5

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.

Anastasiya-Romanova 秀
  • 19,345
Jeff Faraci
  • 9,878
  • 6
    Are you going through some book and just posting all the integrals to this website? – Gerry Myerson Apr 29 '14 at 07:47
  • I haven't got time to look at the details right now, but couldn't you use the product rule a few times to express this in terms of integrands that have explicit evaluation? – dcs24 Apr 29 '14 at 07:48
  • 3
    @GerryMyerson No, that is quite rude though. If you can find all of the integrals I post in a book, please post it. – Jeff Faraci Apr 29 '14 at 08:01
  • 1
    With $\displaystyle{\large y = \sqrt{1 - t \over t}}$ you will arrive to a Beta function. – Felix Marin Apr 29 '14 at 08:07
  • @FelixMarin Thank you friend – Jeff Faraci Apr 29 '14 at 08:08
  • Sorry, just trying to figure out what you're up to. It's kind of weird to post nothing but integrals, and dozens of them. – Gerry Myerson Apr 29 '14 at 11:19
  • 1
    @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... – Jeff Faraci Apr 29 '14 at 16:19
  • Have a look in Gradshteyn and Ryzhik, most certain it is there. I know I have seen similar integrals – N3buchadnezzar Apr 29 '14 at 22:08
  • @N3buchadnezzar Thank you my friend it sure is. I have had Gradshteyn and Ryzhik for a very long time. This question is also in many other books though. However, I like to see new solutions. For ex: look at the solution to this problem in which I checked – Jeff Faraci Apr 30 '14 at 00:21
  • 1
    It's weird that a user named @Integrals posts integrals? It's no mystery - the (wo)man likes integrals. – Bennett Gardiner Apr 30 '14 at 01:10
  • 1
    @BennettGardiner Thank you my friend! and I prefer my gender to be described by being a definite or indefinite integral. Not male/female. – Jeff Faraci Apr 30 '14 at 01:42
  • 1
    That's ok, I find it strange that people question motives when it comes to the problems they like on this site. I also notice @GerryMyerson has only asked 2 questions ever - and fully half of those are definite integrals! – Bennett Gardiner Apr 30 '14 at 11:44
  • @BennettGardiner Yes I am also quite confused as to why sometimes people question my problems in such a manner. This is a math website after all so it is about sharing problems with one another. Thank you my friend- Integrand – Jeff Faraci Apr 30 '14 at 12:56

2 Answers2

4

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 $$-I_2'(1) = \frac{1-n}{2} \frac{\pi}{2} \csc{\frac{\pi}{2}(n+1)} = \frac{1-n}{4} \pi \sec{\frac{\pi n}{2}}$$

Jeff Faraci
  • 9,878
user111187
  • 5,856
3

Make the change of variables $1+y^2 = \frac{1}{u}$ and then use the $\beta$ function. See here.

Community
  • 1
Mhenni Benghorbal
  • 47,431