Mixing
Evaluate the integral $intlimits_{-infty}^infty frac{cos(x)}{x^2+1}dx$.
$begingroup$
Evaluate the integral $displaystyleintlimits_{-infty}^infty frac{cos(x)}{x^2+1} dx$.
Hint: $cos(x) = Re(exp(ix))$
Hi, I am confused that if I need to use the Residue Theorem in order to solve this, and I am not sure where I should start.
integration complex-analysis residue-calculus
edited Jul 26 '17 at 23:17
David G. Stork
10.9k31432
asked Jul 26 '17 at 23:14
Hwi MoonHwi Moon
405
$endgroup$
|
show 4 more comments
$begingroup$
Evaluate the integral $displaystyleintlimits_{-infty}^infty frac{cos(x)}{x^2+1} dx$.
Hint: $cos(x) = Re(exp(ix))$
Hi, I am confused that if I need to use the Residue Theorem in order to solve this, and I am not sure where I should start.
integration complex-analysis residue-calculus
edited Jul 26 '17 at 23:17
David G. Stork
10.9k31432
asked Jul 26 '17 at 23:14
Hwi MoonHwi Moon
405
$endgroup$
$begingroup$
Any idea for a contour?
$endgroup$
– Simply Beautiful Art
Jul 26 '17 at 23:16
$begingroup$
For the function there exist two singularities: +i, -i. -1 lies interior of unit circle, so if C is given as a region that does not contain the singularities, I would apply the Cauchy integral formula. However, the C is not given in the problem and to be honest I really do not know where I should proceed.
$endgroup$
– Hwi Moon
Jul 26 '17 at 23:20
$begingroup$
Add on the integral with integrand $isin(x)/(x^2+1),$ then apply your residue theorems to the integral with integrand $e^{ix}/(x^2+1).$
$endgroup$
– Chickenmancer
Jul 26 '17 at 23:20
$begingroup$
@Chickenmancer I think the intention was to use $cos(x)=Re(e^{ix})$ instead, but that works too.
$endgroup$
– Simply Beautiful Art
Jul 26 '17 at 23:22
3
$begingroup$
The first example in the Wikipedia page seems to be the integral you want, taking $t=1$ and using the hint you mentioned en.wikipedia.org/wiki/Residue_theorem
$endgroup$
– Reveillark
Jul 26 '17 at 23:22
|
show 4 more comments
$begingroup$
Evaluate the integral $displaystyleintlimits_{-infty}^infty frac{cos(x)}{x^2+1} dx$.
Hint: $cos(x) = Re(exp(ix))$
Hi, I am confused that if I need to use the Residue Theorem in order to solve this, and I am not sure where I should start.
integration complex-analysis residue-calculus
edited Jul 26 '17 at 23:17
David G. Stork
10.9k31432
asked Jul 26 '17 at 23:14
Hwi MoonHwi Moon
405
$endgroup$
Evaluate the integral $displaystyleintlimits_{-infty}^infty frac{cos(x)}{x^2+1} dx$.
Hint: $cos(x) = Re(exp(ix))$
Hi, I am confused that if I need to use the Residue Theorem in order to solve this, and I am not sure where I should start.
integration complex-analysis residue-calculus
integration complex-analysis residue-calculus
edited Jul 26 '17 at 23:17
David G. Stork
10.9k31432
asked Jul 26 '17 at 23:14
Hwi MoonHwi Moon
405
edited Jul 26 '17 at 23:17
David G. Stork
10.9k31432
asked Jul 26 '17 at 23:14
Hwi MoonHwi Moon
405
edited Jul 26 '17 at 23:17
David G. Stork
10.9k31432
edited Jul 26 '17 at 23:17
David G. Stork
10.9k31432
edited Jul 26 '17 at 23:17
David G. Stork
10.9k31432
10.9k31432
asked Jul 26 '17 at 23:14
Hwi MoonHwi Moon
405
asked Jul 26 '17 at 23:14
Hwi MoonHwi Moon
405
asked Jul 26 '17 at 23:14
Hwi MoonHwi Moon
405
405
$begingroup$
Any idea for a contour?
$endgroup$
– Simply Beautiful Art
Jul 26 '17 at 23:16
$begingroup$
For the function there exist two singularities: +i, -i. -1 lies interior of unit circle, so if C is given as a region that does not contain the singularities, I would apply the Cauchy integral formula. However, the C is not given in the problem and to be honest I really do not know where I should proceed.
$endgroup$
– Hwi Moon
Jul 26 '17 at 23:20
$begingroup$
Add on the integral with integrand $isin(x)/(x^2+1),$ then apply your residue theorems to the integral with integrand $e^{ix}/(x^2+1).$
$endgroup$
– Chickenmancer
Jul 26 '17 at 23:20
$begingroup$
@Chickenmancer I think the intention was to use $cos(x)=Re(e^{ix})$ instead, but that works too.
$endgroup$
– Simply Beautiful Art
Jul 26 '17 at 23:22
3
$begingroup$
The first example in the Wikipedia page seems to be the integral you want, taking $t=1$ and using the hint you mentioned en.wikipedia.org/wiki/Residue_theorem
$endgroup$
– Reveillark
Jul 26 '17 at 23:22
|
show 4 more comments
$begingroup$
Any idea for a contour?
$endgroup$
– Simply Beautiful Art
Jul 26 '17 at 23:16
$begingroup$
For the function there exist two singularities: +i, -i. -1 lies interior of unit circle, so if C is given as a region that does not contain the singularities, I would apply the Cauchy integral formula. However, the C is not given in the problem and to be honest I really do not know where I should proceed.
$endgroup$
– Hwi Moon
Jul 26 '17 at 23:20
$begingroup$
Add on the integral with integrand $isin(x)/(x^2+1),$ then apply your residue theorems to the integral with integrand $e^{ix}/(x^2+1).$
$endgroup$
– Chickenmancer
Jul 26 '17 at 23:20
$begingroup$
@Chickenmancer I think the intention was to use $cos(x)=Re(e^{ix})$ instead, but that works too.
$endgroup$
– Simply Beautiful Art
Jul 26 '17 at 23:22
3
$begingroup$
The first example in the Wikipedia page seems to be the integral you want, taking $t=1$ and using the hint you mentioned en.wikipedia.org/wiki/Residue_theorem
$endgroup$
– Reveillark
Jul 26 '17 at 23:22
$begingroup$
Any idea for a contour?
