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As you helped me so well last time, I might as well ask a final question! Today I'm trying to prove this:

$$ \int_0^\infty \frac{x^{p}}{ 1+x^{2}}dx = \frac{\pi}{2}\cos\left(p\frac{\pi}{2}\right) $$

For $-1 < p < 1$.

I have no idea how to handle the varying $p$. I've been able to prove the relationship in the case $p = 0$. So now I could try showing this for $-1 < p < 0$ and $0 < p < 1$ but both of those seem to be tricky.

Any tips on dealing with the non-constant $p$? There are poles at $x = i$ and $x = -i$, and if $p < 0$ also at $x = 0$.

I think the best approach would be a semicircle in the top half, maybe with a small inner radius as well in the case $p<0$? Or should I not be trying to prove the relationship for these separate parts but is there a general way to do it?

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You have a branch point at $z=0$, so you need to use the "keyhole" contour. The condition $-1<p<1$ is required for the convergence of the integral. –  Mhenni Benghorbal Aug 10 '12 at 10:58
    
Take a look at this: math.stackexchange.com/questions/121976/… –  Argon Aug 10 '12 at 18:05
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Your statement is false because for $p=1$ the integral diverges but your LHS gives 0! –  Mercy Aug 11 '12 at 9:14
    
@Mercy, I think you meant above "your RHS gives 0" –  DonAntonio Aug 11 '12 at 10:23
    
@DonAntonio Yeah! thanks. –  Mercy Aug 11 '12 at 10:38

3 Answers 3

Let $$ I:(-1,1) \to \mathbb{R},\ I(p)=\int_0^\infty\frac{x^p}{1+x^2}dx. $$ For every $p \in (-1,0]$ one has $-p \in [0,1)$, and setting $t=x^{-1}$ one gets that $$ I(-p)=\int_0^\infty\frac{x^{-p}}{1+x^2}dx=\int_0^\infty\frac{t^{p-2}}{1+t^{-2}}dt=\int_0^\infty\frac{t^p}{1+t^2}=I(p). $$ It is therefore enough to compute $I(p)$ for $p \in [0,1)$.

We have $$ I(0)=\int_0^\infty\frac{1}{1+x^2}dx=\arctan x\big|_0^\infty=\frac{\pi}{2}. $$ From now on $0<p<1$. Denote by $g: \mathbb{C} \to \mathbb{C}$ the principal value of the multiple-valued function $\mathbb{C} \to \mathbb{C}, z \mapsto z^p$, and define $$ f: \mathbb{C} \to \mathbb{C},\ f(z)=\frac{g(z)}{1+z^2}. $$ Let $D_R=\{z=x+iy \in \mathbb{C}:\ |z| \le R,\ y \ge 0\}$ with $R>1$. Thanks to the Residue Theorem we have $$\tag{1} \int_{\partial D_R}f(z)dz=2i\pi\text{Res}(f,i)=\pi g(i)=\pi e^{ip\pi/2}. $$ Since $R>1$, setting $\partial D_R=[-R,R]\times0\cup C_R$, we have $$ \left|\int_{C_R}f(z)dz\right|\le \int_0^\pi R|f(Re^{i\theta})|d\theta \le \pi\frac{R^{p+1}}{R^2-1}, $$ and so $$\tag{2} \lim_{R\to \infty}\int_{C_R}f(z)dz=0. $$ Since $$ \int_{-R}^Rf(z)dz=\int_0^Rf(x)dx+\int_0^Rf(-z)dz=(1+e^{ip\pi})\int_0^Rf(x)dx, $$ it follows from (1) that $$ \int_0^Rf(x)dx=(1+e^{ip\pi})^{-1}\int_{-R}^Rf(z)dz=(1+e^{ip\pi})^{-1}\left[\pi e^{ip\pi/2}-\int_{C_R}f(z)dz\right]. $$ In virtue of (2) we get $$ I(p)=\pi\frac{e^{ip\pi/2}}{1+e^{ip\pi}}=\frac{\pi}{2\cos(p\pi/2)}. $$

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+1 Nice solution and even nicer remarks that, together with joriki's, made me mend my solution. –  DonAntonio Aug 11 '12 at 10:22
    
Thanks Mercy, that's great indeed. I only don't understand the first part, where you show that the integral for negative p is the same as for positive p? I(-z) should be something with x^-p and not x^p right? In fact, every of the 3 or for steps there is a complete mystery to me somehow. Could you elaborate on those? I can't seem to log in from here, so I had to put this up as an extra answer, sorry about that. –  boudewijn Aug 13 '12 at 23:31
    
@boudewijn: your accounts have been merged. Please register to avoid such difficulties in the future. –  Qiaochu Yuan Aug 17 '12 at 16:52

This is a special case of this question to which I gave this answer which says $$ \int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x=\frac{\pi}{m}\csc\left(\pi\frac{n+1}{m}\right) $$ for all real $m>0$ and $-1<n<m-1$.

