Find a difficult integral

Question

How to find the following intergal:

$$\mathcal{I}_\text{n}:=\int_0^\infty\frac{x^\text{n}}{1+x^2}\space\text{d}x\tag1$$

I've no idea where to start, I've tried integration by parts but it lead not to a closed form.

"I''ve no idea where to start!" is not a valid comment on this site. — 5xum, Jan 22 '18 at 12:16
Am I missing something or does it diverge for $n\ge 1$ and for $n=0$ it's $arctan$? — Rab, Jan 22 '18 at 12:21
@RabMakh The integral converges when: $$-1<\Re\left(\text{n}\right)<1$$ — Jan Eerland, Jan 22 '18 at 12:23
@AlexFrancisco In my problem I know that $\text{n}\in\mathbb{R}$. But in the general case, it is possible to have a complex number in $\text{n}$. — Jan Eerland, Jan 22 '18 at 12:26
@Dr.SonnhardGraubner Your expression are conditioned to the value of $n$! — Dog_69, Jan 22 '18 at 12:28
What about a trick that reduces it (via reduction formula) to $$I_n=\frac{x^{n-1}}{n-1}-I_{n-2}$$? — , Jan 22 '18 at 12:28
About the link provided by @ProfessorVector . You have done a more general integral there, how is this one difficult? You trying to find an alternative way of doing this or what? Because this is really strange... — Shashi, Jan 22 '18 at 12:57
@5xum I have evaluated the indefinite form, granted with those limits in the OP my hint may not help (at all). — , Jan 22 '18 at 14:12
See https://www.enotes.com/homework-help/prove-following-reduction-formula-integrate-tan-n-364529 — lab bhattacharjee, Jan 22 '18 at 14:22

Mariusz Iwaniuk · Answer 1 · 2018-01-22T13:20:24.993

Maybe like so. Note that:

$$\int_0^{\infty } \exp (-x t) \sin (t) \, dt=\frac{1}{1+x^2}$$

so,

$$\color{red}{\int_0^{\infty } \frac{x^n}{1+x^2} \, dx}=\\\int_0^{\infty } \left(\int_0^{\infty } x^n \exp (-x t) \sin (t) \, dx\right) \, dt=\\\int_0^{\infty } t^{-1-n} \Gamma (1+n) \sin (t) \, dt=\\-\Gamma (-n) \Gamma (1+n) \sin \left(\frac{n \pi }{2}\right)=\\\color{red}{\frac{1}{2} \pi \sec \left(\frac{n \pi }{2}\right)}$$

for: $\color{red}{-1<\Re(n)<1}$

Addition: $-\Gamma (-n) \Gamma (1+n)=\pi \csc (n \pi )$

score 2 · Accepted Answer · answered Jan 22 '18 at 13:29

I don't have full answer, but I may give an idea by finding pattern and then using Induction.

First, if $n=1$, we have $$ \int \frac{x}{1+x^{2}}dx = \int \frac{1}{2} \frac{d(1+x^{2})}{dx} dx = \frac{1}{2} \ln(1+x^{2}) $$
Second, if $n=2$, we have $$ \int \frac{x^{2}}{1+x^{2}}dx = \int 1 - \frac{1}{1+x^{2}} dx = x - \tan^{-1}(x) $$
Third, if $n=3$, we have $$ \int \frac{x^{3}}{1+x^{2}}dx = x(x-\tan^{-1}(x)) - \int(x-\tan^{-1}(x))dx $$ Note that $\int \tan^{-1}(x)dx = x \tan^{-1}(x) - \frac{\ln(1+x^{2})}{2} $ (reference) so we have $$ \int \frac{x^{3}}{1+x^{2}}dx = \frac{x^{2}}{2}-\frac{\ln(1+x^{2})}{2} $$
Fourth, if $n=4$, we have $$ \int \frac{x^{4}}{1+x^{2}}dx = \int \frac{x^{4}-1+1}{1+x^{2}} dx $$ $$ = \int (x^{2}-1) + \frac{1}{1+x^{2}} dx = \frac{x^{3}}{3} -x + \tan^{-1}(x)$$
Continue until you found a pattern .....

You can see that when $n=1$ and $n=3$ we have only two types : even degree power plus $0.5 \ln(1+x^{2})$. While when $n=2$ and $n=4$, we have only two types : odd degree power plus $\tan^{-1}(x)$. You could found a pattern here, then use induction to prove it.

Hope this helps.

The original integral doesn't even converge when $n>1$... – Shashi Jan 22 '18 at 13:58 — Shashi, Jan 22 '18 at 13:58

Find a difficult integral

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