$endgroup$
– Simply Beautiful Art
Jul 26 '17 at 23:16
$begingroup$
Any idea for a contour?
$endgroup$
– Simply Beautiful Art
Jul 26 '17 at 23:16
$begingroup$
For the function there exist two singularities: +i, -i. -1 lies interior of unit circle, so if C is given as a region that does not contain the singularities, I would apply the Cauchy integral formula. However, the C is not given in the problem and to be honest I really do not know where I should proceed.
$endgroup$
– Hwi Moon
Jul 26 '17 at 23:20
$begingroup$
For the function there exist two singularities: +i, -i. -1 lies interior of unit circle, so if C is given as a region that does not contain the singularities, I would apply the Cauchy integral formula. However, the C is not given in the problem and to be honest I really do not know where I should proceed.
$endgroup$
– Hwi Moon
Jul 26 '17 at 23:20
$begingroup$
Add on the integral with integrand $isin(x)/(x^2+1),$ then apply your residue theorems to the integral with integrand $e^{ix}/(x^2+1).$
$endgroup$
– Chickenmancer
Jul 26 '17 at 23:20
$begingroup$
Add on the integral with integrand $isin(x)/(x^2+1),$ then apply your residue theorems to the integral with integrand $e^{ix}/(x^2+1).$
$endgroup$
– Chickenmancer
Jul 26 '17 at 23:20
$begingroup$
@Chickenmancer I think the intention was to use $cos(x)=Re(e^{ix})$ instead, but that works too.
$endgroup$
– Simply Beautiful Art
Jul 26 '17 at 23:22
$begingroup$
@Chickenmancer I think the intention was to use $cos(x)=Re(e^{ix})$ instead, but that works too.
$endgroup$
– Simply Beautiful Art
Jul 26 '17 at 23:22
3
3
$begingroup$
The first example in the Wikipedia page seems to be the integral you want, taking $t=1$ and using the hint you mentioned en.wikipedia.org/wiki/Residue_theorem
$endgroup$
– Reveillark
Jul 26 '17 at 23:22
$begingroup$
The first example in the Wikipedia page seems to be the integral you want, taking $t=1$ and using the hint you mentioned en.wikipedia.org/wiki/Residue_theorem
$endgroup$
– Reveillark
Jul 26 '17 at 23:22
|
show 4 more comments
4 Answers
4
active
oldest
votes
$begingroup$
We may also see that $$I=int_{-infty}^{infty}frac{cosleft(xright)}{1+x^{2}}dx=2int_{0}^{infty}frac{cosleft(xright)}{1+x^{2}}dx$$ $$ =int_{0}^{infty}frac{e^{ix}+e^{-ix}}{1+x^{2}}dx=frac{e^{-1}}{2}left(int_{0}^{infty}frac{e^{1+ix}+e^{1-ix}}{1+ix}dx+int_{0}^{infty}frac{e^{1+ix}+e^{1-ix}}{1-ix}dxright)$$ $$ =frac{e^{-1}}{2i}left(int_{0}^{infty}frac{1}{x}left(frac{ixe^{1+ix}}{1+ix}+frac{ixe^{1-ix}}{1-ix}right)dx+int_{0}^{infty}frac{1}{x}left(frac{ixe^{1-ix}}{1+ix}+frac{ixe^{1+ix}}{1-ix}right)dxright)$$ and now applying the complex version of Frullani's theorem to the functions $$fleft(xright)=frac{xe^{1-x}}{1-x},,gleft(xright)=frac{xe^{1-x}}{1+x}$$ we get $$I=frac{e^{-1}}{i}logleft(-1right)=color{red}{pi e^{-1}}.$$
answered Jul 27 '17 at 16:25
Marco CantariniMarco Cantarini
29.1k23373
$endgroup$
$begingroup$
(+1) for the nice use of the complex version of Frullani's theorem.
$endgroup$
– Mark Viola
Jul 27 '17 at 16:56
$begingroup$
@MarkViola Thank you Mark! :)
$endgroup$
– Marco Cantarini
Jul 27 '17 at 17:16
add a comment |
$begingroup$
METHODOLOGY $1$: Complex Analysis
Note that the function $frac{e^{iz}}{z^2+1}$ has poles at $pm i$. Then, by Cauchy's Integral Formula we have for $R>1$
$$begin{align}
oint_{C_R}frac{e^{iz}}{z^2+1},dz&=int_{-R}^R frac{e^{ix}}{x^2+1},dx+int_0^pi frac{e^{iRe^{iphi}}}{(Re^{iphi})^1+1},iRe^{iphi},dphitag1\\
&=2pi i frac{e^{i(i)}}{2i}\\
&=pi/e
end{align}$$
As $Rto infty$, the second integral on the right-hand side of $(1)$ approaches $0$. Therefore, we find that
$$int_{-infty}^infty frac{e^{ix}}{x^2+1},dx=frac{pi}{e} tag2$$
Taking the real part of both sides of $(2)$ and exploiting the even symmetry yields
$$bbox[5px,border:2px solid #C0A000]{int_0^infty frac{cos(x)}{x^2+1},dx=frac{pi}{2e}}$$
METHODOLOGY $2$: Real Analysis
Let $f(a)$ be given by the convergent improper integral
$$f(a)=int_0^infty frac{cos(ax)}{x^2+1},dx tag3$$
Since the integral $int_0^infty frac{xsin(ax)}{x^2+1},dx$ is uniformly convergent for $|a|ge delta>0$, we may differentiate under the integral in $(3)$ for $|a|>delta>0$ to obtain
$$begin{align}
f'(a)&=-int_0^infty frac{xsin(ax)}{x^2+1},dx\\
&=-int_0^infty frac{(x^2+1-1)sin(ax)}{x(x^2+1)},dx\\
&=-int_0^infty frac{sin(ax)}{x},dx+int_0^infty frac{sin(ax)}{x(x^2+1)},dx\\
&=-frac{pi}{2}+int_0^infty frac{sin(ax)}{x(x^2+1)},dxtag4
end{align}$$
Again, since the integral $int_0^infty frac{cos(ax)}{x^2+1},dx$ converges uniformly for all $a$, we may differentiate under the integral in $(4)$ to obtain
$$f''(a)=int_0^infty frac{cos(ax)}{x^2+1},dx=f(a)tag 5$$
Solving the second-order ODE in $(5)$ reveals
$$f(a)=C_1 e^{a}+C_2 e^{-a}$$
Using $f(0)=pi/2$ and $f'(0)=-pi/2$, we find that $C_1=0$ and $C_2=frac{pi}{2}$ and hence $f(a)=frac{pi e^{-a}}{2}$. Setting $a=1$ yields the coveted result
$$bbox[5px,border:2px solid #C0A000]{int_0^infty frac{cos(x)}{x^2+1},dx=frac{pi}{2e}}$$
as expected!
