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Hello, all:

I have a few questions concerning a definite integral where I am meeting with limited success. The most important problem concerns:

[tex]\int_{0}^{\infty}\frac{\cos{x}}{x^2+1}dx[/tex]

Contour integration always has been my last resort, but differentiating under the integral sign, I soon found that some of the conditions were violated and simply could not find the right parameterization so that they held.

Relevant equations:

Naturally the residue theorem in complex analysis. I don't really have anything else...

The attempt at a (wrong) solution:

I attack the doubly improper integral over the entire real line (this will give me twice the integral because the integrand is an even function).

I note that there are no branch cuts and that poles exist at i and -i, so I construct the usual semicircular contour and evaluate the integral:

[tex]\oint_{C}\frac{\cos{z}}{z^2+1}dz=\int_{-r}^{r}\frac{\cos{x}}{x^2+1}dx+\int_{0}^{\pi}\frac{\cos{re^{ix}}}{{r^2}e^{2ix}+1}ire^{ix}dx[/tex]

In both cases I take the limit as r goes to infinity. In big-O notation:

[tex]\frac{\cos{re^{ix}}}{{r^2}e^{2ix}+1}ire^{ix}\in \frac{O(r)}{O(r^2)}[/tex]

Which goes to zero as r goes to infinity, so the second integral goes to zero.

Therefore:

[tex]\oint_{C}\frac{\cos{z}}{z^2+1}dz=\int_{-\infty}^{\infty}\frac{\cos{x}}{x^2+1}dx[/tex]

[tex]\frac{1}{2}\oint_{C}\frac{\cos{z}}{z^2+1}dz=\int_{0}^{\infty}\frac{\cos{x}}{x^2+1}dx[/tex]

Then I calculate the complex residue of i (the only pole enclosed in the contour). Since this is a simple pole of a ratio of two functions f(z)=cos(z) and g(z)=(z^2+1), the residue at i is:

[tex]\frac{f(i)}{g'(i)}=\frac{\cos{i}}{2i}=\frac{\cosh{1}}{2i}=\frac{e+e^{-1}}{4i}[/tex]

The residue theorem tells me that [itex]2i\pi[/itex] times the sum of the residues gives me the value of the closed contour integral, so I have:

[tex]\oint_{C}\frac{\cos{z}}{z^2+1}dz=2i\pi\frac{e+e^{-1}}{4i}=\frac{\pi}{2}(e+e^{-1})[/tex]

By the transitive property, this shows that:

[tex]\int_{0}^{\infty}\frac{\cos{x}}{x^2+1}dx=\frac{\pi}{4}(e+e^{-1})[/tex]

When I did the contour integration I must have done something wrong because http://www.wolframalpha.com/input/?i=integrate+cos[x]/[x^2+1]+from+0+to+infinity" gave me this:

[tex]\int_{0}^{\infty}\frac{\cos{x}}{x^2+1}=\frac{\pi}{2e}[/tex]

My question is this: what did I do wrong? Also, is there a clever

All help is appreciated, as always.

QM

I have a few questions concerning a definite integral where I am meeting with limited success. The most important problem concerns:

[tex]\int_{0}^{\infty}\frac{\cos{x}}{x^2+1}dx[/tex]

Contour integration always has been my last resort, but differentiating under the integral sign, I soon found that some of the conditions were violated and simply could not find the right parameterization so that they held.

Relevant equations:

Naturally the residue theorem in complex analysis. I don't really have anything else...

The attempt at a (wrong) solution:

I attack the doubly improper integral over the entire real line (this will give me twice the integral because the integrand is an even function).

I note that there are no branch cuts and that poles exist at i and -i, so I construct the usual semicircular contour and evaluate the integral:

[tex]\oint_{C}\frac{\cos{z}}{z^2+1}dz=\int_{-r}^{r}\frac{\cos{x}}{x^2+1}dx+\int_{0}^{\pi}\frac{\cos{re^{ix}}}{{r^2}e^{2ix}+1}ire^{ix}dx[/tex]

In both cases I take the limit as r goes to infinity. In big-O notation:

[tex]\frac{\cos{re^{ix}}}{{r^2}e^{2ix}+1}ire^{ix}\in \frac{O(r)}{O(r^2)}[/tex]

Which goes to zero as r goes to infinity, so the second integral goes to zero.

Therefore:

[tex]\oint_{C}\frac{\cos{z}}{z^2+1}dz=\int_{-\infty}^{\infty}\frac{\cos{x}}{x^2+1}dx[/tex]

[tex]\frac{1}{2}\oint_{C}\frac{\cos{z}}{z^2+1}dz=\int_{0}^{\infty}\frac{\cos{x}}{x^2+1}dx[/tex]

Then I calculate the complex residue of i (the only pole enclosed in the contour). Since this is a simple pole of a ratio of two functions f(z)=cos(z) and g(z)=(z^2+1), the residue at i is:

[tex]\frac{f(i)}{g'(i)}=\frac{\cos{i}}{2i}=\frac{\cosh{1}}{2i}=\frac{e+e^{-1}}{4i}[/tex]

The residue theorem tells me that [itex]2i\pi[/itex] times the sum of the residues gives me the value of the closed contour integral, so I have:

[tex]\oint_{C}\frac{\cos{z}}{z^2+1}dz=2i\pi\frac{e+e^{-1}}{4i}=\frac{\pi}{2}(e+e^{-1})[/tex]

By the transitive property, this shows that:

[tex]\int_{0}^{\infty}\frac{\cos{x}}{x^2+1}dx=\frac{\pi}{4}(e+e^{-1})[/tex]

When I did the contour integration I must have done something wrong because http://www.wolframalpha.com/input/?i=integrate+cos[x]/[x^2+1]+from+0+to+infinity" gave me this:

[tex]\int_{0}^{\infty}\frac{\cos{x}}{x^2+1}=\frac{\pi}{2e}[/tex]

My question is this: what did I do wrong? Also, is there a clever

*t*-parameterization I can use which doesn't violate the conditions of differentiation under the integral sign? If there is a method to subdue this integral outside of contour integration, I would be pleased to know: a series expansion, differentiation under the integral sign (I presume this would require a differential equation solution, as*e*appears in the result)...All help is appreciated, as always.

QM

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