Does the Following integral admit a closed form?
$$ \int_{0}^{\infty}\dfrac{\sin(x^{2})x^{2}\ln(x)}{e^{x^2}-1}dx $$

What I tried was:

Define another integral $ I(a) $ as:
$$ I(a)= \int_{0}^{\infty}\dfrac{\sin(x^{2})x^{a}}{e^{x^{2}}-1}dx $$

Write it as:
$$ I(a) = \text{Im} \left[ \sum_{r=1}^{\infty} \int_{0}^{\infty} x^{a}e^{-x^{2}(r-\iota)}dx \right] $$

Clearly the required integral is $ I'(2) $.

The above simplifies to:

$$ \text{Im}\left[\frac{\Gamma(\frac{a+1}{2})}{2}\sum_{r=1}^{\infty}\frac{1}{(r-\iota)^{\frac{a+1}{2}}} \right] $$

which further simplifies to :

$$ I(a) = \frac{\Gamma(\frac{a+1}{2})}{2}\sum_{r=1}^{\infty} \frac{\sin(\frac{a+1}{2}\tan^{-1}(\frac{1}{r}))}{(r^{2}+1)^{\frac{a+1}{4}}} $$

Let alone $I'(a) $ I could not evaluate even $I(a)$ in general form The only one which i could solve was $ a=1 $
SO that

$$ I(1) = \int_{0}^{\infty}\dfrac{\sin(x^{2})x}{e^{x^{2}}-1}dx = \frac{1}{2}\left[\frac{e^{2\pi}(\pi -1)+(\pi +1)}{e^{2\pi}-1}\right] $$

Any other approach or hints/suggestions are more than welcome!

  • 1
    $\begingroup$ Setting $u=x^2$ gives $$\int_0^\infty\frac{\sin(x^2)x^2\ln(x)}{e^{x^2}-1}dx=\frac{1}{4}\int_0^\infty\frac{\sin(u)\sqrt{u}\ln(u)}{e^u-1}du,$$ which looks a bit better to me. $\endgroup$ – teadawg1337 Nov 30 '16 at 19:12
  • $\begingroup$ i think essentially your integral is a derivative of a so called Hurwitz Zetafunction... $\endgroup$ – tired Nov 30 '16 at 19:21
  • $\begingroup$ @tired yeah but don't these functions give some gee whiz cool answers in terms of elementary functions?? $\endgroup$ – Kunal Gupta Dec 1 '16 at 10:27
  • 1
    $\begingroup$ @Kunal jack's answer makes this option very unlikely $\endgroup$ – tired Dec 1 '16 at 10:28

$$\begin{eqnarray*}\int_{0}^{+\infty}\frac{\sin(x^2)x^2\log x}{e^{x^2}-1}\,dx &=& \frac{1}{4}\int_{0}^{+\infty}\frac{\sin(z)\sqrt{z}\log(z)}{e^z-1}\,dz\\ &=&\frac{1}{4}\left.\frac{d}{d\alpha}\int_{0}^{+\infty}\frac{\sin(z)z^{\alpha+1/2}}{e^z-1}\,dz\,\right|_{\alpha=0^+}\\&=&\frac{1}{4}\left.\frac{d}{d\alpha}\sum_{n\geq 1}\int_{0}^{+\infty}\sin(z)z^{\alpha+1/2}e^{-nz}\,dz\,\right|_{\alpha=0^+}\\&=&\frac{1}{4}\text{Im}\left.\frac{d}{d\alpha}\sum_{n\geq 1}\int_{0}^{+\infty}z^{\alpha+1/2}e^{(i-n)z}\,dz\,\right|_{\alpha=0^+}\\&=&\frac{1}{4}\text{Im}\left.\frac{d}{d\alpha}\sum_{n\geq 1}\frac{\Gamma\left(\alpha+3/2\right)}{(n-i)^{\alpha+3/2}}\right|_{\alpha=0^+}\\&=&\frac{1}{4}\text{Im}\left[\sum_{n\geq 1}\frac{\Gamma'(3/2)}{(n-i)^{3/2}}+\sum_{n\geq 1}\frac{\Gamma(3/2)\log(n-i)}{(n-i)^{3/2}}\right]\end{eqnarray*}$$ depends on the imaginary part of a Hurwitz zeta function and its derivative at $s=\frac{3}{2}$.
Here we have $\Gamma(3/2)=\tfrac{\sqrt{\pi}}{2}$ and $\Gamma'(3/2)=\Gamma(3/2)\psi(3/2) = \tfrac{\sqrt{\pi}}{2}(2-\log 4-\gamma)$.

  • 2
    $\begingroup$ Fools seldom differ (see my comment)...+1 $\endgroup$ – tired Nov 30 '16 at 19:27

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