answered Jul 27 '17 at 15:14
Mark ViolaMark Viola
131k1275172
$endgroup$
add a comment |
$begingroup$
Use the Fourier transform:
$$frac{2}{pi}int_{-infty}^{infty}frac{e^{inu x}}{x^{2}+1}dx=e^{-left|nuright|}$$
Just set $nu=1$, divide by $frac{2}{pi}$, and take the real part of both sides.
answered Jul 26 '17 at 23:26
MCSMCS
969313
$endgroup$
$begingroup$
It is probably preferred that this be solved using complex analysis techniques.
$endgroup$
– Simply Beautiful Art
Jul 26 '17 at 23:30
1
$begingroup$
Isn't this answer circular? To prove the formula for the Fourier transform, doesn't one have to solve a generalization of the problem proposed?
$endgroup$
– N. S.
Jul 27 '17 at 16:32
$begingroup$
@N.S. I believe that MCS meant to exploit the Fourier inversion formula for $e^{-|nu|}$.
$endgroup$
– Mark Viola
Jul 27 '17 at 16:59
add a comment |
$begingroup$
Another approach: A combination of Feynman's Trick and Laplace Transforms:
begin{equation}
J = int_{-infty}^{infty}frac{cos(x)}{x^2 + 1}:dx
end{equation}
Here let:
begin{equation}
I(t) = int_{-infty}^{infty}frac{cos(xt)}{x^2 + 1}:dx
end{equation}
We see $I(1) = J$ and $I(0) = pi$. Here we take the Laplace Transform w.r.t. '$t$':
begin{align}
mathscr{L}left[I(t)right] &= int_{-infty}^{infty}frac{mathscr{L}left[cos(xt)right]}{x^2 + 1}:dx = int_{-infty}^{infty} frac{s}{s^2 + x^2}cdotfrac{1}{x^2 + 1}:dx \
&=frac{s}{s^2 - 1}int_{-infty}^{infty}left[frac{1}{x^2 + 1} - frac{1}{x^2 + s^2} right]:dx \
&= frac{s}{s^2 - 1} left[ arctan(x) - frac{1}{s}arctanleft(frac{x}{s} right)right]_{-infty}^{infty} \
&= frac{s}{s^2 - 1} left[ left(frac{pi}{2} - frac{1}{s}cdot frac{pi}{2} right) - left(-frac{pi}{2} - frac{1}{s}cdot -frac{pi}{2} right) right] \
&= frac{s}{s^2 - 1} left[ pi - pifrac{1}{s} right] = frac{s}{s^2 - 1} cdot frac{s - 1}{s}pi = frac{pi}{s + 1}
end{align}
We now take the inverse Laplace Transform:
begin{align}
I(t) = mathscr{L}^{-1}left[ frac{pi}{s + 1}right] = pi e^{-t}
end{align}
And finally:
begin{equation}
J = I(1) = pi e^{-1} = frac{pi}{e}
end{equation}
Now this is essentially residue analysis, but I've found it to be a useful technique for integrals of this type.
answered Jan 9 at 0:39
DavidGDavidG
2,2651721
$endgroup$
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%2f2372949%2fevaluate-the-integral-int-limits-infty-infty-frac-cosxx21dx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We may also see that $$I=int_{-infty}^{infty}frac{cosleft(xright)}{1+x^{2}}dx=2int_{0}^{infty}frac{cosleft(xright)}{1+x^{2}}dx$$ $$ =int_{0}^{infty}frac{e^{ix}+e^{-ix}}{1+x^{2}}dx=frac{e^{-1}}{2}left(int_{0}^{infty}frac{e^{1+ix}+e^{1-ix}}{1+ix}dx+int_{0}^{infty}frac{e^{1+ix}+e^{1-ix}}{1-ix}dxright)$$ $$ =frac{e^{-1}}{2i}left(int_{0}^{infty}frac{1}{x}left(frac{ixe^{1+ix}}{1+ix}+frac{ixe^{1-ix}}{1-ix}right)dx+int_{0}^{infty}frac{1}{x}left(frac{ixe^{1-ix}}{1+ix}+frac{ixe^{1+ix}}{1-ix}right)dxright)$$ and now applying the complex version of Frullani's theorem to the functions $$fleft(xright)=frac{xe^{1-x}}{1-x},,gleft(xright)=frac{xe^{1-x}}{1+x}$$ we get $$I=frac{e^{-1}}{i}logleft(-1right)=color{red}{pi e^{-1}}.$$
answered Jul 27 '17 at 16:25
Marco CantariniMarco Cantarini
29.1k23373
$endgroup$
$begingroup$
(+1) for the nice use of the complex version of Frullani's theorem.
$endgroup$
– Mark Viola
Jul 27 '17 at 16:56
$begingroup$
@MarkViola Thank you Mark! :)
$endgroup$
– Marco Cantarini
Jul 27 '17 at 17:16
add a comment |
$begingroup$
We may also see that $$I=int_{-infty}^{infty}frac{cosleft(xright)}{1+x^{2}}dx=2int_{0}^{infty}frac{cosleft(xright)}{1+x^{2}}dx$$ $$ =int_{0}^{infty}frac{e^{ix}+e^{-ix}}{1+x^{2}}dx=frac{e^{-1}}{2}left(int_{0}^{infty}frac{e^{1+ix}+e^{1-ix}}{1+ix}dx+int_{0}^{infty}frac{e^{1+ix}+e^{1-ix}}{1-ix}dxright)$$ $$ =frac{e^{-1}}{2i}left(int_{0}^{infty}frac{1}{x}left(frac{ixe^{1+ix}}{1+ix}+frac{ixe^{1-ix}}{1-ix}right)dx+int_{0}^{infty}frac{1}{x}left(frac{ixe^{1-ix}}{1+ix}+frac{ixe^{1+ix}}{1-ix}right)dxright)$$ and now applying the complex version of Frullani's theorem to the functions $$fleft(xright)=frac{xe^{1-x}}{1-x},,gleft(xright)=frac{xe^{1-x}}{1+x}$$ we get $$I=frac{e^{-1}}{i}logleft(-1right)=color{red}{pi e^{-1}}.$$
answered Jul 27 '17 at 16:25
Marco CantariniMarco Cantarini
29.1k23373
$endgroup$
$begingroup$
(+1) for the nice use of the complex version of Frullani's theorem.
