Hyperbolic integral

Here on facebook timeline Sir Srinivasa Raghava share a integral problem

1. Prove\text{Prove} ∫0∞(tanh⁡(x1)+tanh⁡(x2)+tanh⁡(x3)+tanh⁡(x4))e−xxdx\int_0^{\infty}\left(\tanh\left(\frac{x}{1}\right)+\tanh\left(\frac{x}{2}\right)+\tanh\left(\frac{x}{3}\right)+\tanh\left(\frac{x}{4}\right)\right)\frac{e^{-x}}{x}dx=log⁡(33554432Γ(1312)2Γ(1712)Γ(74)8225π4Γ(712)2Γ(1112)2)13=log⁡(6)=log⁡(2)+log⁡(3)=\log\left(\frac{33554432\Gamma\left(\frac{13}{12}\right)^2\Gamma\left(\frac{17}{12}\right)\Gamma\left(\frac{7}{4}\right)^8}{225\pi^4\Gamma\left(\frac{7}{12}\right)^2\Gamma\left(\frac{11}{12}\right)^2}\right)^{\frac{1}{3}}=\log(6)=\log(2)+\log(3)

So this inspires me to generalize integral in closed form ∫0∞(tanh⁡(x1)+tanh⁡(x2)+tanh⁡(x3)+⋯+tanh⁡(xn))e−xxdx\int_{0}^{\infty} \left( \tanh\left(\frac{x}{1}\right)+\tanh\left(\frac{x}{2}\right)+ \tanh\left(\frac{x}{3}\right)+ \cdots +\tanh\left(\frac{x}{n}\right) \right)\frac{e^{-x}}{x}dx =12∫01(Hx2+n−24−Hx2+n−44)dx=∑1≤k≤nlog⁡(Γ(k4)Γ(k+44)Γ(k+24)2)=\frac{1}{2}\int_0^{1}\left(H_{\frac{x}{2}+\frac{n-2}{4}}-H_{\frac{x}{2}+\frac{n-4}{4}}\right)dx=\sum_{1\leq k\leq n}\log\left(\frac{\Gamma\left(\frac{k}{4}\right)\Gamma\left(\frac{k+4}{4}\right)}{\Gamma\left(\frac{k+2}{4}\right)^2}\right)

2.Worthy analytical results.lim⁡n→0+n∫0∞tanh⁡(xn)e−xxdx=1lim⁡n→∞+(1+∫0∞tanh⁡(xn)e−xxdx)n=elim⁡m→∞+lim⁡n→∞+∏k=1m(1+∫0∞tanh⁡(xn)e−xdxx)nk1m=eγ\lim_{n\to 0^+} {n}\int_0^{\infty}\tanh \left({xn}\right) \frac{e^{-x}}{x} dx = 1\\\lim_{n\to\infty^+}\left(1+\int_0^{\infty} \tanh \left(\frac{x}{n}\right) \frac{e^{-x}}{x} dx\right)^n =e\\ \lim_{m\to\infty^+}\lim_{n\to\infty^+}\prod_{k=1}^{m}\left(1+\int_0^{\infty} \tanh \left(\frac{x}{n}\right) \frac{e^{-x}dx}{x}\right)^{\frac{n}{k}}\frac{1}{m} =e^{\gamma} Here γ\gamma and ee is Euler-Mascheroni constant and Euler's number.

#Calculus

Note by Naren Bhandari
1 year, 1 month ago

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