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Prove That : $$ \int_{0}^{\infty} \ln x\left[\ln \left( \dfrac{x+1}{2} \right) - \dfrac{1}{x+1} - \psi \left( \dfrac{x+1}{2} \right) \right] \mathrm{d}x = \dfrac{\ln^2 2}{2}+\ln2\cdot\ln\pi-1 $$

where $\psi(z)$ denotes the Digamma Function.

This integral arose from my attempt to find an alternate solution to Problem 5, i.e,

$$ {\large\int}_0^\infty\frac{\ln\left(x^2+1\right)\,\arctan x}{e^{\pi x}-1}dx=\frac{\ln^22}2+\ln2\cdot\ln\pi-1 $$


Here's my try : We have the identity,

$$ \displaystyle \int_0^\infty \frac{\ln y}{(y+a)^2 + b^2}\,\mathrm{d}y \; = \; \tfrac{1}{2b}\,\tan^{-1}\tfrac{b}{a}\,\ln(a^2+b^2) $$

Since substituting $y \mapsto \dfrac{a^2+b^2}{y}$ gives,

$$\displaystyle \int_0^\infty \frac{\ln y}{(y+a)^2+b^2} \, \mathrm{d}y =\ln(a^2+b^2)\int_0^\infty \frac{dy}{y^2+2ay+a^2+b^2} - \int_0^\infty \frac{\ln y}{(y+a)^2+b^2} \, \mathrm{d}y $$

$$ \implies \displaystyle \int_0^\infty \frac{\ln y}{(y+a)^2 + b^2}\,\mathrm{d}y \; = \; \tfrac{1}{2b}\,\tan^{-1}\tfrac{b}{a}\,\ln(a^2+b^2) $$

Putting $a=1$ and $b=x$, we have,

$ \displaystyle \int_0^\infty \frac{2x \ln y}{(y+1)^2 + x^2}\,\mathrm{d}y \; = \; \,\,\ln(1+x^2) \ \tan^{-1}x \tag{1} $

Now, we have to prove,

$$\displaystyle {\large\int}_0^\infty\frac{\ln\left(x^2+1\right)\,\tan^{-1}x}{e^{\pi x}-1} \mathrm{d}x=\frac{\ln^22}2+\ln2\cdot\ln\pi-1$$

Let,

$$\displaystyle \text{I} = {\large\int}_0^\infty\frac{\ln\left(x^2+1\right)\,\tan^{-1}x}{e^{\pi x}-1} \mathrm{d}x $$

$$\displaystyle = \int_{0}^{\infty} \int_{0}^{\infty} \frac{2x \ln y}{[(y+1)^2 + x^2][e^{\pi x} - 1]} \mathrm{d}x \ \mathrm{d}y \quad (\text{From 1}) \tag{2}$$

The inner integral is of the form,

$$ \displaystyle \text{J} = \int_{0}^{\infty} \dfrac{x}{(x^2+a^2)(e^{\pi x} - 1)} \ \mathrm{d}x \ ; \ a = (y+1)$$

I have proved here that,

$\displaystyle \int_{0}^{\infty} \dfrac{\log(1-e^{-2a\pi x})}{1+x^2} \mathrm{d}x = \pi \left[\dfrac{1}{2} \log (2a\pi ) + a(\log a - 1) - \log(\Gamma(a+1)) \right] \tag{3}$

Differentiating both sides w.r.t. $a$, substituting $ a \mapsto \frac{a}{2} $ and $ x \mapsto \frac{x}{a} $, we get,

$\displaystyle \int_{0}^{\infty} \dfrac{x}{(x^2+a^2)(e^{\pi x} - 1)} \ \mathrm{d}x = \dfrac{1}{2} \left[ \dfrac{1}{a} + \ln \left( \dfrac{a}{2} \right) - \psi \left( \dfrac{a}{2} + 1 \right) \right] \tag{4}$

Putting $(4)$ in $(2)$, we have,

$$ \displaystyle \text{I} = \int_{0}^{\infty} \left[ \dfrac{\ln y}{y+1} + \ln y \ln \left( \dfrac{y+1}{2} \right) - \ln y \ \psi \left( \dfrac{y+1}{2} + 1 \right) \right] \mathrm{d}y $$

$ = \displaystyle \int_{0}^{\infty} \ln x\left[\ln \left( \dfrac{x+1}{2} \right) - \dfrac{1}{x+1} - \psi \left( \dfrac{x+1}{2} \right) \right] \mathrm{d}x \tag{*}$


Since the original question has already been proved in the link, so $(*)$ must be equal to the stated closed form. It also matches numerically.

I'm looking for some method to evaluate $(*)$ independent of Problem 5.

Any help will be greatly appreciated.

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1 Answer 1

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Two Auxiliary Identities

First, we shall establish two simple identities.

$\textbf{Identity }(*)$ $$\int^1_0\left[\frac{1}{\log{x}}+\frac{1}{1-x}\right]x^{s-1}\ dx=\log{s}-\psi_0(s)\tag{*}$$

$\text{Proof Outline: }$ Differentiate with respect to $s$, recognise the integral representation of the trigamma function, then integrate back.

