$$I = \int_{0}^{\infty} \frac{x}{\sqrt{e^{2\pi\sqrt{x}}-1}}\,dx$$.

Solving an Integral

Question

$$I = \int_{0}^{\infty} \frac{x}{\sqrt{e^{2\pi\sqrt{x}}-1}}\,dx$$. I tried to solve by substituting $t = 2\pi\sqrt x \implies t^2 = 4{\pi}^2x \implies tdt = 2\pi^2dx$ $$I = \frac1{8\pi^4}\int_0^{\infty}\frac{t^3}{\sqrt{e^t-1}}dt.$$ But now I'm unable to solve from here.

Answer

$$I = \int_0^{\infty}\frac{x}{\sqrt{e^{2\pi\sqrt x}-1}}dx$$ Let $$x = \frac{t^2}{4\pi^2} \implies dx = \frac{tdt}{2\pi^2}$$ Thus, $$I =\frac1{8\pi^4} \int_0^{\infty}\frac{t^3}{\sqrt{e^t-1}}dt$$ Let's consider a more general integral. $$f(s) = \int_0^{\infty} \frac{e^{-st}}{\sqrt{e^t-1}}dt$$ $$\frac {df(s)}{ds^3}=f_3(s) = -\int_0^{\infty}\frac{t^3e^{-st}}{\sqrt{e^t-1}}dt$$ $$\color{blue}{I = -\frac 1{8\pi^4}\left(\frac {df(s)}{ds^3}\right)_{s=0}}\text{...................#1}$$ Now, $$u =e^{-t} \implies \ln u = -t \implies -\frac{du}{u} = dt$$

Variable Transformation

              +---------+--------+
       t -->> |    0    |   ∞    |
              +---------+--------+
       u -->> |    1    |   0    |
              +---------+--------+

From this, the substitution relationships are:

$$\begin{align*} f(s) & = \int_0^1 \frac{u^{s-1}}{\sqrt{\frac 1{u}-1}}du\\ & = \int_0^1 u^{s-\frac 1{2}}(1-u)^{-\frac1{2}}du\\ & = \beta \left(s+\frac 1{2}, \frac 1{2} \right)\\ & = \frac{\Gamma(s+\frac 1{2})\Gamma(\frac 1{2})}{\Gamma(s+1)}\\ & \implies \color{red}{f_1(s)} = \color{blue}{f(s)}\color{green}{\left(\psi_0\left(s+\frac 1{2}\right) - \psi_0(s+1)\right)} = \color{blue}{f}\color{green}{\times g}\\ \end{align*}$$

What is ψ-function

Derivative of β-function

$$\begin{align*}f_1 = & f\times g\\ & \implies f_2 = f_1g + fg_1 = (fg)g+fg_1 = fg^2+fg_1\\ & \implies f_3 = f_1g^2+2fgg_1+f_1g_1+fg_2\\ & \implies f_3(s) = fg^3 + 3fgg_1 + fg_2 = \color{blue}{f(s)\left[g^3(s)+3g(s)g_1(s) + g_2(s)\right] = f_3(s)}\text{.................#2}\\ \end{align*}$$ let's find zeros of function $$\color{green}{f(s=0)} = \beta\left(\frac 1{2}, \frac 1{2}\right) = \frac {\sqrt\pi\sqrt\pi}{1} = \color{green}{\pi}$$
  • $g(0)$
    • $$\begin{align*} \color{green}{g(s)} & = \left(\psi_0(s+ \frac 1{2}) - \psi_0(s+1)\right) \\ &\implies g(s=0) = \left(\color{red}{\psi_0(\frac 1{2})} - \psi(1)\right) \\ &= (\color{red}{(-\gamma-2\ln2)}+\gamma) = \color{green}{-2\ln(2)}\\ \end{align*}$$
  • $g_1(0)$
    • $$\begin{align*} \color{green}{g_1(s)} &= \left(\psi_1(s+ \frac 1{2}) - \psi_1(s+1)\right)\\ &\implies g_1(0) = \left(\color{red}{\psi_1(\frac 1{2})} - \psi_1(1)\right)\\ & = \left(\color{red}{\frac{\pi^2}{2}}-\frac{\pi^2}{6}\right) = \color{green}{\frac{\pi^2}{3}}\\ \end{align*}$$
  • $g_2(0)$
    • $$\begin{align*} \color{green}{g_2(s)} = &\left(\psi_2(s+ \frac 1{2}) - \psi_2(s+1)\right)\\ &\implies g_2(0) = \left(\color{red}{\psi_2( \frac 1{2})} - \psi_2(1)\right)\\ & = \color{red}{-14\zeta(3)} + 2\zeta(3) = \color{green}{-12\zeta(3)} \end{align*}$$ From above $$\begin{align*}I = & -\frac 1{8\pi^4} f_3(0) = -\frac 1{8\pi^4}\left(\pi\left((-2\ln 2)^3 + 3(-2\ln 2)\times \frac{\pi^2}{3} - 12\zeta(3)\right)\right)\\ & = \frac {\ln^3(2)}{\pi^3} + \frac{\ln(2)}{4\pi} + \frac {3\zeta(3)}{2\pi^3} \\ \end{align*}$$