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I'm trying to take $\int_{0}^{\infty}\frac{x^{1/3}\textrm{ln}(x)}{x^2+1}dx$. I've chosen my contour to be a semicircle in the upper half of the plane with radius $R$ with an indent at the origin of radius $\epsilon$, and the branch cut has been shifted to $(-\pi/2,3\pi/2]$. I have already justified why the inner semicircle and the outer semicircle vanish as $R \rightarrow \infty$ and $\epsilon \rightarrow 0$. However, for some reason, I'm having trouble calculating the residue at $i$. I'm supposed to show the final integral is $\pi^2/6$.

I get that the integral over the entire contour (i.e. $2\pi i$ times the only residue in my contour, $i$) is supposed to be: $\frac{z^{1/3}\textrm{Log}(z)}{z+i}$ evaluated at $z=i$ since it's a simple pole, which turns into $$\frac{e^{\textrm{Log}(i)/3}\textrm{Log}(i)}{2i}$$ $$\frac{e^{(\textrm{ln}(1)+i\pi/2)/3}(\textrm{ln}(1)+i\pi/2)}{2i}$$ $$\frac{e^{i\pi/6}i\pi/2}{2i}$$ $$\frac{\pi\sqrt{3}}{8}+\frac{\pi i}{8}$$ but the real component when multiplied by $2\pi i$ ends up being $-\pi^2/4$. The other semicircle components vanish, so I'm left with the parts on the real axis on $[-R,-\epsilon]\cap[\epsilon,R]$ . And then since I've taken epsilon to $0$ and $R$ to $\infty$, I should get the value for the integral when divided by 2, since the index goes from $0$ to $\infty$ instead of $-\infty$ to $\infty$. However, this is wrong, and I can't figure out what I messed up!

Also, I'm supposed to use this previous calculation to calculate $\int_{0}^{\infty}\frac{x^{1/3}}{x^2+1}dx$, which I'm not sure how to do directly from my past work. Thanks for any help!

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  • $\begingroup$ would it be possible to show all your working on here? (perhaps for the parts that vanish just write $\int_{c_1} dz$ one mistake people make is that they don't equate the sum of the individual contours with the sum of the residues ie $$\int_{\Gamma} f(z) dz = 2 \pi i \cdot \sum res~f(f,z_i) = \sum_{i} \int_{\gamma_i} f(z)dz$$ where $\bigcup_{i}^{n} \gamma_i = \Gamma$ $\endgroup$
    – Vaas
    Apr 30 '20 at 21:39
  • $\begingroup$ out of curiosity; have you got it so that $|f(x)x^{\alpha-1}|\leq \frac{M_1}{|z^{1+\delta_1}|}$ for $|x| > R$ and $|f(x)x^{\alpha-1}| \leq M_2|z|^{\delta_2 - 1}$ for $|x|<r$ i mean since you've got the arcs to vanish i imagine you've got something along these lines it's just with out youre working its hard to see what rationale you've used If you have then remember that $$\int_{0}^{\infty} x^{\alpha-1} f(x) dx = \frac{2 \pi i}{1-e^{2\pi i (\alpha-1)}} \sum_{k}^{n} Res(f(z)z^{\alpha-1},z_k)$$ $\endgroup$
    – Vaas
    Apr 30 '20 at 21:52
  • $\begingroup$ @Vaas I added some more detail to my work... I'm confused, the only residue I'm taking here is the single one at $i$, right? Also for the arcs I just did standard ML estimates (taking the max and multiplying by $\pi R$ or $\pi \epsilon$ and saying the limit of the product is 0.) $\endgroup$
    – user780610
    Apr 30 '20 at 22:24
  • $\begingroup$ Since the function isnt even, i'm not sure you can do $$\int_{-a}^{a} f(z) dz = 2 \int_{0}^{a} f(z) dz$$ $\endgroup$
    – Vaas
    Apr 30 '20 at 22:31
  • $\begingroup$ you've extended into the complex domain, and you've got a contour in the upper half plane giving $$\int_{\Gamma} f(z) = 2 \pi i \sum_{i=1}^{n}Res(f,z_i) = \int_{\gamma_1}fdz+\int_{\gamma_2}fdz+\int_{\gamma_3}fdz+\int_{\gamma_4} dz$$ Then your integral, 0 to infinity would be $$\int_{\gamma_1}f(z) dz = 2 \pi i \sum_{i=1}^{n} Res(f,z_i)- \left(\int_{\gamma_2}fdz+\int_{\gamma_3}fdz+\int_{\gamma_4}fdz\right)$$ Where $\gamma_{1}$ extends from -r to R, $\gamma_2$ from R arching to -R, $\gamma_3$ from -R to -r and finally $\gamma_4$ from -r to r. that....should do it $\endgroup$
    – Vaas
    Apr 30 '20 at 22:43
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I'm hoping this is correct (big confidence eh?)

i'll skip the majority of it as you seem to have done the majority of the working out. like i said try $$\int_{\Gamma}f(z) = 2 \pi i Res(f,i) = \sum_{i=1}^{4} \int_{\gamma_i}f(z) dz$$ where $$\bigcup_{i=1}^{4} \gamma_i = \Gamma$$

