MHB Integral Prove: $\frac {n}{n^{2} - 1}$ for $n > 1$

[alyafey22]

Prove that for $$n >1$$ we have

$$\int^{\infty}_{0}\frac {dx}{\left(x + \sqrt {1 + x^{2}}\right)^{n}} = \frac {n}{n^{2} - 1}$$​

This appeared in the Bee integration contest.

 

[polygamma]

Spoiler

$ \displaystyle \int_{0}^{\infty} \frac{1}{(x+\sqrt{1+x^{2}})^{n}} \ dx $Let $u = \sinh x $.$ \displaystyle = \int_{0}^{1} \frac{1}{(\sinh x + \cosh x)^{n}} \ \cosh x \ dx = \int_{0}^{\infty} e^{-nx} \cosh x \ dx $

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

$ \displaystyle = \frac{1}{2} \frac{1}{n-1} + \frac{1}{2} \frac{1}{n+1} = \frac{n}{n^{2}-1} $

 

[chisigma]

Prove that for $$n >1$$ we have

$$\int^{\infty}_{0}\frac {dx}{\left(x + \sqrt {1 + x^{2}}\right)^{n}} = \frac {n}{n^{2} - 1}$$​

This appeared in the Bee integration contest.

Spoiler

Setting $\displaystyle x = \sinh t$ the integral becomes...

$\displaystyle I = \int_{0}^{\infty} \frac{\cosh t}{(\sinh t + \cosh t)^{n}}\ dt\ (1)$ ... and because is...

$\displaystyle \int \frac{\cosh t}{(\sinh t + \cosh t)^{n}}\ dt = - \frac{(\sinh t + \cosh t)^{- n}\ (n \ \cosh t + \sinh t)}{n^{2}-1}\ (2)$

... the result is immediate...

Kind regards

$\chi$ $\sigma$

 

[Saitama]

Prove that for $$n >1$$ we have

$$\int^{\infty}_{0}\frac {dx}{\left(x + \sqrt {1 + x^{2}}\right)^{n}} = \frac {n}{n^{2} - 1}$$​

This appeared in the Bee integration contest.

Hello again! :)

Spoiler

Use the substitution $x+\sqrt{1+x^2}=e^t$. A bit of simplification gives $\displaystyle x=\frac{e^t-e^{-t}}{2} \Rightarrow dx=\frac{e^t+e^{-t}}{2}dt$

Hence, the integral is

$$\int_0^{\infty} \frac{e^t+e^{-t}}{2e^{nt}}dt$$

$$=\frac{1}{2}\int_0^{\infty}\left( e^{-t(n-1)}+e^{-(n+1)}\right)dt$$
Its easy to solve the above integral so I skip the steps. Solving the integral gives:

$$\frac{1}{2}\left(\frac{1}{n+1}+\frac{1}{n-1}\right)=\frac{n}{n^2-1}$$