# Integral(s) involving differential equations or contour integration

• {???}
In summary: I guess I'm a noobie again :-PIn summary, the conversation is about a definite integral that the person is having difficulty solving using different methods. They attempted to use contour integration but ran into issues with the conditions for differentiation under the integral sign. They also mention using the residue theorem and ask for help in finding a solution outside of 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 http://www.wolframalpha.com/input/?i=integrate+cos[x]/[x^2+1]+from+0+to+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

Last edited by a moderator:
{?} said:
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 http://www.wolframalpha.com/input/?i=integrate+cos[x]/[x^2+1]+from+0+to+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 \.

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

{?} said:
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.

micromass said:
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

## 1. What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It is used to model various physical phenomena and is an important tool in many branches of science and engineering.

## 2. What is an integral involving a differential equation?

An integral involving a differential equation is an integral that contains a differential equation in its integrand. It is used to solve differential equations and find the relationship between two or more variables.

## 3. What is contour integration?

Contour integration is a method of evaluating integrals by integrating along a curve in the complex plane. It is used to solve integrals that are difficult to solve using traditional methods.

## 4. How is contour integration used in solving differential equations?

Contour integration is used in solving differential equations by transforming the original equation into an equivalent complex-valued integral, which can then be evaluated using contour integration techniques. This method is particularly useful for solving differential equations with singularities or complex solutions.

## 5. What are some applications of integral involving differential equations or contour integration?

The applications of integral involving differential equations or contour integration are vast and varied. They are used in physics, engineering, economics, and many other fields to model and solve complex problems. Some examples include the calculation of electric fields and potential, the prediction of fluid flow, and the analysis of financial markets.

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