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I need to find a closed form for this integral: $$\mathcal{I}=\int_{-\infty}^\infty\frac{\arctan e^x}{\cosh x}\cdot\frac{\tanh\frac{x}2}{x}dx.$$ A numerical integration results in an approximation $\mathcal{I}\approx1.0887930451518...$, and WolframAlpha suggests a possible closed form for this number: $$\mathcal{I}\stackrel?=\frac\pi2\ln2$$

Is it the correct value of this integral? If so, how to prove it?

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Replacing $x \mapsto -x$, we have

$$ I = \int_{-\infty}^{\infty} \frac{\frac{\pi}{2} - \arctan e^{x}}{\cosh x} \, \tanh\left(\frac{x}{2}\right) \, \frac{dx}{x}. $$

Averaging,

\begin{align*} I &= \frac{\pi}{4} \int_{-\infty}^{\infty} \frac{\tanh(x/2)}{\cosh x} \, \frac{dx}{x} \\ &= \pi \int_{0}^{\infty} \frac{e^{-x}(1 - e^{-x})}{(1 + e^{-2x})(1 + e^{-x})} \, \frac{dx}{x} \\ &= \pi \int_{0}^{\infty} \left( \frac{e^{-x}}{1+ e^{-x}} - \frac{e^{-2x}}{1 + e^{-2x}} \right) \, \frac{dx}{x}. \tag{1} \end{align*}

Note that the last integral belongs to the class of integrals called Frullani integral:

$$ \int_{0}^{\infty} \frac{f(bx) - f(ax)}{x} \, dx. \tag{2} $$

Heuristically, we can evaluate (2) as

\begin{align*} \int_{0}^{\infty} \frac{f(bx) - f(ax)}{x} \, dx &= \int_{0}^{\infty} \int_{a}^{b} f'(xt) \, dtdx = \int_{a}^{b} \int_{0}^{\infty} f'(xt) \, dxdt \\ &= \int_{a}^{b} \frac{f(\infty) - f(0)}{t} dt = \{ f(\infty) - f(0) \} \log\left(\frac{b}{a}\right) \end{align*}

and this is justified via Fubini's theorem if $f$ is absoultely continuous and $f'$ is either integrable or non-negative on $[0, \infty)$, which is clearly the case for

$$f(x) = \frac{e^{-x}}{1 + e^{-x}}$$

with $(a, b) = (1, 2)$. Thus it follows that

\begin{align*} \int_{0}^{\infty} \left( \frac{e^{-x}}{1+ e^{-x}} - \frac{e^{-2x}}{1 + e^{-2x}} \right) \, \frac{dx}{x} &= \frac{1}{2}\log 2. \end{align*}

Plugging this back to (1), we obtain the desired result.

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  • $\begingroup$ What is meant by " averaging " ? I forgot. Why can we remove the arctan(exp(x)) term ? $\endgroup$ – mick Jun 8 '14 at 22:15
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    $\begingroup$ @mick, That means we are calculating $I = (I + I)/2$, where we use the original integral for the first $I$ and the transformed integral for the second $I$. $\endgroup$ – Sangchul Lee Jun 10 '14 at 5:08
  • $\begingroup$ Huh ? I dont understand. $\endgroup$ – mick Jun 10 '14 at 11:24
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    $\begingroup$ @mick, I mean, we have used $$ \int_{-\infty}^{\infty} f(x) \, dx = \int_{-\infty}^{\infty} \frac{f(x) + f(-x)}{2} \, dx. $$ $\endgroup$ – Sangchul Lee Jun 10 '14 at 15:27

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