Integral(s) involving differential equations or contour integration

[{???}]

Hello, all:
I have a few questions concerning a definite integral where I am meeting with limited success. The most important problem concerns:
\int_{0}^{\infty}\frac{\cos{x}}{x^2+1}dx
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:
\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
In both cases I take the limit as r goes to infinity. In big-O notation:
\frac{\cos{re^{ix}}}{{r^2}e^{2ix}+1}ire^{ix}\in \frac{O(r)}{O(r^2)}
Which goes to zero as r goes to infinity, so the second integral goes to zero.
Therefore:
\oint_{C}\frac{\cos{z}}{z^2+1}dz=\int_{-\infty}^{\infty}\frac{\cos{x}}{x^2+1}dx
\frac{1}{2}\oint_{C}\frac{\cos{z}}{z^2+1}dz=\int_{ 0}^{\infty}\frac{\cos{x}}{x^2+1}dx
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:
\frac{f(i)}{g'(i)}=\frac{\cos{i}}{2i}=\frac{\cosh{ 1}}{2i}=\frac{e+e^{-1}}{4i}
The residue theorem tells me that 2i\pi times the sum of the residues gives me the value of the closed contour integral, so I have:
\oint_{C}\frac{\cos{z}}{z^2+1}dz=2i\pi\frac{e+e^{-1}}{4i}=\frac{\pi}{2}(e+e^{-1})
By the transitive property, this shows that:
\int_{0}^{\infty}\frac{\cos{x}}{x^2+1}dx=\frac{\pi }{4}(e+e^{-1})

When I did the contour integration I must have done something wrong because WolframAlpha (http://www.wolframalpha.com/input/?i=integrate+cos%5Bx%5D%2F%5Bx%5E2%2B1%5D+from+0+t o+infinity) gave me this:
\int_{0}^{\infty}\frac{\cos{x}}{x^2+1}=\frac{\pi}{ 2e}
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

[micromass]

Hello, all:
I have a few questions concerning a definite integral where I am meeting with limited success. The most important problem concerns:
\int_{0}^{\infty}\frac{\cos{x}}{x^2+1}dx
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:
\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
In both cases I take the limit as r goes to infinity. In big-O notation:
\frac{\cos{re^{ix}}}{{r^2}e^{2ix}+1}ire^{ix}\in \frac{O(r)}{O(r^2)}
Which goes to zero as r goes to infinity, so the second integral goes to zero.
Therefore:
\oint_{C}\frac{\cos{z}}{z^2+1}dz=\int_{-infty}^{\infty}\frac{\cos{x}}{x^2+1}dx
\frac{1}{2}\oint_{C}\frac{\cos{z}}{z^2+1}dz=\int_{ 0}^{\infty}\frac{\cos{x}}{x^2+1}dx
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:
\frac{f(i)}{g'(i)}=\frac{\cos{i}}{2i}=\frac{\cosh{ 1}}{2i}=\frac{e+e^{-1}}{4i}
The residue theorem tells me that 2i\pi times the sum of the residues gives me the value of the closed contour integral, so I have:
\oint_{C}\frac{\cos{z}}{z^2+1}dz=2i\pi\frac{e+e^{-1}}{4i}=\frac{\pi}{2}(e+e^{-1})
By the transitive property, this shows that:
\int_{-infty}^{\infty}\frac{\cos{x}}{x^2+1}dx=\frac{\pi}{ 4}(e+e^{-1})

When I did the contour integration I must have done something wrong because WolframAlpha (http://www.wolframalpha.com/input/?i=integrate+cos%5Bx%5D%2F%5Bx%5E2%2B1%5D+from+0+t o+infinity) gave me this:
\int_{0}^{\infty}\frac{\cos{x}}{x^2+1}=\frac{\pi}{ 2e}
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

Fixed LaTeX. You have to use / in the [ /tex] brackets, not \.

[{???}]

Thanks, micromass. I'm completely new to this LaTeX equation thing...I've been on Math Equation Object for so long now...

[micromass]

In big-O notation:
\frac{\cos{re^{ix}}}{{r^2}e^{2ix}+1}ire^{ix}\in \frac{O(r)}{O(r^2)}
Which goes to zero as r goes to infinity, so the second integral goes to zero.


This is where your mistake lies. You have \cos(re^{ix}), but this is the complex cosine and it can grow exponentially large!! I.e. you don't have |\cos(re^{ix})|\leq 1 anymore.

In general, this method won't work because the second integral does not go to 0.

What to do then? Well, it turns out that we can consider (with C being the semicircular contour):

\int_{C}{\frac{e^{iz}dz}{z^2+1}}

This can be split up into two integrals, and in this case, the second integral will go to 0. So we have

\int_{C}{\frac{e^{iz}dz}{z^2+1}}\rightarrow \int_{-\infty}^{+\infty}{\frac{e^{ix}dx}{x^2+1}}

And your integral can be found as the real part of this.

[{???}]

This is where your mistake lies. You have \cos(re^{ix}), but this is the complex cosine and it can grow exponentially large!! I.e. you don't have |\cos(re^{ix})|\leq 1 anymore.

In general, this method won't work because the second integral does not go to 0.

What to do then? Well, it turns out that we can consider (with C being the semicircular contour):

\int_{C}{\frac{e^{iz}dz}{z^2+1}}

This can be split up into two integrals, and in this case, the second integral will go to 0. So we have

\int_{C}{\frac{e^{iz}dz}{z^2+1}}\rightarrow \int_{-\infty}^{+\infty}{\frac{e^{ix}dx}{x^2+1}}

And your integral can be found as the real part of this.
Ahh! Thanks, that does make sense now. In my haste such a simple observation eluded me! You can see clearly why I am not cut out for contour integration. I will attack the problem with this new integrand now.

Thanks again,
QM