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I wishing to calculating the integral$$ I:=\int_0^\infty \frac{y^{n}}{(1+y^2)^2}dy \qquad n=0,1,2 $$ I am looking for real analytic solutions thanks, the closed form is cosecant function so it is very nice.

My input can be from a complex analytic method only: Note, we have double poles at $y=\pm i$. We close contour around a complex function $f(z)=z^n(1+z^2)^{-2}$ in the upper half plane to obtain $$ \text{Res}_{z=i}=\frac{1}{4i}, \ (n=2)\quad \text{Res}_{z=i}=0, \ (n=1)\quad \text{Res}_{z=i}=\frac{1}{2i}, \ (n=0). $$ Now we can write $2\pi i \cdot \text{Res}_{z=i} \ \forall \ n=0,1,2 $. And then we can finish the problem. But I wish to calculate I by real analysis methods, thanks.

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    $\begingroup$ Are you going through some book and just posting all the integrals to this website? $\endgroup$ – Gerry Myerson Apr 29 '14 at 7:47
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    $\begingroup$ @GerryMyerson No, that is quite rude though. If you can find all of the integrals I post in a book, please post it. $\endgroup$ – Jeff Faraci Apr 29 '14 at 8:01
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    $\begingroup$ With $\displaystyle{\large y = \sqrt{1 - t \over t}}$ you will arrive to a Beta function. $\endgroup$ – Felix Marin Apr 29 '14 at 8:07
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    $\begingroup$ @GerryMyerson I am retired and studied mathematics when growing up quite competitively so now one of my hobbies is integration. I post many integrals because this is a math website and I hope that it is of interest to everybody on here. They are not for homework and MOST are not in books. It's just a hobby and I apologize if it makes you annoyed. It is all in the love of integration and sharing math problems... $\endgroup$ – Jeff Faraci Apr 29 '14 at 16:19
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    $\begingroup$ It's weird that a user named @Integrals posts integrals? It's no mystery - the (wo)man likes integrals. $\endgroup$ – Bennett Gardiner Apr 30 '14 at 1:10
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Let $$I_m(a) = \int_0^{\infty} \frac{y^n}{a+y^m}dy$$. Then $-I_2'(1) = \int_0^{\infty} \frac{y^n}{(1+y^2)^2}$ is your integral. Write $$I_m(a) = \frac{1}{a} \int_0^{\infty} \frac{y^n}{1+(y/a^{1/m})^m}dy \\= \frac{1}{a} a^\frac{n+1}{m}\int_0^{\infty} \frac{u^n}{1+u^m}du\\ = a^{\frac{n+1}{m}-1} \frac{\pi}{m \sin{\frac{\pi (n+1)}{m}}}.$$ Here I used an obvious substitution and an integral that can be found using the Beta function. This gives $$-I_2'(1) = \frac{1-n}{2} \frac{\pi}{2} \csc{\frac{\pi}{2}(n+1)} = \frac{1-n}{4} \pi \sec{\frac{\pi n}{2}}$$

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  • $\begingroup$ Thanks a lot. Did you check your result numerically? Very nice solution and nice technique $\endgroup$ – Jeff Faraci Apr 29 '14 at 8:03
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    $\begingroup$ I checked it for m = 2, n = $\sqrt(3)$ and it gave the (exact) right answer. $\endgroup$ – user111187 Apr 29 '14 at 8:04
  • $\begingroup$ Yes, you are correct. i have verified too now. Very nice...Thank you $\endgroup$ – Jeff Faraci Apr 29 '14 at 8:05
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Make the change of variables $1+y^2 = \frac{1}{u}$ and then use the $\beta$ function. See here.

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