$endgroup$
– Mark Viola
Jul 27 '17 at 16:56
$begingroup$
@MarkViola Thank you Mark! :)
$endgroup$
– Marco Cantarini
Jul 27 '17 at 17:16
add a comment |
$begingroup$
We may also see that $$I=int_{-infty}^{infty}frac{cosleft(xright)}{1+x^{2}}dx=2int_{0}^{infty}frac{cosleft(xright)}{1+x^{2}}dx$$ $$ =int_{0}^{infty}frac{e^{ix}+e^{-ix}}{1+x^{2}}dx=frac{e^{-1}}{2}left(int_{0}^{infty}frac{e^{1+ix}+e^{1-ix}}{1+ix}dx+int_{0}^{infty}frac{e^{1+ix}+e^{1-ix}}{1-ix}dxright)$$ $$ =frac{e^{-1}}{2i}left(int_{0}^{infty}frac{1}{x}left(frac{ixe^{1+ix}}{1+ix}+frac{ixe^{1-ix}}{1-ix}right)dx+int_{0}^{infty}frac{1}{x}left(frac{ixe^{1-ix}}{1+ix}+frac{ixe^{1+ix}}{1-ix}right)dxright)$$ and now applying the complex version of Frullani's theorem to the functions $$fleft(xright)=frac{xe^{1-x}}{1-x},,gleft(xright)=frac{xe^{1-x}}{1+x}$$ we get $$I=frac{e^{-1}}{i}logleft(-1right)=color{red}{pi e^{-1}}.$$
answered Jul 27 '17 at 16:25
Marco CantariniMarco Cantarini
29.1k23373
$endgroup$
We may also see that $$I=int_{-infty}^{infty}frac{cosleft(xright)}{1+x^{2}}dx=2int_{0}^{infty}frac{cosleft(xright)}{1+x^{2}}dx$$ $$ =int_{0}^{infty}frac{e^{ix}+e^{-ix}}{1+x^{2}}dx=frac{e^{-1}}{2}left(int_{0}^{infty}frac{e^{1+ix}+e^{1-ix}}{1+ix}dx+int_{0}^{infty}frac{e^{1+ix}+e^{1-ix}}{1-ix}dxright)$$ $$ =frac{e^{-1}}{2i}left(int_{0}^{infty}frac{1}{x}left(frac{ixe^{1+ix}}{1+ix}+frac{ixe^{1-ix}}{1-ix}right)dx+int_{0}^{infty}frac{1}{x}left(frac{ixe^{1-ix}}{1+ix}+frac{ixe^{1+ix}}{1-ix}right)dxright)$$ and now applying the complex version of Frullani's theorem to the functions $$fleft(xright)=frac{xe^{1-x}}{1-x},,gleft(xright)=frac{xe^{1-x}}{1+x}$$ we get $$I=frac{e^{-1}}{i}logleft(-1right)=color{red}{pi e^{-1}}.$$
answered Jul 27 '17 at 16:25
Marco CantariniMarco Cantarini
29.1k23373
answered Jul 27 '17 at 16:25
Marco CantariniMarco Cantarini
29.1k23373
answered Jul 27 '17 at 16:25
Marco CantariniMarco Cantarini
29.1k23373
answered Jul 27 '17 at 16:25
Marco CantariniMarco Cantarini
29.1k23373
29.1k23373
$begingroup$
(+1) for the nice use of the complex version of Frullani's theorem.
$endgroup$
– Mark Viola
Jul 27 '17 at 16:56
$begingroup$
@MarkViola Thank you Mark! :)
$endgroup$
– Marco Cantarini
Jul 27 '17 at 17:16
add a comment |
$begingroup$
(+1) for the nice use of the complex version of Frullani's theorem.
$endgroup$
– Mark Viola
Jul 27 '17 at 16:56
$begingroup$
@MarkViola Thank you Mark! :)
$endgroup$
– Marco Cantarini
Jul 27 '17 at 17:16
$begingroup$
(+1) for the nice use of the complex version of Frullani's theorem.
$endgroup$
– Mark Viola
Jul 27 '17 at 16:56
$begingroup$
(+1) for the nice use of the complex version of Frullani's theorem.
$endgroup$
– Mark Viola
Jul 27 '17 at 16:56
$begingroup$
@MarkViola Thank you Mark! :)
$endgroup$
– Marco Cantarini
Jul 27 '17 at 17:16
$begingroup$
@MarkViola Thank you Mark! :)
$endgroup$
– Marco Cantarini
Jul 27 '17 at 17:16
add a comment |
$begingroup$
METHODOLOGY $1$: Complex Analysis
Note that the function $frac{e^{iz}}{z^2+1}$ has poles at $pm i$. Then, by Cauchy's Integral Formula we have for $R>1$
$$begin{align}
oint_{C_R}frac{e^{iz}}{z^2+1},dz&=int_{-R}^R frac{e^{ix}}{x^2+1},dx+int_0^pi frac{e^{iRe^{iphi}}}{(Re^{iphi})^1+1},iRe^{iphi},dphitag1\\
&=2pi i frac{e^{i(i)}}{2i}\\
&=pi/e
end{align}$$
As $Rto infty$, the second integral on the right-hand side of $(1)$ approaches $0$. Therefore, we find that
$$int_{-infty}^infty frac{e^{ix}}{x^2+1},dx=frac{pi}{e} tag2$$
Taking the real part of both sides of $(2)$ and exploiting the even symmetry yields
$$bbox[5px,border:2px solid #C0A000]{int_0^infty frac{cos(x)}{x^2+1},dx=frac{pi}{2e}}$$
METHODOLOGY $2$: Real Analysis
Let $f(a)$ be given by the convergent improper integral
$$f(a)=int_0^infty frac{cos(ax)}{x^2+1},dx tag3$$
Since the integral $int_0^infty frac{xsin(ax)}{x^2+1},dx$ is uniformly convergent for $|a|ge delta>0$, we may differentiate under the integral in $(3)$ for $|a|>delta>0$ to obtain
$$begin{align}
f'(a)&=-int_0^infty frac{xsin(ax)}{x^2+1},dx\\
&=-int_0^infty frac{(x^2+1-1)sin(ax)}{x(x^2+1)},dx\\
&=-int_0^infty frac{sin(ax)}{x},dx+int_0^infty frac{sin(ax)}{x(x^2+1)},dx\\
&=-frac{pi}{2}+int_0^infty frac{sin(ax)}{x(x^2+1)},dxtag4
end{align}$$
Again, since the integral $int_0^infty frac{cos(ax)}{x^2+1},dx$ converges uniformly for all $a$, we may differentiate under the integral in $(4)$ to obtain
$$f''(a)=int_0^infty frac{cos(ax)}{x^2+1},dx=f(a)tag 5$$
Solving the second-order ODE in $(5)$ reveals
$$f(a)=C_1 e^{a}+C_2 e^{-a}$$
Using $f(0)=pi/2$ and $f'(0)=-pi/2$, we find that $C_1=0$ and $C_2=frac{pi}{2}$ and hence $f(a)=frac{pi e^{-a}}{2}$. Setting $a=1$ yields the coveted result
$$bbox[5px,border:2px solid #C0A000]{int_0^infty frac{cos(x)}{x^2+1},dx=frac{pi}{2e}}$$
as expected!
