# Differentiation wrt parameter $\int_0^\infty \sin^2(x)\cdot(x^2(x^2+1))^{-1}dx$

Use differentiation with respect to parameter obtaining a differential equation to solve

$$\int_0^\infty \frac{\sin^2(x)}{x^2(x^2+1)}dx$$ No complex variables, only this approach. Interesting integral and it should have a nice ODE. I have not found the right way yet. we have singularities at $x=\pm i$.

• Are we able to use $\int_0^\infty \frac{\cos(kx)}{x^2+1}dx= \frac{\pi}{2e^k}$ as a given? – Meow Feb 26 '14 at 21:37
• @Alyosha No. because that is just a fourier transform which is easy to do because of the 2 simple poles, but no complex methods. – Jeff Faraci Feb 26 '14 at 21:50

Consider for $a>0$ $$I(a)=\int_0^\infty\frac{\sin^2(ax)}{x^2(x^2+1)}dx$$ Differentiate it twice. Since $$\int_0^\infty\frac{\cos(kx)}{x^2+1}dx=\frac{\pi}{2e^k}$$ for $k>0$ we get $I''(a)=\pi e^{-2a}$. Note that $I'(0)=I(0)=0$, so after solving respective IVP we get $$I(a)=\frac{\pi}{4}(-1+2a+e^{-2a})$$ It is remains to substitute $a=1$.

• This is the approach I was looking for. Very nice, thanks – Jeff Faraci Feb 27 '14 at 3:24
• @Jeff. there were some typos, see edis. – Norbert Feb 27 '14 at 5:24
• @Norbert $I(a)$ is an even function of $a$. – Felix Marin May 30 '14 at 15:52

Well, as an alternative (and I promise, none of the dreaded complex analysis stuff), we could use Parseval's theorem for Fourier transforms:

For example, the FT of $(\sin{x}/x)^2$ is

$$\int_{-\infty}^{\infty} dx \: \frac{\sin^2{x}}{x^2} e^{i k x} = \begin{cases} \\\pi \left (1 - \frac{|k|}{2} \right ) & |k| \le 2 \\ 0 & |k| > 2 \end{cases}$$

The FT of $1/(1+x^2)$ is

$$\int_{-\infty}^{\infty} dx \: \frac1{1+x^2} e^{i k x} = \pi \, e^{-|k|}$$

By Parseval's theorem,

\begin{align}\int_0^{\infty} dx \: \frac{\sin^2{x}}{x^2} \frac1{1+x^2} &= \frac{\pi}{2} \int_0^2 dk \, \left (1 - \frac{k}{2} \right ) e^{-k}\\ &= \frac{\pi}{2} \left (1-e^{-2}- \frac12 (1-3 e^{-2}) \right )\\ &= \frac{\pi}{4} \left (1+\frac1{e^2} \right )\end{align}

• this is nice approach thanks – Jeff Faraci Feb 26 '14 at 21:56

Well, as an alternative (like Mr. Ron Gordon did). \begin{align} \int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}dx&=\int_0^\infty\left[\frac{\sin^2x}{x^2}-\frac{\sin^2x}{1+x^2}\right]dx\\ &=\int_0^\infty\frac{\sin^2x}{x^2}dx-\frac{1}{2}\int_0^\infty\frac{1-\cos2x}{1+x^2}dx\\ &=\frac{\pi}{2}-\frac{1}{2}\int_0^\infty\frac{1}{1+x^2}dx+\frac{1}{2}\int_0^\infty\frac{\cos2x}{1+x^2}dx\\ &=\frac{\pi}{2}-\frac{1}{2}\frac{\pi}{2}+\frac{1}{2}\frac{\pi}{2e^2}\\ &=\frac{\pi}{4}+\frac{\pi}{4e^2} \end{align} where I use these links: $\displaystyle\int_0^\infty\frac{\sin^2x}{x^2}dx$ and $\displaystyle\int_0^\infty\frac{\cos2x}{1+x^2}dx$ to help me out.

Unfortunately, this is not differentiation with respect to parameter method (the Feynman way) but I still love this method. (>‿◠)✌

• ...is that symbol giving us the proverbial? – JP McCarthy May 30 '14 at 10:59
• @JpMcCarthy What did you mean? Please use a basic word for 'proverbial'? Sorry, English is not my native language – Anastasiya-Romanova 秀 May 30 '14 at 11:26
• Is the last symbol a V finger salute?? – JP McCarthy May 30 '14 at 11:32
• @JpMcCarthy In my country it means 'peace' symbol. It often uses when we want to express our idea/ mind without making the other to be offended (ô‿ô) – Anastasiya-Romanova 秀 May 30 '14 at 13:10

How about this method? Denote the integral by $I$. Let $$I(a,b)=\int_0^\infty \frac{\sin(ax)\sin(bx)}{x^2(1+x^2)}dx.$$ Then $I(0,b)=I(a,0)=0$, $I(1,1)=I$ and \begin{eqnarray} \frac{\partial^2I}{\partial a\partial b}&=&\int_0^\infty\frac{\cos(ax)\cos(bx)}{1+x^2}dx\\ &=&\frac{1}{2}\int_0^\infty\frac{\cos((a+b)x)}{1+x^2}dx+\frac{1}{2}\int_0^\infty\frac{\cos((a-b)x)}{1+x^2}dx\\ &=&\frac{\pi}{4}(e^{-(a+b)}+e^{-|a-b|}) \end{eqnarray} where we use $$\int_0^\infty\frac{\cos(\alpha x)}{1+x^2}dx=\frac{\pi}{2}e^{-|\alpha|}.$$ Thus $$I=\frac{\pi}{4}\int_0^1\int_0^1(e^{-(a+b)}+e^{-|a-b|})dadb=\frac{(1+e^2)\pi}{4e^2}.$$