Problem without words (Calculus Challenge 9)!

Proposition : $\int_{0}^{\infty} \dfrac{\log(1-e^{-2a\pi x})}{1+x^2} \mathrm{d}x = \pi \left[\dfrac{1}{2} \log (2a\pi ) + a(\log a - 1) - \log(\Gamma(a+1)) \right]$

Proof : Let $\displaystyle \text{I} (a) = \int_{0}^{\infty} \dfrac{\log(1-e^{-2a\pi x})}{1+x^2} \mathrm{d}x$

$\displaystyle = -\sum_{r=1}^{\infty} \dfrac{1}{r} \int_{0}^{\infty} \dfrac{e^{-2ar \pi x}}{1+x^2} \mathrm{d}x$

$\displaystyle = -\sum_{r=1}^{\infty} \dfrac{1}{r} \int_{0}^{\infty} \int_{0}^{\infty} e^{-x(2ar \pi + y)} \sin y \ \mathrm{d}y \ \mathrm{d}x$

$\displaystyle = -\sum_{r=1}^{\infty} \dfrac{1}{r} \int_{0}^{\infty} \dfrac{\sin y}{2ar\pi + y} \mathrm{d}y$

$\displaystyle = - \sum_{r=1}^{\infty} \dfrac{1}{r} \int_{0}^{\infty} \dfrac{\sin x}{2ar\pi + x} \mathrm{d}x$

Substitute $\displaystyle x \mapsto 2 r \pi x$

$\displaystyle \implies \text{I}(a) = - \sum_{r=1}^{\infty} \dfrac{1}{r} \int_{0}^{\infty} \dfrac{\sin 2 r \pi x}{x + a} \mathrm{d}x$

$\displaystyle = -\int_{0}^{\infty} \dfrac{1}{x+a} \sum_{r=1}^{\infty} \dfrac{\sin(2 r \pi x)}{r} \mathrm{d}x$

$\displaystyle = -\pi \int_{0}^{\infty} \dfrac{1}{x+a} \left(\dfrac{1}{2} - \{x\} \right) \mathrm{d}x \quad \left( \because \sum_{r=1}^{\infty} \dfrac{\sin (2 r \pi x)}{r} = \dfrac{\pi}{2} - \pi \{x\} \right)$

$\displaystyle = -\pi \left[ \int_{0}^{\infty} \dfrac{\mathrm{d}x}{2(x+a)} - \int_{0}^{\infty} \dfrac{\{x\}}{x+a} \mathrm{d}x \right]$

$\displaystyle = -\pi \lim_{n \to \infty} \left[ \dfrac{1}{2} \log \left(\dfrac{a+n}{a}\right) - \text{A}(n) \right]$

where $\displaystyle \text{A}(n) = \int_{0}^{n} \dfrac{\{x\} }{x+a} \mathrm{d}x$

Now,

$\displaystyle \text{A}(n) = \int_{0}^{n} \dfrac{\{x\} }{x+a} \mathrm{d}x$

$\displaystyle = \sum_{k=0}^{\infty} \int_{k}^{k+1} \dfrac{x-k}{x+a} \mathrm{d}x$

$\displaystyle = \sum_{k=0}^{\infty} \left[1 - (k+a)\log \left(\dfrac{k+a+1}{k+a}\right) \right]$

$\displaystyle = n - \sum_{k=0}^{\infty} \left[ (k+a+1) \log (k+a+1) - (k+a) \log (k+a) - \log (k+a+1) \right]$

$\displaystyle = n + a\log a - (a+n) \log (a+n) +\log(a \cdot (a+1) \cdot \ldots \cdot (a+n)) - \log a$

$\displaystyle \implies \text{I} (a) = -\pi \lim_{n \to \infty} \left[ \dfrac{1}{2} \log \left(\dfrac{a+n}{a}\right) - n - a\log a + (a+n) \log (a+n) - \log(a \cdot (a+1) \cdot \ldots \cdot (a+n)) + \log a \right]$

Note that $\displaystyle \lim_{n \to \infty} \dfrac{ n! n^t}{t \cdot (t+1) \cdot \ldots \cdot (t+n)} = \Gamma(t)$

$\displaystyle \implies \text{I} (a) = -\pi \lim_{n \to \infty} \left[ \dfrac{1}{2} \log \left(\dfrac{a+n}{a}\right) - n - a\log a + (a+n) \log (a+n) - a\log(n) - \log (n!) + \log(\Gamma(a)) + \log a \right]$

Simplifying using Stirling's Approximation and $\displaystyle \lim_{n \to \infty } n \log \left(1 + \dfrac{a}{n} \right) = a$ , we have,

$\displaystyle \text{I} (a) = -\pi \left[\log(\Gamma(a+1)) - \dfrac{1}{2} \log(2a \pi) - a (\log(a) -1) \right]$

$\displaystyle = \pi \left[ \dfrac{1}{2} \log(2a \pi) + a (\log(a) -1) - \log(\Gamma(a+1)) \right] \quad \square$

Corollary : $\displaystyle \int_{0}^{\infty} \dfrac{\log(1+e^{-2a \pi x})}{1+x^2} \mathrm{d}x = \pi \left[ \log(\Gamma(a)) + \left(2a - \dfrac{1}{2}\right) \log 2 - a(\log a -1) - \log (\Gamma(2a)) \right]$

Proof : Let $\displaystyle \text{G} (a) = \int_{0}^{\infty} \dfrac{\log(1+e^{-2a \pi x})}{1+x^2} \mathrm{d}x$

Note that,

$\displaystyle \text{G}(a) = \int_{0}^{\infty} \dfrac{\log \left( \dfrac{1-e^{-4a \pi x}}{1-e^{-2a \pi x}} \right)}{1+x^2} \mathrm{d}x$

$\displaystyle = \int_{0}^{\infty} \dfrac{\log (1-e^{-4a \pi x}) }{1+x^2} \mathrm{d}x - \int_{0}^{\infty} \dfrac{\log (1-e^{-2a \pi x})}{1+x^2} \mathrm{d}x$

$\displaystyle = \text{I} (2a) - \text{I} (a)$

$\displaystyle =\pi \left[\log(\Gamma(2a+1)) - \dfrac{1}{2} \log(4a \pi) - 2a (\log(2a) -1) \right] - \pi \left[\log(\Gamma(a+1)) - \dfrac{1}{2} \log(2a \pi) - a (\log(a) -1) \right]$

$\displaystyle = \pi \left[ \log(\Gamma(a)) + \left(2a - \dfrac{1}{2}\right) \log 2 - a(\log a -1) - \log (\Gamma(2a)) \right] \quad \square$

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$\therefore \displaystyle \left \lfloor 10^5 \int_{0}^{\infty}{\frac{-\ln(1+e^{-\pi x})}{1+x^2} \ dx} \right \rfloor = \boxed{-22735}$

This is not a solution. It can be seen as a hint or more precisely, Calculus Challenge 9

Prove the following 2 integrals -

$\displaystyle \int_{0}^{\infty}{\ln\left(1 - e^{-2a\pi x}\right) \frac{dx}{1+x^2}} = -\pi\left[\frac{1}{2} \ln 2a\pi + a (\ln a - 1) - \ln \Gamma(a+1)\right]$

$\displaystyle \int_{0}^{\infty}{\ln\left(1 + e^{-2a\pi x}\right) \frac{dx}{1+x^2}} = \pi\left[\ln\Gamma(2a) - \ln \Gamma(a) + a(1 - \ln a) - \left(2a - \frac{1}{2}\right)\ln 2\right]$

$\frac{\pi}{2}(1-\log \pi)$