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Problem Of The Week # 331 - Nov 13, 2018

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Euge

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Jun 20, 2014
1,925
Here is this week's POTW:

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Show that

$$\int_0^\infty \frac{x^\alpha \log x}{x^2 + 1}\, dx = \frac{\pi^2}{4} \frac{\sin(\pi \alpha/2)}{\cos^2(\pi \alpha/2)}\quad (0 < \alpha < 1)$$


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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
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Jun 20, 2014
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Hello MHB community,

Since this POTW is more difficult than usual, I will give another week for user’s to attempt a solution. Please write a careful analysis in your solution, not simply a calculation.

As a side note, transforming the integral via the transformation $x\mapsto 1/x$, the result can be shown to be valid more generally for $-1 < \alpha < 1$.
 
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Euge

MHB Global Moderator
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Jun 20, 2014
1,925
Congratulations to Opalg , Ackbach for correct, carefully written solutions. Honorable mention goes to Dhamnekar Winod .

1. Here is Opalg 's solution:

Preliminary calculation: $1+e^{i\pi\alpha} = 1 + \cos(\pi\alpha) + i\sin(\pi\alpha) = 2\cos^2(\pi\alpha/2) + 2i\sin(\pi\alpha/2)\cos(\pi\alpha/2) = 2e^{i\pi\alpha/2}\cos(\pi\alpha/2)$.

First main calculation: \(\displaystyle \int_0^\infty \frac{x^\alpha}{x^2+1}dx = \frac{\pi}{2\cos(\pi\alpha/2)}.\)

To see that, integrate the function $f(z) = \frac{z^\alpha}{z^2+1}$ round a contour consisting of the segment $[-R,R]$ of the real axis and the semicircle in the upper half-plane with ends at $R$ and $-R$, and then let $R\to\infty$. (The function has a branch point at the origin but is zero there. If you want to be super-careful, you can avoid the branch point by taking a small semi-circular detour round the origin, staying in the upper half-plane so as to stay on the same branch of the function. The integral along the detour will go to $0$ as its radius goes to $0$.)

The integral consists of three segments: \(\displaystyle I_1= \int_0^\infty \frac{x^\alpha}{x^2+1}dx\) is the part on the positive real axis. For $I_2$, the integral round the semicircle, let $z = Re^{i\theta}$. Then \(\displaystyle I_2 = \int_0^\pi \frac{R^\alpha e^{i\alpha\theta}}{R^2e^{2i\theta} + 1}iRe^{i\theta}d\theta.\) The size of the integrand is of order $R^{\alpha-1}$, so $I_2\to0$ as $R\to\infty$.

For $I_3$, the integral along the negative real axis, put $z = xe^{i\pi} = -x$. Then \(\displaystyle I_3 = \int_{-\infty}^0 \frac{x^\alpha e^{i\pi\alpha}}{x^2+1}(-dx) = e^{i\pi\alpha}I_1.\)

The only singularity of $f(z)$ inside the contour is at $z=i = e^{i\pi/2}$. Since $z^2+1 = (z-i)(z+i)$, the residue there is \(\displaystyle \left.\frac{z^\alpha}{z+1}\right|_{z=i} = \frac{e^{i\pi\alpha/2}}{2i}.\)

By Cauchy's theorem, $I_1+I_2+I_3 = 2\pi i\, \text{res}_if(z)$, so $(1+e^{i\pi\alpha})I_1 = \pi e^{i\pi\alpha/2}$. It follows from the preliminary calculation that \(\displaystyle I_1 = \frac{\pi}{2\cos(\pi\alpha/2)}\), as required.

Second main calculation is to find \(\displaystyle \int_0^\infty \frac{x^\alpha \log x}{x^2+1}dx\). For this, proceed exactly as for the first main calculation, using the function $g(z) = \dfrac{z^\alpha\log z}{z^2+1}$. The integral $J_1$ along the positive real axis is the answer that we are looking for. For $J_2$, the integral round the semicircle, the extra factor $\log R$ in the integrand does not stop the integral from going to zero, because $R^{\alpha-1}\log R \to0$.

For $J_3$, the integral along the negative real axis, the substitution $z=xe^{i\pi}$ gives $\log z = \log x + i\pi$, and so \(\displaystyle J_3 = \int_{-\infty}^0 \frac{x^\alpha e^{i\pi\alpha}(\log x + i\pi)}{x^2+1}(-dx) = e^{i\pi\alpha}(J_1 + i\pi I_1).\)

The residue of $g(z)$ at $i$ is \(\displaystyle \left.\frac{z^\alpha\log z}{z+i}\right|_{z=i} = \frac{e^{i\pi\alpha/2}(i\pi/2)}{2i}.\) Cauchy's theorem then gives \(\displaystyle (1 + e^{i\pi\alpha})J_1 + i\pi e^{i\pi\alpha}I_1 = \frac{i\pi^2}2e^{i\pi\alpha/2}.\)