From your own working out you have it that $\gamma_2, \gamma_3 \longrightarrow 0,$ as $r \longrightarrow 0 ~\&~R \longrightarrow \infty$

so that leaves you with $$\int_{\Gamma}f(z) dz = 2 \pi i \lim_{z \longrightarrow i} (z-i)\frac{z^{1/3} ln(z)}{z^2+1} = 2 \pi i\left(\frac{\pi\sqrt{3}}{8} + \frac{i \pi}{8}\right) = \frac{i \pi^2 \sqrt{3}}{4} - \frac{\pi^2}{4} = \int_{\gamma_4} f(z) dz$$

what youre missing is this next bit. $$\int_{\gamma_4} f(z) dz = \lim_{(r,R)\longrightarrow (0,\infty)}\int_{-R}^{-r}\frac{z^{1/3}ln(z)}{z^2+1} dz = \frac{\pi^2(-5+3i \sqrt{3})}{12}$$ Moving $\int_{\gamma_4} f(z) dz$ to the left side gives $$2\pi i Res(f,i) - \int_{\gamma_4}f(z) dz = \frac{i \pi^2 \sqrt{3}}{4} - \frac{\pi^2}{4} - \left(\frac{\pi^2(-5+3i \sqrt{3})}{12}\right)$$ and considering the real part only gives $$\frac{-\pi^2}{4} + \frac{5\pi^2}{12} = \frac{5\pi^2}{12} + \frac{3\pi^2}{12} = \frac{\pi^2}{6}$$ as required.

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  • $\begingroup$ Oops, didn't realize you posted this as I was writing up my own solution haha! $\endgroup$
    – user780610
    May 1 '20 at 1:26
  • $\begingroup$ Heh no worries, out of curiosity which if the two methods do you find easier? $\endgroup$
    – Vaas
    May 1 '20 at 2:13
  • $\begingroup$ @user780610 also dont forget to mark the question as answered or it'll linger on the unanswered section (i believe thats how it works) $\endgroup$
    – Vaas
    May 1 '20 at 3:14
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Okay, here is a full answer to my own question, if anyone ever looks at this and wants some closure. My mistake earlier (as Vaas pointed out) was that I treated it like an even function when it was in fact odd.

$\int_{0}^{\infty}\frac{x^{1/3}\textrm{ln}(x)}{x^2+1}dx$ has poles at $\pm i$, but only $i$ is inside the contour we consider, which is as follows:

Here is a link to a diagram of the contour (since I don't have enough reputation to embed it, haha).

So we have $$\int_C=\int_{C_1}+\int_{C_\epsilon}+\int_{C_2}+\int_{C_R}$$ We know $\int_C$ will be $2\pi i \textrm{Res}f(i)$, which turns out to be $ie^{i\pi/6}\pi^2/4$.

For $\int_{C_\epsilon}$ and $\int_{C_R}$ we perform basic ML estimates, multiplying a bound on the integrand by the length of the arc to get a bound on the integral. For example we get that $\int_{C_\epsilon}\leq \pi\epsilon\frac{\epsilon^{1/3}(\textrm{ln}\epsilon+\pi)}{1-\epsilon^2}$, where the length of the arc is $\pi\epsilon$. This goes to $0$ as $\epsilon \rightarrow 0$. The same can by done for $\int_{C_R}$.

Now consider (with the limit) $\int_{C_1}+\int_{C_2}=\int_{0}^{\infty}\frac{x^{1/3}\textrm{ln}(x)}{x^2+1}dx+\int_{-\infty}^{0}\frac{(-x)^{1/3}\textrm{Log}(-x)}{x^2+1}dx$, since the function is odd and we can substitute in $-x$ in the negative component to find it in terms of the positive component. This becomes: $$\int_{0}^{\infty}\frac{x^{1/3}\textrm{ln}(x)}{x^2+1}dx+\int_{0}^{\infty}\frac{e^{i\pi/3}(x)^{1/3}(\textrm{ln}(x)+i\pi)}{x^2+1}dx$$ $$=(1+e^{i\pi/3})\int_{0}^{\infty}\frac{x^{1/3}\textrm{ln}(x)}{x^2+1}dx+i\pi e^{i\pi/3}\int_{0}^{\infty}\frac{(x)^{1/3}}{x^2+1}dx$$

So now we must calculate $\int_{0}^{\infty}\frac{(x)^{1/3}}{x^2+1}dx$ which we would do by the same argument as above (same vanishing arguments, same contour, etc.). We get

$$(1+e^{i\pi/3})\int_{0}^{\infty}\frac{x^{1/3}}{x^2+1}dx=2\pi i \textrm{Res}=\pi e^{i\pi/6}$$ $$\int_{0}^{\infty}\frac{x^{1/3}}{x^2+1}dx=\frac{\pi e^{i\pi/6}}{1+e^{i\pi/3}}=\frac{\pi}{\sqrt{3}}$$

Subbing back in, we now have $$\int_{0}^{\infty}\frac{x^{1/3}\textrm{ln}(x)}{x^2+1}dx=\frac{2\pi i \textrm{Res}-i\pi e^{i\pi/3}\frac{\pi}{\sqrt{3}}}{1+e^{i\pi/3}}=\frac{\frac{ie^{i\pi/6}\pi^2}{4}-i\pi e^{i\pi/3}\frac{\pi}{\sqrt{3}}}{1+e^{i\pi/3}}=\frac{\pi^2}{6}$$ after a lot of simplication.

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