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This integral seems to produce a neat closed form

We seek to prove that

$$2e^2\int_{0}^{\infty} {x\ln x\over 1-e^{2e\pi{x}}}\mathrm dx=\ln A\tag1$$

Where A is Glaisher Kinkelin Constant

An attempt:

Recall $(2)$

$${x\over e^x-1}=\sum_{n=0}^{\infty}{B_nx^n\over n!}\tag2$$

Then $(1)$ becomes

$$-2e^2\sum_{n=0}^{\infty}{B_n\over n!}\cdot{(2e\pi)^n}\color{red}{\int_{0}^{\infty}x^{n+1}\ln x\mathrm dx}\tag3$$

red part diverges.

Or rewrite $(1)$ as

$$e^2\int_{0}^{\infty}\left({x\ln x\over 1-e^{e\pi{x}}}+{x\ln x\over 1+e^{e\pi{x}}}\right)\mathrm dx\tag4$$

Recall $(5)$

$$\zeta(s)={1\over \Gamma(s)}\int_{0}^{\infty}{x^{s-1}\over e^x-1}\mathrm dx\tag5$$

Go this far, no idea where to go from here.

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  • 1
    $\begingroup$ By regularization of a divergent series the limit defining $A$ is related with $\zeta'(-1)$. By the reflection formula for the $\zeta$ function we have that $\zeta'(-1)$ is related with $\zeta'(2)$, and by a substitution and differentiation under the integral sign your integral just depends on $\zeta'(2)$. $\endgroup$ – Jack D'Aurizio Mar 30 '17 at 13:07
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In $(1)$, let $2e\pi x=u$ to get

$$I=2e^2\int_0^\infty\frac{x\ln(x)}{1-e^{2e\pi x}}\ dx=\frac1{2\pi^2}\int_0^\infty\frac{u\ln(u)-u\ln(2e\pi)}{1-e^u}\ du$$

Then notice that

$$\int_0^\infty\frac{x^{s-1}}{e^x-1}\ dx=\zeta(s)\Gamma(s)$$

By differentiating under the integral, we also find that

$$\int_0^\infty\frac{x^{s-1}\ln(x)}{e^x-1}\ dx=\frac d{ds}\zeta(s)\Gamma(s)=\zeta'(s)\Gamma(s)+\zeta(s)\Gamma'(s)$$

And thus,

$$I=-\frac1{2\pi^2}(\zeta'(2)\Gamma(2)+\zeta(2)\Gamma'(2)\ln(2e\pi))=\ln(A)$$

You may find derivatives of the Riemann zeta function here and the derivative of the Gamma function is merely the digamma function.

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I will note $I$ the quantity you want to evaluate. I guess the first think to do is a sustitution $y=ex$. You get $$I = 2 \int_0^{+\infty}\frac{y(\ln(y)-1)}{1-e^{2\pi y}}dy=2 \left(\int_0^{+\infty}\frac{y}{e^{2\pi y}-1}dy-\int_0^{+\infty}\frac{y\ln(y)}{e^{2\pi y}-1}dy \right).$$ According to your link, we have $$\int_0^{+\infty}\frac{y\ln(y)}{e^{2\pi y}-1}dy=\frac{1}{24}-\frac{1}{2}\ln(A).$$ So, we just have to prove that $$\int_0^{+\infty}\frac{y}{e^{2\pi y}-1}dy=\frac{1}{24}$$ to get the conclusion. To compute this, set $u = 2\pi y$. You get $du = 2\pi dy$. Hence $$ \int_0^{+\infty}\frac{y}{e^{2\pi y}-1}dy =\frac{1}{4\pi^2}\int_0^{+\infty} \frac{u}{e^u-1}du=\frac{1}{4\pi^2}\Gamma(2)\zeta(2).$$ Hence the conclusion, since $\zeta(2) = \frac{\pi^2}{6}.$

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  • $\begingroup$ For the last integral, why not use $u = 2\pi y$ and then use $(5)$ in OP's post? $\endgroup$ – N3buchadnezzar Mar 30 '17 at 12:44
  • $\begingroup$ Yes of course , I'll edit. $\endgroup$ – C. Dubussy Mar 30 '17 at 12:55
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Sketch of the proof

\begin{align*} -2e^2\int_{0}^{\infty} {x e^{-2e\pi x}\log x\over 1-e^{-2e\pi{x}}}\mathrm dx &= -2e^2\sum_{n=0}^\infty\int^\infty_0 xe^{-2e\pi x (n+1)} \log(x)\mathrm dx \tag{1} \\&=-2e^2\sum_{n= 1}^\infty\frac{\psi(2)-\log(2e\pi n)}{(2e\pi)^2 n^2} \tag{2} \\&= \frac{\log(2e\pi)\zeta(2)-\psi(2)\zeta(2)-\zeta'(2)}{2\pi ^2}\tag{3} \\&= \frac{1}{12}-\zeta'(-1)\tag{4} \\&=\log A\tag{5} \end{align*}


