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Consider this integral

$$\int_{0}^{\infty}\ln (x)\sin\left[\pi\left(x^2-{1\over 4}\right)\right]\mathrm dx={\pi\over 8}\tag1$$

How can we prove $(1)?$

An attempt:

Very difficult to even make an attempt of just varying $(1)$

Rewrite $(1)$ as

$$\int_{0}^{\infty}\ln(x)[\sin(\pi x^2)-\cos(\pi x^2)]\mathrm dx\tag2$$

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Hint. By using the Euler gamma evaluation, $$ \int_0^\infty x^s\cdot e^{-(a+i\pi)x^2}dx=\frac12 \frac{\Gamma \left(\frac{s+1}{2}\right)}{(a+i\pi)^{ s/2+1/2}},\quad \text{Re}(s)>-1,\,a>0,\tag1 $$ one gets, as $a \to 0^+$, $$ \int_0^{\infty} x^s\cos\left(\pi x^2\right)\,dx=\frac{\pi^{-s/2-1/2}}{2} \cos \left(\frac{\pi(s+1)}{4} \right)\cdot\Gamma \left(\frac{s+1}{2}\right),\quad-1<\Re(s)<1, \tag2 $$$$ \int_0^{\infty} x^s\sin\left(\pi x^2\right)\,dx=\frac{\pi^{-s/2-1/2}}{2} \sin \left(\frac{\pi (s+1)}{4} \right)\cdot\Gamma \left(\frac{s+1}{2}\right),\quad-1<\Re(s)<1, \tag3 $$ then differentiating $(2)$ and $(3)$ with respect to $s$ and making $s=0$ gives

$$ \begin{align} \int_{0}^{\infty}\ln (x)\cos\left[\pi\left(x^2-{1\over 4}\right)\right]\mathrm dx&=-\frac{1}{4} (\gamma +\log (4 \pi )),\tag4 \\\\ \int_{0}^{\infty}\ln (x)\sin\left[\pi\left(x^2-{1\over 4}\right)\right]\mathrm dx&={\pi\over 8}.\tag5 \end{align} $$

where $\gamma$ is the Euler-Mascheroni constant.

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  • $\begingroup$ Very elegant method. $\endgroup$ – MrYouMath Apr 3 '17 at 13:34
  • $\begingroup$ @MrYouMath Thank you. $\endgroup$ – Olivier Oloa Apr 3 '17 at 14:08

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