bessel-function
- integrate back w.r.t. $alpha$
Use $tfrac{2}{pi z}int dz K_0(z)= L_0(z)
K_0(z)+L_{-1}(z)K_1(z)+C$, $L_b(z)$ is a so called Struve
function - Find Integration constant $C$ by noting that the inital integral is clearly $0$ for $alpha=infty$
PS: Your contour Approach fails since you didn't take the branchcut $[i,iinfty]$ into account :/
$endgroup$
- Differentiate w.r.t $alpha$
- Sub $z=R sinh(x)$
- do the integral over x $int_{mathbb{R}}dxexp(-ccosh(x))=K_0(c)$, $K_0(c)$ is a modified Bessel-function
- integrate back w.r.t. $alpha$
Use $tfrac{2}{pi z}int dz K_0(z)= L_0(z)
K_0(z)+L_{-1}(z)K_1(z)+C$, $L_b(z)$ is a so called Struve
function - Find Integration constant $C$ by noting that the inital integral is clearly $0$ for $alpha=infty$
PS: Your contour Approach fails since you didn't take the branchcut $[i,iinfty]$ into account :/
answered Dec 10 '18 at 21:08
BesselsbabeBesselsbabe
211
211
add a comment |
add a comment |
$begingroup$
When you integrate using residue theorem, you often integrate over a semicircumference with radius that goes to infinity. Remember that you have to integrate over ALL the semicircumference. Often the integral on the semicircumference (curved part) is 0, in this case it probably is not. You have to sum both integrals (real line AND semicircumference) to use residue theorem. Actually you can use other curves, like rectangles, the important thing is that you have to consider all the pieces of the curve.
$endgroup$
$begingroup$
The semicircumference should go to 0 I thought, because my integrand has an upper bound by $1/z^2$, so it decays faster than than $1/z$, so this upper circumference integral should be zero.
$endgroup$
– Ewoud
Dec 10 '18 at 20:33
$begingroup$
Yeah I think I remember a result saying something like that. I’d suggest you to give it a try, though. Also, did you remember you had to multiply by the derivative of the parametrization?
$endgroup$
– tommy1996q
Dec 10 '18 at 20:36
add a comment |
$begingroup$
When you integrate using residue theorem, you often integrate over a semicircumference with radius that goes to infinity. Remember that you have to integrate over ALL the semicircumference. Often the integral on the semicircumference (curved part) is 0, in this case it probably is not. You have to sum both integrals (real line AND semicircumference) to use residue theorem. Actually you can use other curves, like rectangles, the important thing is that you have to consider all the pieces of the curve.
$endgroup$
$begingroup$
The semicircumference should go to 0 I thought, because my integrand has an upper bound by $1/z^2$, so it decays faster than than $1/z$, so this upper circumference integral should be zero.
$endgroup$
– Ewoud
Dec 10 '18 at 20:33
$begingroup$
Yeah I think I remember a result saying something like that. I’d suggest you to give it a try, though. Also, did you remember you had to multiply by the derivative of the parametrization?
$endgroup$
– tommy1996q
Dec 10 '18 at 20:36
add a comment |
$begingroup$
When you integrate using residue theorem, you often integrate over a semicircumference with radius that goes to infinity. Remember that you have to integrate over ALL the semicircumference. Often the integral on the semicircumference (curved part) is 0, in this case it probably is not. You have to sum both integrals (real line AND semicircumference) to use residue theorem. Actually you can use other curves, like rectangles, the important thing is that you have to consider all the pieces of the curve.
$endgroup$
When you integrate using residue theorem, you often integrate over a semicircumference with radius that goes to infinity. Remember that you have to integrate over ALL the semicircumference. Often the integral on the semicircumference (curved part) is 0, in this case it probably is not. You have to sum both integrals (real line AND semicircumference) to use residue theorem. Actually you can use other curves, like rectangles, the important thing is that you have to consider all the pieces of the curve.
edited Dec 10 '18 at 20:24
answered Dec 10 '18 at 20:18
tommy1996qtommy1996q
591415
591415
$begingroup$
The semicircumference should go to 0 I thought, because my integrand has an upper bound by $1/z^2$, so it decays faster than than $1/z$, so this upper circumference integral should be zero.
$endgroup$
– Ewoud
Dec 10 '18 at 20:33
$begingroup$
Yeah I think I remember a result saying something like that. I’d suggest you to give it a try, though. Also, did you remember you had to multiply by the derivative of the parametrization?
$endgroup$
– tommy1996q
Dec 10 '18 at 20:36
add a comment |
$begingroup$
The semicircumference should go to 0 I thought, because my integrand has an upper bound by $1/z^2$, so it decays faster than than $1/z$, so this upper circumference integral should be zero.
$endgroup$
– Ewoud
Dec 10 '18 at 20:33
$begingroup$
Yeah I think I remember a result saying something like that. I’d suggest you to give it a try, though. Also, did you remember you had to multiply by the derivative of the parametrization?
$endgroup$
– tommy1996q
Dec 10 '18 at 20:36
$begingroup$
The semicircumference should go to 0 I thought, because my integrand has an upper bound by $1/z^2$, so it decays faster than than $1/z$, so this upper circumference integral should be zero.
$endgroup$
– Ewoud
Dec 10 '18 at 20:33
$begingroup$
The semicircumference should go to 0 I thought, because my integrand has an upper bound by $1/z^2$, so it decays faster than than $1/z$, so this upper circumference integral should be zero.
$endgroup$
– Ewoud
Dec 10 '18 at 20:33
$begingroup$
Yeah I think I remember a result saying something like that. I’d suggest you to give it a try, though. Also, did you remember you had to multiply by the derivative of the parametrization?
$endgroup$
– tommy1996q
Dec 10 '18 at 20:36
$begingroup$
Yeah I think I remember a result saying something like that. I’d suggest you to give it a try, though. Also, did you remember you had to multiply by the derivative of the parametrization?
$endgroup$
– tommy1996q
Dec 10 '18 at 20:36
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034403%2fthe-integral-int-infty-infty-frac1r2z2e-alpha-sqrtr2z2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
You could try differentiating $I$ with respect to $alpha$ twice and see where that takes you.
$endgroup$
– Zachary
Dec 10 '18 at 20:25
$begingroup$
I got a result using modified Struve functions and Modified Bessel functions: $$-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+frac{pi}{R}$$ I will write an answer, shortly.
$endgroup$
– projectilemotion
Dec 10 '18 at 20:57