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I would like to know if I solved this improper integral right:

$$\int_0^{\infty}\frac{1}{(x^2+y)^n}dx$$

for $y\gt 0$

My solution:

$$\int_0^{\infty}\frac{1}{(x^2+y)^n} \, dx=\lim_{M\rightarrow \infty}\int_0^M1\cdot\frac{1}{(x^2+y)^n} \, dx$$

now I used integration by parts:

$$\left[ \frac{x}{(x^2+y)^n} \right]_0^M-\int_0^M\frac{-2nx}{(x^2+y)^{n+1}} \, dx$$

what is inside the square brackets is $0$ so we get that the integral is:

$$\left[-\frac{1}{(x^2+y)^n}\right]_0^M=\frac 1 {y^n}$$

I'm not sure I could use integration by parts so that's is my main concern.

If I made a mistake please let me now.

edit: I know I made a mistake, what it the right way to solve?

Michael Hardy
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segevp
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  • You know that $\int_0^\infty \frac{dx}{x^2+1} = \frac{\pi}{2}$ ...right? So follow this in your steps to find your mistake. – GEdgar Aug 06 '17 at 01:20
  • You can factor out the variable $y$ after some substitution. Leaving you with some form of a beta integral. – Zaid Alyafeai Aug 06 '17 at 01:21
  • The mistake in your derivation comes from your integration by parts. If $dv = dx/(x^2+y)^n$, then what is $v$? It's the same problem as you started with, so you haven't accomplished anything. – eyeballfrog Aug 06 '17 at 01:22
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    @eyeballfrog thank you, I won't do math at 4AM again. – segevp Aug 06 '17 at 01:24
  • https://math.stackexchange.com/questions/453783/show-that-int-infty-infty-fracdxx21n1-frac-2n-pi2/453812#453812 – Zaid Alyafeai Aug 06 '17 at 01:28
  • Note that using $x = \sqrt{y} ,t$ we get

    $$y^{\frac{1}{2}-n}\int^\infty_0 \frac{dt}{(1+t^2)^{n}}$$

     – Zaid Alyafeai Aug 06 '17 at 01:32

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By my essay, putting $a=y$ yields the result $$ \boxed{\int_{-\infty}^{\infty} \frac{d x}{\left(x^{2}+y\right)^{n+1}}=\frac{(2n-1)!!\pi}{2^{n} n !}y^{-\frac{2 n+1}{2}}} $$

Lai
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