show that $\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)^{n+1}}=\frac {(2n)!\pi}{2^{2n}(n!)^2}$

Question

show that:

$$\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)^{n+1}}=\frac {(2n)!\pi}{2^{2n}(n!)^2}$$

where $n=0,1,2,3,\ldots$.

is there any help?

thanks for all

Answer 1

Write $${\vartheta _n} = \int_{ - \infty }^{ + \infty } {\frac{1}{{{{\left( {1 + {x^2}} \right)}^n}}}} \frac{{dx}}{{1 + {x^2}}}$$

Put $x=\tan\vartheta$. Then $${\vartheta _n} = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{{\cos }^{2n}}\vartheta } d\vartheta $$

so $${\vartheta _n} = 2\int_0^{\frac{\pi }{2}} {{{\cos }^{2n}}\vartheta } d\vartheta $$

We can come up with a recursion for $\vartheta_n$ using integration by parts, namely $${\vartheta _n} = \frac{{2n - 1}}{{2n}}{\vartheta _{n - 1}}$$

This means that $$\prod\limits_{k = 1}^n {\frac{{{\vartheta _k}}}{{{\vartheta _{k - 1}}}}} = \prod\limits_{k = 1}^n {\frac{{2k - 1}}{{2k}}} $$

so by telescopy $$\frac{{{\vartheta _n}}}{{{\vartheta _0}}} = \prod\limits_{k = 1}^n {\frac{{2k - 1}}{{2k}}} $$ but ${\vartheta _0} = \pi $ so $$\begin{align} {\vartheta _n} &= \pi \prod\limits_{k = 1}^n {\frac{{2k - 1}}{{2k}}} \cr &= \pi \prod\limits_{k = 1}^n {\frac{{2k - 1}}{{2k}}} \frac{{2k}}{{2k}} \cr &= \pi \frac{{\left( {2n} \right)!}}{{{2^{2n}}n{!^2}}}=\frac{\pi}{4^n}\binom{2n}{n} \end{align} $$

as desired.

Answer 2

Let $I_k=\int_{-\infty}^\infty\frac{dx}{(x^2+1)^{k+1}}$. An easy integration shows that $I_1=\frac{\pi}{2}$.

Evaluate $I_k$ using integration by parts, letting $du=dx$ and $v=\frac{1}{(x^2+1)^{k+1}}$. We get $u=x$ and $dv=-(k+1)(2x)\frac{1}{(x^2+1)^{k+2}}$. Thus $$I_k=\left.\frac{x}{(x^2+1)^{k+1}}\right|_{-\infty}^\infty+\int_{-\infty}^\infty\frac{2(k+1)x^2}{(x^2+1)^{k+2}}\,dx=\int_{-\infty}^\infty\frac{2(k+1)x^2}{(x^2+1)^{k+2}}\,dx.$$ Rewrite the top of the last integrand as $2(k+1)(x^2+1-1)$. Then we obtain $$I_k=2(k+1)I_k -2(k+1)I_{k+1}.$$ We can rewrite this as $$I_{k+1}=\frac{2k+1}{2(k+1)}I_k.$$ Now the only thing that remains is to show that if $A_k=\frac{(2k)!\pi}{2^{2k}(k!)^2}$ then $A_k$ satisfies the same initial condition as $I_k$ (easy) and the same recurrence relation (straightforward). Since $I_1=A_1$ and $I_k$ and $A_k$ satisfy the same recurrence, the two sequences must be the same.

Answer 3

By the evenness of the function we have

$$2\int_{0}^{\infty} \frac{dx}{(x^2+1)^{n+1}}$$

By the transformation $x^2\to x $ we have

$$\int_{0}^{\infty} \frac{x^{-\frac{1}{2}}}{(x+1)^{n+1}} \, dx$$

which reduces to

$$\frac{\Gamma\left(\frac{1}{2}\right) \Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(n+1\right)}$$

It can be simplifed to

$$\frac{\Gamma\left(\frac{1}{2}\right)2^{1-2n} \Gamma(2n) \sqrt{\pi}}{\Gamma\left(n+1\right) \Gamma(n)}$$

which comes to be

$$\frac{\pi (2n!)}{2^{2n}(n!)^2}$$

Answer 4

Let,$$I_n=\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)^{n+1}}.$$ Put $x=\tan t$. So, $$I_n=2\displaystyle\int_0^{\pi /2} (\cos t)^{2n}\ \mathrm {dt}=2\displaystyle\int_0^{\pi /2} (\sin t)^{2n}\ \mathrm {dt}.$$ By integration by parts we will have, $I_n=\frac{2n-1}{2n}I_{n-1}$.

So, $\frac{I_n}{I_0}=\prod\limits_{k=1}^n {2k-1 \over 2k}=\frac{(2n)!}{4^n(n!)^2}$ also, $I_0= \pi$. Hence our answer.

Answer 5

It is easier to find the integral in a more general form below:

$$I_{n}(a)=\int_{-\infty}^{+\infty} \frac{d x}{\left(x^{2}+a\right)^{n+1}} \quad \text {, where } n=0,1,2,3, \cdots \textrm{ and }a>0.$$

We first start with the easiest one $I_0(a)$, $$\int_{-\infty}^{+\infty} \frac{d x}{x^{2}+a}=\frac{1}{\sqrt{a}} \left[ \tan ^{-1} \frac{x}{\sqrt{a}}\right]_{-\infty}^{\infty}=\frac{\pi}{\sqrt{a}} $$

In order to arrive at the power of (n+1), I plan to differentiate $I_{0}(a)$ w.r.t $a$ by $n$ times. $$ \frac{d^{n}}{da^{n}}\int_{-\infty}^{+\infty} \frac{d x}{x^{2}+a}=\pi \frac{d^{n}}{da^{n}}\left(a^{-\frac{1}{2}}\right) $$ $$ \int_{-\infty}^{\infty} \frac{(-1)^{n} \cdot n !}{\left(x^{2}+a\right)^{n+1}} d x=\pi\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdots\left(-\frac{2 n-1}{2}\right) a^{-\frac{2 n+1}{2}} $$ $$ \boxed{\int_{-\infty}^{\infty} \frac{d x}{\left(x^{2}+a\right)^{n+1}}=\frac{(2n-1)!!\pi}{2^{n} n !}a^{-\frac{2 n+1}{2}}} $$ Putting $a=1$ yields the result $$ \int_{-\infty}^{\infty} \frac{dx }{\left(x^{2}+1\right)^{n+1}}=\frac{(2 n-1) ! ! \pi}{2^{n} n !} \cdot \frac{2\cdot4 \cdot 6 \cdots(2 n)}{2 \cdot 4 \cdot 6 \cdots(2 n)}= \frac{(2 n) ! \pi}{2^{2 n}(n !)^{2}}. \quad \blacksquare $$