Integral question tricky

Question

We're asked to evaluate the integral $\int_{-\infty }^{\infty }\dfrac{dx}{% (x^{2}+1)^{3}}.$. For some reason, the methods I tried lead nowhere :(

Any suggestions? Thank you!

Answer 1

You asked for suggestions so this is what you will get:

try to use the residual theorem and use the facts that:

• $\left(x^{2}+1\right)^{3}=\left(x+i\right)^{3}\left(x-i\right)^{3}$

• the integral of your function on a curve in $\mathbb{C}$ which is "very very far away" from $i$ and $-i$ and which is "connecting $-\infty$ and $+\infty$" equals zero.

I do not want to go into more detail because this seems to be a homework. :-P

Answer 2

A simple method is using trigonometric substitution that $x=\tan(y)$. Then the integral have been transform by \begin{equation} \int_{-\pi/2}^{\pi/2}\frac{1/\cos(y)^2}{1/\cos(y)^6}dy=\int_{-\pi/2}^{\pi/2}\cos(y)^4dy=2\int_0^{\pi/2}\cos(y)^4dy=\frac{3\pi}{8} \end{equation}

Answer 3

The integrals $$I_n=\int_{-\infty }^{\infty }\dfrac{dx}{% (x^{2}+1)^{n}}.$$ are not as tricky as you could think. These have to asked several times on this site and received nice answers.

If,as already suggested in many other answers, you change variable (setting $x=\tan (\theta )$) and proceed as it has been done in previous answers to posts, they reduce to $$ I_n= \int_{0}^{\pi/2} \cos^{{2n-2}}(\theta) d\theta $$ and using multiple angle formula you arrive to $$I_n=\frac{\sqrt{\pi } \Gamma \left(n-\frac{1}{2}\right)}{\Gamma (n)}$$