answered Jul 27 '17 at 15:14
Mark ViolaMark Viola
131k1275172
$endgroup$
add a comment |
$begingroup$
METHODOLOGY $1$: Complex Analysis
Note that the function $frac{e^{iz}}{z^2+1}$ has poles at $pm i$. Then, by Cauchy's Integral Formula we have for $R>1$
$$begin{align}
oint_{C_R}frac{e^{iz}}{z^2+1},dz&=int_{-R}^R frac{e^{ix}}{x^2+1},dx+int_0^pi frac{e^{iRe^{iphi}}}{(Re^{iphi})^1+1},iRe^{iphi},dphitag1\\
&=2pi i frac{e^{i(i)}}{2i}\\
&=pi/e
end{align}$$
As $Rto infty$, the second integral on the right-hand side of $(1)$ approaches $0$. Therefore, we find that
$$int_{-infty}^infty frac{e^{ix}}{x^2+1},dx=frac{pi}{e} tag2$$
Taking the real part of both sides of $(2)$ and exploiting the even symmetry yields
$$bbox[5px,border:2px solid #C0A000]{int_0^infty frac{cos(x)}{x^2+1},dx=frac{pi}{2e}}$$
METHODOLOGY $2$: Real Analysis
Let $f(a)$ be given by the convergent improper integral
$$f(a)=int_0^infty frac{cos(ax)}{x^2+1},dx tag3$$
Since the integral $int_0^infty frac{xsin(ax)}{x^2+1},dx$ is uniformly convergent for $|a|ge delta>0$, we may differentiate under the integral in $(3)$ for $|a|>delta>0$ to obtain
$$begin{align}
f'(a)&=-int_0^infty frac{xsin(ax)}{x^2+1},dx\\
&=-int_0^infty frac{(x^2+1-1)sin(ax)}{x(x^2+1)},dx\\
&=-int_0^infty frac{sin(ax)}{x},dx+int_0^infty frac{sin(ax)}{x(x^2+1)},dx\\
&=-frac{pi}{2}+int_0^infty frac{sin(ax)}{x(x^2+1)},dxtag4
end{align}$$
Again, since the integral $int_0^infty frac{cos(ax)}{x^2+1},dx$ converges uniformly for all $a$, we may differentiate under the integral in $(4)$ to obtain
$$f''(a)=int_0^infty frac{cos(ax)}{x^2+1},dx=f(a)tag 5$$
Solving the second-order ODE in $(5)$ reveals
$$f(a)=C_1 e^{a}+C_2 e^{-a}$$
Using $f(0)=pi/2$ and $f'(0)=-pi/2$, we find that $C_1=0$ and $C_2=frac{pi}{2}$ and hence $f(a)=frac{pi e^{-a}}{2}$. Setting $a=1$ yields the coveted result
$$bbox[5px,border:2px solid #C0A000]{int_0^infty frac{cos(x)}{x^2+1},dx=frac{pi}{2e}}$$
as expected!
answered Jul 27 '17 at 15:14
Mark ViolaMark Viola
131k1275172
$endgroup$
add a comment |
$begingroup$
METHODOLOGY $1$: Complex Analysis
Note that the function $frac{e^{iz}}{z^2+1}$ has poles at $pm i$. Then, by Cauchy's Integral Formula we have for $R>1$
$$begin{align}
oint_{C_R}frac{e^{iz}}{z^2+1},dz&=int_{-R}^R frac{e^{ix}}{x^2+1},dx+int_0^pi frac{e^{iRe^{iphi}}}{(Re^{iphi})^1+1},iRe^{iphi},dphitag1\\
&=2pi i frac{e^{i(i)}}{2i}\\
&=pi/e
end{align}$$
As $Rto infty$, the second integral on the right-hand side of $(1)$ approaches $0$. Therefore, we find that
$$int_{-infty}^infty frac{e^{ix}}{x^2+1},dx=frac{pi}{e} tag2$$
Taking the real part of both sides of $(2)$ and exploiting the even symmetry yields
$$bbox[5px,border:2px solid #C0A000]{int_0^infty frac{cos(x)}{x^2+1},dx=frac{pi}{2e}}$$
METHODOLOGY $2$: Real Analysis
Let $f(a)$ be given by the convergent improper integral
$$f(a)=int_0^infty frac{cos(ax)}{x^2+1},dx tag3$$
Since the integral $int_0^infty frac{xsin(ax)}{x^2+1},dx$ is uniformly convergent for $|a|ge delta>0$, we may differentiate under the integral in $(3)$ for $|a|>delta>0$ to obtain
$$begin{align}
f'(a)&=-int_0^infty frac{xsin(ax)}{x^2+1},dx\\
&=-int_0^infty frac{(x^2+1-1)sin(ax)}{x(x^2+1)},dx\\
&=-int_0^infty frac{sin(ax)}{x},dx+int_0^infty frac{sin(ax)}{x(x^2+1)},dx\\
&=-frac{pi}{2}+int_0^infty frac{sin(ax)}{x(x^2+1)},dxtag4
end{align}$$
Again, since the integral $int_0^infty frac{cos(ax)}{x^2+1},dx$ converges uniformly for all $a$, we may differentiate under the integral in $(4)$ to obtain
$$f''(a)=int_0^infty frac{cos(ax)}{x^2+1},dx=f(a)tag 5$$
Solving the second-order ODE in $(5)$ reveals
$$f(a)=C_1 e^{a}+C_2 e^{-a}$$
Using $f(0)=pi/2$ and $f'(0)=-pi/2$, we find that $C_1=0$ and $C_2=frac{pi}{2}$ and hence $f(a)=frac{pi e^{-a}}{2}$. Setting $a=1$ yields the coveted result
$$bbox[5px,border:2px solid #C0A000]{int_0^infty frac{cos(x)}{x^2+1},dx=frac{pi}{2e}}$$
as expected!