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It's even more recommended to know how to handle with general case. (+1) –  Chris's sis Aug 11 '12 at 12:04

If we put $$I:=\int_0^\infty\frac{x^p}{1+x^2}dx$$ making the following variable change we get $$x=-u\Longrightarrow dx=-du\Longrightarrow \int_{-\infty}^0\frac{x^p}{1+x^2}dx=\int_\infty^0\frac{(-1)^pu^p}{1+u^2}(-du)=(-1)^pI$$ so that $$\int_{-\infty}^\infty\frac{x^p}{1+x^2}dx=\left(1+(-1)^p\right)I=\left(1+e^{p\pi i}\right)I\Longleftrightarrow$$

$$\Longleftrightarrow \,\,(***)\,\,\,I=\Re\left(\frac{1}{1+e^{p\pi i}}\right)\int_{-\infty}^\infty\frac{x^p}{1+x^2}dx=\frac{1}{2}\int_{-\infty}^\infty\frac{x^p}{1+x^2}dx$$

For $\,p\geq 0\,$ let us try the following contour, with $\,1<R\in\Bbb R\,$: $$C:=[-R,R]\cup\left(\gamma_R:=\left\{z\in\Bbb C\;:\;z=Re^{it}\,\,,0\leq t\leq\pi\right\}\right)$$

Putting $$f(z):=\frac{z^p}{1+z^2}\Longrightarrow Res_{z=i}(f)=\lim_{z\to i}(z-i)f(z)=\frac{i^p}{2i}=\frac{e^{p\pi i/2}}{2i}$$ so by CIT we get $$\oint_Cf(z)\,dz=2\pi i\frac{1}{2i}e^{p\pi i/2}=\pi \left(\cos\frac{p\pi}{2}+i\sin\frac{p\pi}{2}\right)$$

Now, by the estimation lemma, we get $$\left|\int_{\gamma_R}f(z)\,dz\right|\leq \max_{z\in\gamma_R}\frac{|z|^p}{|1+z^2|}R\pi\leq \frac{\pi R^{p+1}}{1-R^2}\xrightarrow [R\to\infty]{} 0\,\,,\,\text{since}\,\,p+1<2$$

Thus, taking the limit above, we get $$(@@@)\,\,\,\pi \left(\cos\frac{p\pi}{2}+i\sin\frac{p\pi}{2}\right)=\lim_{R\to\infty}\oint_Cf(z)dz=\int_{-\infty}^\infty\frac{x^p}{1+x^2}dx\Longrightarrow$$ $$\int_{-\infty}^\infty\frac{x^p}{1+x^2}dx=\pi\cos\frac{p\pi}{2}\Longleftrightarrow$$

$$\stackrel{\text{from (***) above}}\Longleftrightarrow I=\frac{\pi\cos p\frac{\pi}{2}}{2}\,\,\,\,\,Q.E.D.$$

Now, for $\,-1< p<0\,$ all the above remains mutatis mutandi but there's also the matter of the pole at $\,z=0\,$ ,which I can't handle as I can't manage to find out what is this pole's multiplicity. Yet I'm almost sure the residue here is zero, but can't prove it.

Added: As did proposes in a comment below, let's use the following change of variables: $$x=\frac{1}{u}\Longrightarrow dx=-\frac{du}{u^2}\Longrightarrow I(p):=\int_\infty^0\frac{u^{-p}}{1+\frac{1}{u^2}}\left(-\frac{du}{u^2}\right)=:I(-p)$$ so that all we did above indeed remains valid when $-1< p<0\,$

Added trying to address Joriki's point in the comments below: We got, 3 lines before $\,(***)\,$, that $$\int_{-\infty}^0\frac{x^p}{1+x^2}dx=e^{ip\pi}I\Longrightarrow \int_{-\infty}^\infty\frac{x^p}{1+x^2}dx=\int_{-\infty}^0\frac{x^p}{1+x^2}dx+\int_0^\infty\frac{x^p}{1+x^2}dx=$$ $$=(1+e^{ip\pi})I$$ and as Joriki remarks from this it follows $$I=\frac{1}{1+e^{ip\pi}}\int_{-\infty}^\infty\frac{x^p}{1+x^2}dx$$ without taking the real part in the RHS , and thus in $\,(@@@)\,$ we actually have $$\int_{-\infty}^\infty\frac{x^p}{1+x^2}dx=\pi\left(\cos p\frac{\pi}{2}+i\sin p\frac{\pi}{2}\right)=\pi e^{ip\pi}$$ and from here we get the same result as Mercy got below: $$I=\frac{\pi e^{ip\pi}}{1+e^{ip\pi}}=\frac{\pi}{2}\sec p\frac{\pi}{2}$$

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About the case $-1\lt p\lt0$: first show that $I(p)=I(-p)$ by the change of variable $x\to1/x$. –  Did Aug 10 '12 at 22:01
    
+1 That was clever. I'll add it to my answer. –  DonAntonio Aug 11 '12 at 1:05
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@DonAntonio: I don't understand -- the integral from $0$ to $\infty$ is real, but the other one isn't. Surely $A=kI$ and $I=\Re(1/k)A$ can't both be right if $k$ isn't real? –  joriki Aug 11 '12 at 9:30
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@DonAntonio Sorry, I was wrong about that; still the given integral is NOT equal to $\pi/2\cos(p\pi/2)$, otherwise you get $\infty$ of the LHS and $0$ on the RHS! –  Mercy Aug 11 '12 at 9:41
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A minor point: Note that "below" is in the eye of the beholder; people can sort answers differently, and the sort criteria "active" and "votes" can change over time. –  joriki Aug 11 '12 at 10:26

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