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Use the function $ f(z) = \dfrac{(\log z)^2}{z^2+1} ; ( |z|>0, - \frac{- \pi}{2} < \arg(z)< \frac{3 \pi}{2})$ to show that $\int_{0}^{\infty} \dfrac{(\ln x)^2}{x^2+1}\, dx = \dfrac{\pi^3}{8}$ .

I have solved it. I'm having problem in showing $ \int_{C_{\rho}} f(z)\, dz \rightarrow0$ as $\rho \rightarrow 0$. I'm using mobile so I haven't written the whole solution here. The link to the work I did is : I did this much work. Photos are in gdrive and here: page 1, page 2, page 3, page 4.

robjohn
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user505797
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  • Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. I will post an answer after more context is given. – robjohn Mar 12 '18 at 09:43
  • Related: https://math.stackexchange.com/questions/850442/an-interesting-identity-involving-powers-of-pi-and-values-of-etas – Jack D'Aurizio Mar 12 '18 at 15:01

1 Answers1

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Without Contour Integration

Note that $$ \begin{align} \int_0^1x^n\log(x)^2\,\mathrm{d}x &=\frac1{n+1}\int_0^1\log(x)^2\,\mathrm{d}x^{n+1}\\ &=-\frac2{n+1}\int_0^1x^n\log(x)\,\mathrm{d}x\\ &=-\frac2{(n+1)^2}\int_0^1\log(x)\,\mathrm{d}x^{n+1}\\ &=\frac2{(n+1)^2}\int_0^1x^n\,\mathrm{d}x\\[3pt] &=\frac2{(n+1)^3} \end{align} $$ Therefore, using the value of $\beta(3)$ from this answer, $$ \begin{align} \int_0^\infty\frac{\log(x)^2}{x^2+1}\,\mathrm{d}x &=2\int_0^1\frac{\log(x)^2}{x^2+1}\,\mathrm{d}x\\ &=2\sum_{k=0}^\infty\int_0^1(-1)^kx^{2k}\log(x)^2\,\mathrm{d}x\\ &=\sum_{k=0}^\infty(-1)^k\frac4{(2k+1)^3}\\[6pt] &=4\beta(3)\\[9pt] &=\frac{\pi^3}8 \end{align} $$

Using Contour Integration

Let $$ \begin{align} \gamma&=\left[(-1+i)\epsilon,Re^{i\sin^{-1}\left(\frac\epsilon{R}\right)}\right]\,\cup\,\color{#C00}{Re^{i\left[\sin^{-1}\left(\frac\epsilon{R}\right),2\pi-\sin^{-1}\left(\frac\epsilon{R}\right)\right]}}\\ &\cup\,\left[Re^{-i\sin^{-1}\left(\frac\epsilon{R}\right)},(-1-i)\epsilon\right]\,\cup\,\color{#090}{[(-1-i)\epsilon,(-1+i)\epsilon]} \end{align} $$ then, because the integral over the large red circular arc is bounded by $\color{#C00}{2\pi R\frac{(\log(R)+2\pi)^3}{R^2-1}}$ and the integral over the small green segment is bounded by $\color{#090}{2\epsilon\frac{(2\pi-\log(\epsilon))^3}{1-\epsilon^2}}$, both of which vanish as $R\to\infty$ and $\epsilon\to0$, $$ \begin{align} \int_\gamma\frac{\log(z)^3}{z^2+1}\,\mathrm{d}z &=\int_0^\infty\frac{\log(x)^3-(\log(x)+2\pi i)^3}{x^2+1}\,\mathrm{d}x\\ &=\int_0^\infty\frac{-6\pi i\log(x)^2+12\pi^2\log(x)+8\pi^3i}{x^2+1}\,\mathrm{d}x\\ &=2\pi i\left[\vphantom{\left(\frac{\pi i}2\right)^3}\right.\underbrace{\left(\frac{\pi i}2\right)^3\left(\frac1{2i}\right)}_{\text{residue at }i}+\underbrace{\left(\frac{3\pi i}2\right)^3\left(-\frac1{2i}\right)}_{\text{residue at }-i}\left.\vphantom{\left(\frac{\pi i}2\right)^3}\right]\\ &=\frac{13\pi^4i}4 \end{align} $$ Thus, $$ -6\pi i\int_0^\infty\frac{\log(x)^2}{x^2+1}\,\mathrm{d}x +12\pi^2\int_0^\infty\frac{\log(x)}{x^2+1}\,\mathrm{d}x =-\frac{3\pi^4i}4 $$ which means $$ \int_0^\infty\frac{\log(x)^2}{x^2+1}\,\mathrm{d}x=\frac{\pi^3}8 $$ and $$ \int_0^\infty\frac{\log(x)}{x^2+1}\,\mathrm{d}x=0 $$

robjohn
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