answered Jul 27 '17 at 15:14
Mark ViolaMark Viola
131k1275172
$endgroup$
METHODOLOGY $1$: Complex Analysis
Note that the function $frac{e^{iz}}{z^2+1}$ has poles at $pm i$. Then, by Cauchy's Integral Formula we have for $R>1$
$$begin{align}
oint_{C_R}frac{e^{iz}}{z^2+1},dz&=int_{-R}^R frac{e^{ix}}{x^2+1},dx+int_0^pi frac{e^{iRe^{iphi}}}{(Re^{iphi})^1+1},iRe^{iphi},dphitag1\\
&=2pi i frac{e^{i(i)}}{2i}\\
&=pi/e
end{align}$$
As $Rto infty$, the second integral on the right-hand side of $(1)$ approaches $0$. Therefore, we find that
$$int_{-infty}^infty frac{e^{ix}}{x^2+1},dx=frac{pi}{e} tag2$$
Taking the real part of both sides of $(2)$ and exploiting the even symmetry yields
$$bbox[5px,border:2px solid #C0A000]{int_0^infty frac{cos(x)}{x^2+1},dx=frac{pi}{2e}}$$
METHODOLOGY $2$: Real Analysis
Let $f(a)$ be given by the convergent improper integral
$$f(a)=int_0^infty frac{cos(ax)}{x^2+1},dx tag3$$
Since the integral $int_0^infty frac{xsin(ax)}{x^2+1},dx$ is uniformly convergent for $|a|ge delta>0$, we may differentiate under the integral in $(3)$ for $|a|>delta>0$ to obtain
$$begin{align}
f'(a)&=-int_0^infty frac{xsin(ax)}{x^2+1},dx\\
&=-int_0^infty frac{(x^2+1-1)sin(ax)}{x(x^2+1)},dx\\
&=-int_0^infty frac{sin(ax)}{x},dx+int_0^infty frac{sin(ax)}{x(x^2+1)},dx\\
&=-frac{pi}{2}+int_0^infty frac{sin(ax)}{x(x^2+1)},dxtag4
end{align}$$
Again, since the integral $int_0^infty frac{cos(ax)}{x^2+1},dx$ converges uniformly for all $a$, we may differentiate under the integral in $(4)$ to obtain
$$f''(a)=int_0^infty frac{cos(ax)}{x^2+1},dx=f(a)tag 5$$
Solving the second-order ODE in $(5)$ reveals
$$f(a)=C_1 e^{a}+C_2 e^{-a}$$
Using $f(0)=pi/2$ and $f'(0)=-pi/2$, we find that $C_1=0$ and $C_2=frac{pi}{2}$ and hence $f(a)=frac{pi e^{-a}}{2}$. Setting $a=1$ yields the coveted result
$$bbox[5px,border:2px solid #C0A000]{int_0^infty frac{cos(x)}{x^2+1},dx=frac{pi}{2e}}$$
as expected!
answered Jul 27 '17 at 15:14
Mark ViolaMark Viola
131k1275172
edited Oct 19 '17 at 18:22
answered Jul 27 '17 at 15:14
Mark ViolaMark Viola
131k1275172
answered Jul 27 '17 at 15:14
Mark ViolaMark Viola
131k1275172
answered Jul 27 '17 at 15:14
Mark ViolaMark Viola
131k1275172
131k1275172
add a comment |
add a comment |
$begingroup$
Use the Fourier transform:
$$frac{2}{pi}int_{-infty}^{infty}frac{e^{inu x}}{x^{2}+1}dx=e^{-left|nuright|}$$
Just set $nu=1$, divide by $frac{2}{pi}$, and take the real part of both sides.
answered Jul 26 '17 at 23:26
MCSMCS
969313
$endgroup$
$begingroup$
It is probably preferred that this be solved using complex analysis techniques.
$endgroup$
– Simply Beautiful Art
Jul 26 '17 at 23:30
1
$begingroup$
Isn't this answer circular? To prove the formula for the Fourier transform, doesn't one have to solve a generalization of the problem proposed?
$endgroup$
– N. S.
Jul 27 '17 at 16:32
$begingroup$
@N.S. I believe that MCS meant to exploit the Fourier inversion formula for $e^{-|nu|}$.
$endgroup$
– Mark Viola
Jul 27 '17 at 16:59
add a comment |
$begingroup$
Use the Fourier transform:
$$frac{2}{pi}int_{-infty}^{infty}frac{e^{inu x}}{x^{2}+1}dx=e^{-left|nuright|}$$
Just set $nu=1$, divide by $frac{2}{pi}$, and take the real part of both sides.
answered Jul 26 '17 at 23:26
MCSMCS
969313
$endgroup$
$begingroup$
It is probably preferred that this be solved using complex analysis techniques.
$endgroup$
– Simply Beautiful Art
Jul 26 '17 at 23:30
1
$begingroup$
Isn't this answer circular? To prove the formula for the Fourier transform, doesn't one have to solve a generalization of the problem proposed?
$endgroup$
– N. S.
Jul 27 '17 at 16:32
$begingroup$
@N.S. I believe that MCS meant to exploit the Fourier inversion formula for $e^{-|nu|}$.
