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The integral cant be expressed in "standard mathematical functions" -Wolfram I'm asked to determine if its convergent or divergent (I can do that via comparison theorem) and its convergent (and wolfram agrees) but them I'm asked to evaluate because it's convergent. Am I missing something? Is there a way to evaluate if I cant express the integral?

Simon S
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Nawar Ismail
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  • As for "evaluating" it, the definite integral (approximately $0.43049780563463779025$ probably can't be expressed in closed form. You could write it as a series: $2 \sum_{k=0}^\infty (-1)^{k+1} Ei(-2k-1)/(2k+1)$. – Robert Israel Jul 13 '15 at 20:17
  • $\displaystyle\int_{\color{red}0}^\infty\text{sech }x\cdot\ln x~dx ~=~ \frac\pi2\cdot\ln\frac{4~\pi^3}{\Gamma\bigg(\dfrac14\bigg)^4}$ – Lucian Jul 13 '15 at 21:25

2 Answers2

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It is convergent for sure since:

$$ 0\leq \int_{1}^{+\infty}\frac{2\log x}{e^{x}+e^{-x}}\,dx\leq \int_{1}^{+\infty}2(x-1)e^{-x}\,dx = \frac{2}{e}.$$ The upper bound can be improved up to $\frac{\sqrt{\pi}}{e}$ if we exploit $\log x\leq\sqrt{x-1}$ for any $x\geq 1$.

Jack D'Aurizio
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  • Not sech of logx, its sechx multiplied by logx – Nawar Ismail Jul 13 '15 at 19:44
  • Quote: "I'm asked to determine if its convergent or divergent (I can do that via comparison theorem) and its convergent". – Did Jul 13 '15 at 19:55
  • @NawarIsmail What is used here is \begin{align} sech(x) = \frac{1}{\cosh(x)} = \frac{2}{e^{x} + e^{-x}} \end{align} and $\log(x) = \log_{e}(x) = \ln(x)$. What Jack has presented is demonstration that the integral is convergent and less than $2 , e^{-1}$ in the event a closed form is obtained. – Leucippus Jul 13 '15 at 19:56
  • How did you obtain $2(x-1)e^-x$? – Nawar Ismail Jul 13 '15 at 20:02
  • @Leucippus Nawar was responding to the previous version of this answer. – anon Jul 13 '15 at 20:03
  • @NawarIsmail: $\log x\leq(x-1)$ for any $x\geq 1$ is equivalent to $\log(u+1)\leq u$ for anu $u\geq 0$ or to $e^{u}\geq 1+u$, pretty well-known. The other piece is just $e^{x}+e^{-x}\geq e^{x}$. – Jack D'Aurizio Jul 13 '15 at 20:04
  • Thank you guys very much! Very helpful, but so is there no way to obtain an actual value? – Nawar Ismail Jul 13 '15 at 20:12
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    @NawarIsmail: numerical integration returns $0.43049780563463779\ldots$. Interestingly, the integral over $\mathbb{R}^+$ equals $$ \frac{\pi}{2}\log\left(\frac{4\pi^3}{\Gamma\left(\frac{1}{4}\right)^{4}}\right)$$ and the integral over $[0,1]$ can be computed from a Taylor expansion centered at $x=1$. – Jack D'Aurizio Jul 13 '15 at 20:19
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One method to attempt to evaluate the integral is as follows.

Using \begin{align} sech(x) = 2 \, \sum_{k=0}^{\infty} (-1)^{k} \, e^{-(2k+1) \, x} \end{align} then \begin{align} I(x) &= \int_{1}^{\infty} \frac{t^{x-1} \, dt}{\cosh(t)} \\ &= 2 \, \sum_{k=0}^{\infty} (-1)^{k} \, \int_{1}^{\infty} t^{x-1} \, e^{-(2k+1) t} \, dt \\ &= 2 \, \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^{x}} \, \int_{2k+1}^{\infty} e^{-u} \, u^{x-1} \, du \mbox{ where } u = (2k+1) t \\ &= 2 \sum_{k=0}^{\infty} \frac{(-1)^{k} \, \Gamma(x, 2k+1)}{(2k+1)^{x}}, \end{align} where $\Gamma(x,\alpha)$ is one of the incomplete gamma functions. Now taking the derivative of both sides with respect to $x$ and setting $x=1$ leads to the desired integral being \begin{align} \int_{1}^{\infty} \frac{\ln(t) \, dt}{\cosh(t)} = 2 \, \sum_{k=0}^{\infty} \frac{(-1)^{k} \, \Gamma'(1,2k+1)}{2k+1} - 2 \, \sum_{k=0}^{\infty} \frac{(-1)^{k} \, \ln(2k+1) \, \Gamma(1,2k+1)}{2k+1}. \end{align}

Leucippus
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