Hyperbolic integral

Here on facebook timeline Sir Srinivasa Raghava share a integral problemProve\text{Prove} 0(tanh(x1)+tanh(x2)+tanh(x3)+tanh(x4))exxdx\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))exxdx\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 =1201(Hx2+n24Hx2+n44)dx=1knlog(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)

Important Observation and results \text{Important Observation and results } Based on observation from the general closed form its leads me with results as follows


1. If nmod(4)=0n\bmod(4)=0 and nmod(4)=3n\bmod(4)=3 then the integral always has the form of ln(pq)\displaystyle \ln\left(\frac{p}{q}\right) and ln(p1πq1)\displaystyle \ln\left(\frac{p_1\pi}{q_1}\right) with p>q,p1>q1p>q,p_1>q_1 such that g.c.d(p,q),g.c.d(p1,q1)=1\text{g.c.d}(p,q),\text{g.c.d}(p_1,q_1)=1 respectively. ie I1 =01k4ntanh(xk)exxdx=lnp+lnqI2=01k4n1tanh(xk)exxdx=lnπ+lnp1+lnq1\begin{aligned}I_1 \ =\int_0^{\infty}\sum_{1\leq k\leq 4n} \frac{\tanh \left(\frac{x}{k}\right)e^{-x}}{x} dx =\ln p +\ln q \\ I_2 =\int_0^{\infty}\sum_{1\leq k\leq 4n-1} \frac{\tanh \left(\frac{x}{k}\right)e^{-x}}{x} dx =\ln\pi +\ln p_1 +\ln q_1 \end{aligned} these result can be proved from the main general closed form

2.Worthy analytical results.limn0+n0tanh(xn)exxdx=1limn+(1+0tanh(xn)exxdx)n=elimm+limn+k=1m(1+0tanh(xn)exdxx)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.

Note by Naren Bhandari
3 months ago

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