$\textbf{Identity }(**)$ $$\int^\infty_0e^{-ax}\log{x}\ dx=-\frac{\gamma+\log{a}}{a}\tag{**}$$

$\text{Proof Outline: }$ Differentiate the result $\int^\infty_0x^{s-1}e^{-ax}\ dx=a^{-s}\Gamma(s)$ with respect to $s$ and set $s=1$.


The Integral In Question

Applying these two identities and switching the order of integration gives us \begin{align} I &:=\int^\infty_0\log{x}\left[\log\left(\frac{x+1}{2}\right)-\psi_0\left(\frac{x+1}{2}\right)-\frac{1}{x+1}\right]\ dx\\ &=\int^\infty_0\log{x}\int^1_0\left[\frac{1}{\log{t}}+\frac{1}{1-t}-\frac{1}{2}\right]t^{\frac{x-1}{2}}\ dt\ dx\\ &=\int^1_0\frac{1}{\sqrt{t}}\left[\frac{1}{\log{t}}+\frac{1}{1-t}-\frac{1}{2}\right]\int^\infty_0 \exp\left[-\left(-\frac{\log{t}}{2}\right)x\right]\log{x}\ dx\ dt\\ &=2\int^1_0\frac{\gamma+\log\left(-\tfrac{\log{t}}{2}\right)}{\sqrt{t}\log{t}}\left[\frac{1}{\log{t}}+\frac{1}{1-t}-\frac{1}{2}\right]\ dt\\ &=2\int^\infty_0\frac{e^{-u}(\gamma+\log{u})}{u}\left[\frac{1}{2u}-\frac{1}{1-e^{-2u}}+\frac{1}{2}\right]\ du \end{align} Let us define $$I(s)=2\int^\infty_0u^{s-1}e^{-u}(\gamma+\log{u})\left[\frac{1}{2}+\frac{1}{2u}-\frac{1}{1-e^{-2u}}\right]\ du$$ so that $I=I(0)$. We have \begin{align} I(s) &=\color{red}{\gamma\Gamma(s)+\gamma\Gamma(s-1)}+\color{blue}{\Gamma'(s)+\Gamma'(s-1)}-2\sum^\infty_{n=0}\int^\infty_0 u^{s-1}e^{-(2n+1)u}(\gamma+\log{u})\ du\\ &=\color{red}{\frac{\gamma\Gamma(1+s)}{s}\frac{s}{s-1}}\color{blue}{-\frac{s\Gamma(s)\psi_0(s)}{1-s}-\frac{\Gamma(s)}{(1-s)^2}}-2\sum^\infty_{n=0}\frac{\Gamma(s)\left(\gamma+\psi_0(s)-\log(2n+1)\right)}{(2n+1)^s}\\ &=\color{red}{\frac{\gamma\Gamma(1+s)}{s-1}}\color{blue}{-\frac{s\Gamma(s)\psi_0(s)}{1-s}-\frac{\Gamma(s)}{(1-s)^2}}\color{green}{-2(1-2^{-s})\Gamma(s)\zeta(s)(\gamma+\psi_0(s))}\\ &\color{purple}{\ \ \ \ -2\Gamma(s)\frac{d}{ds}(1-2^{-s})\zeta(s)}\\ \end{align} With the expansions \begin{align} \Gamma(s)&\sim_0\frac{1}{s}-\gamma+\mathcal{O}(s)\\ \gamma+\psi_0(s)&\sim_0 -\frac{1}{s}+\mathcal{O}(s)\\ (1-2^{-s})\zeta(s)&\sim_0\left(s\log{2}-\frac{s^2\log^2{2}}{2}+\mathcal{O}(s^3)\right)\left(-\frac{1}{2}-\frac{s\log(2\pi)}{2}+\mathcal{O}(s^2)\right)\\ &\sim_0 -\frac{s\log{2}}{2}-\left(\frac{\log^2{2}}{4}+\frac{\log 2\log{\pi}}{2}\right)s^2+\mathcal{O}(s^3) \end{align} we obtain \begin{align} I(s) &\sim_0\color{red}{-\gamma}\color{blue}{-1+\gamma}\color{green}{-\frac{\log{2}}{s}+\gamma\log{2}-\frac{\log^2{2}}{2}-\log{2}\log\pi}+\color{purple}{\frac{\log{2}}{s}-\gamma\log{2}}\\ &\color{purple}{\ \ \ \ \ \ \ +\log^2{2}+2\log{2}\log\pi}+\mathcal{O}(s)\\ &\sim_0 -1+\log{2}\log{\pi}+\frac{\log^2{2}}{2}+\mathcal{O}(s) \end{align} as was to be shown.


Note:

One can obtain the values of the derivatives of the zeta function at $0$ by differentiating Riemann's functional equation and using the Laurent series of the zeta function at $1$.

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    $\begingroup$ (+1) Nice answer. I'd like to add a bounty for the question since I feel that there can be alternate answers too. $\endgroup$
    – MathGod
    Dec 21, 2016 at 15:17

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