$endgroup$
– Mark Viola
Jul 27 '17 at 16:59
add a comment |
$begingroup$
Use the Fourier transform:
$$frac{2}{pi}int_{-infty}^{infty}frac{e^{inu x}}{x^{2}+1}dx=e^{-left|nuright|}$$
Just set $nu=1$, divide by $frac{2}{pi}$, and take the real part of both sides.
answered Jul 26 '17 at 23:26
MCSMCS
969313
$endgroup$
Use the Fourier transform:
$$frac{2}{pi}int_{-infty}^{infty}frac{e^{inu x}}{x^{2}+1}dx=e^{-left|nuright|}$$
Just set $nu=1$, divide by $frac{2}{pi}$, and take the real part of both sides.
answered Jul 26 '17 at 23:26
MCSMCS
969313
edited Jul 27 '17 at 21:32
answered Jul 26 '17 at 23:26
MCSMCS
969313
answered Jul 26 '17 at 23:26
MCSMCS
969313
answered Jul 26 '17 at 23:26
MCSMCS
969313
969313
$begingroup$
It is probably preferred that this be solved using complex analysis techniques.
$endgroup$
– Simply Beautiful Art
Jul 26 '17 at 23:30
1
$begingroup$
Isn't this answer circular? To prove the formula for the Fourier transform, doesn't one have to solve a generalization of the problem proposed?
$endgroup$
– N. S.
Jul 27 '17 at 16:32
$begingroup$
@N.S. I believe that MCS meant to exploit the Fourier inversion formula for $e^{-|nu|}$.
$endgroup$
– Mark Viola
Jul 27 '17 at 16:59
add a comment |
$begingroup$
It is probably preferred that this be solved using complex analysis techniques.
$endgroup$
– Simply Beautiful Art
Jul 26 '17 at 23:30
1
$begingroup$
Isn't this answer circular? To prove the formula for the Fourier transform, doesn't one have to solve a generalization of the problem proposed?
$endgroup$
– N. S.
Jul 27 '17 at 16:32
$begingroup$
@N.S. I believe that MCS meant to exploit the Fourier inversion formula for $e^{-|nu|}$.
$endgroup$
– Mark Viola
Jul 27 '17 at 16:59
$begingroup$
It is probably preferred that this be solved using complex analysis techniques.
$endgroup$
– Simply Beautiful Art
Jul 26 '17 at 23:30
$begingroup$
It is probably preferred that this be solved using complex analysis techniques.
$endgroup$
– Simply Beautiful Art
Jul 26 '17 at 23:30
1
1
$begingroup$
Isn't this answer circular? To prove the formula for the Fourier transform, doesn't one have to solve a generalization of the problem proposed?
$endgroup$
– N. S.
Jul 27 '17 at 16:32
$begingroup$
Isn't this answer circular? To prove the formula for the Fourier transform, doesn't one have to solve a generalization of the problem proposed?
$endgroup$
– N. S.
Jul 27 '17 at 16:32
$begingroup$
@N.S. I believe that MCS meant to exploit the Fourier inversion formula for $e^{-|nu|}$.
$endgroup$
– Mark Viola
Jul 27 '17 at 16:59
$begingroup$
@N.S. I believe that MCS meant to exploit the Fourier inversion formula for $e^{-|nu|}$.
$endgroup$
– Mark Viola
Jul 27 '17 at 16:59
add a comment |
$begingroup$
Another approach: A combination of Feynman's Trick and Laplace Transforms:
begin{equation}
J = int_{-infty}^{infty}frac{cos(x)}{x^2 + 1}:dx
end{equation}
Here let:
begin{equation}
I(t) = int_{-infty}^{infty}frac{cos(xt)}{x^2 + 1}:dx
end{equation}
We see $I(1) = J$ and $I(0) = pi$. Here we take the Laplace Transform w.r.t. '$t$':
begin{align}
mathscr{L}left[I(t)right] &= int_{-infty}^{infty}frac{mathscr{L}left[cos(xt)right]}{x^2 + 1}:dx = int_{-infty}^{infty} frac{s}{s^2 + x^2}cdotfrac{1}{x^2 + 1}:dx \
&=frac{s}{s^2 - 1}int_{-infty}^{infty}left[frac{1}{x^2 + 1} - frac{1}{x^2 + s^2} right]:dx \
&= frac{s}{s^2 - 1} left[ arctan(x) - frac{1}{s}arctanleft(frac{x}{s} right)right]_{-infty}^{infty} \
&= frac{s}{s^2 - 1} left[ left(frac{pi}{2} - frac{1}{s}cdot frac{pi}{2} right) - left(-frac{pi}{2} - frac{1}{s}cdot -frac{pi}{2} right) right] \
&= frac{s}{s^2 - 1} left[ pi - pifrac{1}{s} right] = frac{s}{s^2 - 1} cdot frac{s - 1}{s}pi = frac{pi}{s + 1}
end{align}
We now take the inverse Laplace Transform:
begin{align}
I(t) = mathscr{L}^{-1}left[ frac{pi}{s + 1}right] = pi e^{-t}
end{align}
And finally:
begin{equation}
J = I(1) = pi e^{-1} = frac{pi}{e}
end{equation}
Now this is essentially residue analysis, but I've found it to be a useful technique for integrals of this type.
answered Jan 9 at 0:39
DavidGDavidG
2,2651721
$endgroup$
add a comment |
$begingroup$
Another approach: A combination of Feynman's Trick and Laplace Transforms:
begin{equation}
J = int_{-infty}^{infty}frac{cos(x)}{x^2 + 1}:dx
end{equation}
Here let:
begin{equation}
I(t) = int_{-infty}^{infty}frac{cos(xt)}{x^2 + 1}:dx
end{equation}
We see $I(1) = J$ and $I(0) = pi$. Here we take the Laplace Transform w.r.t. '$t$':
begin{align}
mathscr{L}left[I(t)right] &= int_{-infty}^{infty}frac{mathscr{L}left[cos(xt)right]}{x^2 + 1}:dx = int_{-infty}^{infty} frac{s}{s^2 + x^2}cdotfrac{1}{x^2 + 1}:dx \
&=frac{s}{s^2 - 1}int_{-infty}^{infty}left[frac{1}{x^2 + 1} - frac{1}{x^2 + s^2} right]:dx \
&= frac{s}{s^2 - 1} left[ arctan(x) - frac{1}{s}arctanleft(frac{x}{s} right)right]_{-infty}^{infty} \
&= frac{s}{s^2 - 1} left[ left(frac{pi}{2} - frac{1}{s}cdot frac{pi}{2} right) - left(-frac{pi}{2} - frac{1}{s}cdot -frac{pi}{2} right) right] \
&= frac{s}{s^2 - 1} left[ pi - pifrac{1}{s} right] = frac{s}{s^2 - 1} cdot frac{s - 1}{s}pi = frac{pi}{s + 1}
end{align}
We now take the inverse Laplace Transform:
begin{align}
I(t) = mathscr{L}^{-1}left[ frac{pi}{s + 1}right] = pi e^{-t}
end{align}
And finally:
begin{equation}
J = I(1) = pi e^{-1} = frac{pi}{e}
end{equation}
Now this is essentially residue analysis, but I've found it to be a useful technique for integrals of this type.