Proof of (1)

Use the series expansion

$$\frac{1}{1-e^{2\pi e x}}={e^{-2e\pi x}\over 1-e^{-2e\pi{x}}} = \sum_{n=0}^\infty e^{-2\pi e x (n+1)}=\sum_{n=1}^\infty e^{-2\pi e x n}$$


Proof of (3)

Start by the definition of the gamma function $$\int^\infty_0 x^{v-1} e^{-sx}\,dx = \frac{1}{s^v}\int^\infty_0 x^{v-1} e^{-x}\,dx= \frac{\Gamma(v)}{s^v}$$

By differentiation we have

$$\int^\infty_0 x^{v-1} e^{-sx} \log(x)\,dx = \frac{\Gamma(v)\psi(v)-\log(s)}{s^v}$$

Let $v=2$

$$\int^\infty_0 x e^{-sx} \log(x)\,dx = \frac{\psi(2)-\log(s)}{s^2}$$


Proof of (4)

Use the functional equation

$$\zeta(s)=2^s\pi^{s-1}\sin \left( \frac{s\pi}2\right) \Gamma(1-s)\zeta(1-s)$$

By taking the derivative

$$\zeta'(s)=2^s\pi^{s-1}\log(2\pi)\sin \left( \frac{s\pi}2\right) \Gamma(1-s)\zeta(1-s)+\frac{\pi}{2}2^s\pi^{s-1}\cos\left( \frac{s\pi}2\right) \Gamma(1-s)\zeta(1-s)-2^s\pi^{s-1}\sin \left( \frac{s\pi}2\right) \Gamma(1-s)\psi(1-s)\zeta(1-s)-2^s\pi^{s-1}\sin \left( \frac{s\pi}2\right) \Gamma(1-s)\zeta'(1-s)$$

Take $s \to -1$

$$\zeta'(-1)=-2^{-1}\pi^{-2}\log(2\pi)\zeta(2)+2^{-1}\pi^{-2}\Gamma(2)\psi(2)\zeta(2)+2^{-1}\pi^{-2} \Gamma(2)\zeta'(2)$$

$$2\pi^2\zeta'(-1)=-\log(2\pi)\zeta(2)+\psi(2)\zeta(2)+ \zeta'(2)$$

By some rearrangements and adding $\log(e)=1$

$$\frac{\log(2e\pi)\zeta(2)-\psi(2)\zeta(2)-\zeta'(2)}{2\pi ^2} = \frac{1}{12}-\zeta'(-1)$$


Proof of (5)

Start by the definition

$$\zeta(s) = \lim_{m \to \infty} \left( \sum_{k=1}^{m} k^{-s} - \frac{m^{1-s}}{1-s} - \frac{m^{-s}}{2} + \frac{sm^{-s-1}}{12} \right) \ , \ \text{Re}(s) >-3.$$

Differentiate with respect to $s$

$$\zeta'(s) = \lim_{m \to \infty} \left(- \sum_{k=1}^{m} k^{-s}\log(k) - \frac{m^{1-s}}{(1-s)^2} +\frac{m^{1-s}}{1-s}\log(m)+\frac{m^{-s}}{2}\log(m)\\ + \frac{m^{-s-1}}{12}-\frac{m^{-s-1}}{12}\log(m)\right) $$

Now let $s \to -1$

$$\zeta'(-1) = \lim_{m \to \infty} \left(- \sum_{k=1}^{m} k\log(k) - \frac{m^{2}}{4} +\frac{m^{2}}{2}\log(m)+\frac{m}{2}\log(m)\\ + \frac{1}{12}-\frac{1}{12}\log(m)\right) $$

Take the exponential of both sides

$$e^{\zeta'(-1)}= e^{1/12}\lim_{m \to \infty} \frac{m^{m^2/2+m/2-1/12} e^{-m^2/4}}{ e^{\sum_{k=1}^{m} k\log(k)}} = e^{1/12}\lim_{m \to \infty} \frac{m^{m^2/2+m/2-1/12} e^{-m^2/4}}{ e^{\sum_{k=1}^{m} k\log(k)}} = \frac{ e^{1/12}}{A} $$

We conclude that

$$\zeta'(-1) = \frac{1}{12}-\log A$$

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  • $\begingroup$ Thank you for the detailed explanation @zaid. $\endgroup$ – gymbvghjkgkjkhgfkl Apr 2 '17 at 4:28
  • $\begingroup$ @Bui, you are welcome. If yo have any doubt dont histate to ask. $\endgroup$ – Zaid Alyafeai Apr 2 '17 at 4:34
  • $\begingroup$ Could you please give a reference/link to the derivation of the first equation in Proof of (5)? $\endgroup$ – Hans Jun 15 '17 at 21:43

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