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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?

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  • $\begingroup$ 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. $\endgroup$ – GEdgar Aug 6 '17 at 1:20
  • $\begingroup$ You can factor out the variable $y$ after some substitution. Leaving you with some form of a beta integral. $\endgroup$ – Zaid Alyafeai Aug 6 '17 at 1:21
  • $\begingroup$ 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. $\endgroup$ – eyeballfrog Aug 6 '17 at 1:22
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    $\begingroup$ @eyeballfrog thank you, I won't do math at 4AM again. $\endgroup$ – segevp Aug 6 '17 at 1:24
  • $\begingroup$ math.stackexchange.com/questions/453783/… $\endgroup$ – Zaid Alyafeai Aug 6 '17 at 1:28

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