answered Jan 9 at 0:39
DavidGDavidG
2,2651721
$endgroup$
add a comment |
$begingroup$
Another approach: A combination of Feynman's Trick and Laplace Transforms:
begin{equation}
J = int_{-infty}^{infty}frac{cos(x)}{x^2 + 1}:dx
end{equation}
Here let:
begin{equation}
I(t) = int_{-infty}^{infty}frac{cos(xt)}{x^2 + 1}:dx
end{equation}
We see $I(1) = J$ and $I(0) = pi$. Here we take the Laplace Transform w.r.t. '$t$':
begin{align}
mathscr{L}left[I(t)right] &= int_{-infty}^{infty}frac{mathscr{L}left[cos(xt)right]}{x^2 + 1}:dx = int_{-infty}^{infty} frac{s}{s^2 + x^2}cdotfrac{1}{x^2 + 1}:dx \
&=frac{s}{s^2 - 1}int_{-infty}^{infty}left[frac{1}{x^2 + 1} - frac{1}{x^2 + s^2} right]:dx \
&= frac{s}{s^2 - 1} left[ arctan(x) - frac{1}{s}arctanleft(frac{x}{s} right)right]_{-infty}^{infty} \
&= frac{s}{s^2 - 1} left[ left(frac{pi}{2} - frac{1}{s}cdot frac{pi}{2} right) - left(-frac{pi}{2} - frac{1}{s}cdot -frac{pi}{2} right) right] \
&= frac{s}{s^2 - 1} left[ pi - pifrac{1}{s} right] = frac{s}{s^2 - 1} cdot frac{s - 1}{s}pi = frac{pi}{s + 1}
end{align}
We now take the inverse Laplace Transform:
begin{align}
I(t) = mathscr{L}^{-1}left[ frac{pi}{s + 1}right] = pi e^{-t}
end{align}
And finally:
begin{equation}
J = I(1) = pi e^{-1} = frac{pi}{e}
end{equation}
Now this is essentially residue analysis, but I've found it to be a useful technique for integrals of this type.
answered Jan 9 at 0:39
DavidGDavidG
2,2651721
$endgroup$
Another approach: A combination of Feynman's Trick and Laplace Transforms:
begin{equation}
J = int_{-infty}^{infty}frac{cos(x)}{x^2 + 1}:dx
end{equation}
Here let:
begin{equation}
I(t) = int_{-infty}^{infty}frac{cos(xt)}{x^2 + 1}:dx
end{equation}
We see $I(1) = J$ and $I(0) = pi$. Here we take the Laplace Transform w.r.t. '$t$':
begin{align}
mathscr{L}left[I(t)right] &= int_{-infty}^{infty}frac{mathscr{L}left[cos(xt)right]}{x^2 + 1}:dx = int_{-infty}^{infty} frac{s}{s^2 + x^2}cdotfrac{1}{x^2 + 1}:dx \
&=frac{s}{s^2 - 1}int_{-infty}^{infty}left[frac{1}{x^2 + 1} - frac{1}{x^2 + s^2} right]:dx \
&= frac{s}{s^2 - 1} left[ arctan(x) - frac{1}{s}arctanleft(frac{x}{s} right)right]_{-infty}^{infty} \
&= frac{s}{s^2 - 1} left[ left(frac{pi}{2} - frac{1}{s}cdot frac{pi}{2} right) - left(-frac{pi}{2} - frac{1}{s}cdot -frac{pi}{2} right) right] \
&= frac{s}{s^2 - 1} left[ pi - pifrac{1}{s} right] = frac{s}{s^2 - 1} cdot frac{s - 1}{s}pi = frac{pi}{s + 1}
end{align}
We now take the inverse Laplace Transform:
begin{align}
I(t) = mathscr{L}^{-1}left[ frac{pi}{s + 1}right] = pi e^{-t}
end{align}
And finally:
begin{equation}
J = I(1) = pi e^{-1} = frac{pi}{e}
end{equation}
Now this is essentially residue analysis, but I've found it to be a useful technique for integrals of this type.
answered Jan 9 at 0:39
DavidGDavidG
2,2651721
answered Jan 9 at 0:39
DavidGDavidG
2,2651721
answered Jan 9 at 0:39
DavidGDavidG
2,2651721
answered Jan 9 at 0:39
DavidGDavidG
2,2651721
2,2651721
add a comment |
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%2f2372949%2fevaluate-the-integral-int-limits-infty-infty-frac-cosxx21dx%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
$begingroup$
Any idea for a contour?
$endgroup$
– Simply Beautiful Art
Jul 26 '17 at 23:16
$begingroup$
For the function there exist two singularities: +i, -i. -1 lies interior of unit circle, so if C is given as a region that does not contain the singularities, I would apply the Cauchy integral formula. However, the C is not given in the problem and to be honest I really do not know where I should proceed.
$endgroup$
– Hwi Moon
Jul 26 '17 at 23:20
$begingroup$
Add on the integral with integrand $isin(x)/(x^2+1),$ then apply your residue theorems to the integral with integrand $e^{ix}/(x^2+1).$
$endgroup$
– Chickenmancer
Jul 26 '17 at 23:20
$begingroup$
@Chickenmancer I think the intention was to use $cos(x)=Re(e^{ix})$ instead, but that works too.
$endgroup$
– Simply Beautiful Art
Jul 26 '17 at 23:22
3
$begingroup$
The first example in the Wikipedia page seems to be the integral you want, taking $t=1$ and using the hint you mentioned en.wikipedia.org/wiki/Residue_theorem
$endgroup$
– Reveillark
Jul 26 '17 at 23:22