Now use the two previous calculations to see that \[ 2e^{i\pi\alpha/2}\cos(\pi\alpha/2)J_1 = \frac{i\pi^2}2e^{i\pi\alpha/2} - \frac{i\pi^2e^{i\pi\alpha}}{2\cos(\pi\alpha/2)},\] \[ 2\cos(\pi\alpha/2)J_1 = \frac{i\pi^2}2 - \frac{i\pi^2e^{i\pi\alpha/2}}{2\cos(\pi\alpha/2)} = \frac{i\pi^2}2 - \frac{i\pi^2\bigl(\cos(\pi\alpha/2 + i\sin(\pi\alpha/2)\bigr)}{2\cos(\pi\alpha/2)} = \frac{\pi^2}2\frac{\sin(\pi\alpha/2)}{\cos(\pi\alpha/2)}, \] and so \(\displaystyle J_1 = \frac{\pi^2}4\frac{\sin(\pi\alpha/2)}{\cos^2(\pi\alpha/2)}.\)


2. Here is Ackbach 's solution.

We consider the complex integral equivalent
$$\int_0^{\infty}\frac{z^{\alpha}\ln(z)}{z^2+1}\,dz,$$
and make this part of a contour as follows (let $z=re^{i\theta}$):
\begin{align*}
\gamma_1:&\;\varepsilon<r\le R, \; \theta=0 \\
\gamma_2:&\;0<\theta<\pi, \; r=R \\
\gamma_3:&\;-R<r<-\varepsilon, \; \theta=\pi \\
\gamma_4:&\;r=\varepsilon, \; \pi >\theta >0.
\end{align*}
We define
\begin{align*}
z^{\alpha}&=r^{\alpha}e^{i\alpha \operatorname{arg}(z)}, \\
\ln(z)&=\operatorname{Log}(r)+i\operatorname{arg}(z),
\end{align*}
where
$$-\frac{\pi}{2}<\operatorname{arg}(z)<\frac{3\pi}{2}.$$
That is, we choose the branch cut for these two functions along the negative imaginary axis. Let
$$f(z)=\frac{z^{\alpha}\ln(z) }{z^2+1}.$$
Then the only pole of $f$ inside the contour described above is at $z=i$. Hence, we have that
$$\sum_{j=1}^4\int_{\gamma_j}f(z)\,dz=2\pi i \operatorname{Res}[f,i], $$
by the Residue Theorem. It follows that
$$\int_{\gamma_1}f(z)\,dz=2\pi i \operatorname{Res}[f,i]-\sum_{j=2}^4\int_{\gamma_j}f(z)\,dz=2\pi i\left(\frac{\pi e^{i\alpha\pi/2}}{4}\right)-\sum_{j=2}^4\int_{\gamma_j}f(z)\,dz=\frac{i\pi^2 e^{i\alpha\pi/2}}{2}-\sum_{j=2}^4\int_{\gamma_j}f(z)\,dz.$$
The next step is to show that
$$\int_{\gamma_2}f(z)\,dz=\int_{\gamma_4}f(z)\,dz=0.$$
Both of these will follow by the ML inequality. On $\gamma_4,$ we have that
$$|f(z)|\le \varepsilon^\alpha \sqrt{\ln^2(\varepsilon) + \pi^2}/(1 - \varepsilon^2)\lesssim \varepsilon^\alpha \ln(1/\varepsilon) \lesssim \varepsilon^{(\alpha - 1)}.$$
By ML, $\displaystyle\int_{\gamma_4}$ is $O(\varepsilon^{\alpha}),$ which goes to zero as $\varepsilon\to 0.$ Similarly, on $\gamma_2,$ $f$ is $O(R^{\alpha-1}),$ which goes to zero as $R\to\infty,$ since $\alpha<1.$ So we now have
$$\int_{\gamma_1}f(z)\,dz=\frac{i\pi^2 e^{i\alpha\pi/2}}{2}-\int_{\gamma_3}f(z)\,dz.$$
We can show (I'll omit the tedious calculations) that:
$$\int_{\gamma_1}f(z)\,dz=\int_{0}^{\infty}\frac{x^{\alpha}\ln(x)}{x^2+1}\,dx\quad\text{and}\quad\int_{\gamma_3}f(z)\,dz=e^{i\alpha\pi}\int_{0}^{\infty}\frac{x^{\alpha}\ln(x)}{x^2+1}\,dx+i\pi e^{i\alpha\pi}\int_{0}^{\infty}\frac{x^{\alpha}}{x^2+1}\,dx.$$
We'll pull the first one on the RHS to the LHS to obtain
$$(1+e^{i\alpha\pi})\int_{0}^{\infty}\frac{x^{\alpha}\ln(x)}{x^2+1}\,dx=\frac{i\pi^2 e^{i\alpha\pi/2}}{2}-i\pi e^{i\alpha\pi}\int_{0}^{\infty}\frac{x^{\alpha}}{x^2+1}\,dx. $$
Next we multiply by $e^{-i\alpha\pi}$ and take the real part to obtain
$$(1+\cos(\alpha\pi))\int_{0}^{\infty}\frac{x^{\alpha}\ln(x)}{x^2+1}\,dx=\frac{\pi^2\sin(\alpha\pi/2)}{2}. $$
Trig identities will yield the final result.

Note: many thanks to Euge for lots of help on this one. I learned not a few tricks!


Note that while the above solutions use complex analysis, the integral may be evaluated by real variable methods. One can use the properties of the Gamma and Beta functions, a method employed by Dhamnenkar.
 
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