Extended Fresnel integral

Question

I want to see the method for computing the following integrals: $$\int_0^\infty\ln(x)\sin(x^2)dx$$ and $$\int_0^\infty\ln(x)\cos(x^2)dx$$ I believe I have seen these in this forum before, but I cannot find them again using the search function. Can someone help me compute these integrals, or maybe just share the original link? Thank you.

Answer 1

Let us consider, for $\alpha\in(-1,0)$, $$ S(\alpha)=\lim_{T\to +\infty}\int_{0}^{T}x^\alpha \sin(x^2)\,dx = \frac{1}{2}\lim_{T\to +\infty}\int_{0}^{T}x^{\frac{\alpha-1}{2}}\sin(x)\,dx. $$ By the self-adjointness of the Laplace transform the RHS equals $$ \frac{1}{2\,\Gamma\left(\frac{1-\alpha}{2}\right)}\int_{0}^{+\infty}\frac{ds}{(s^2+1)s^{\frac{\alpha+1}{2}}} $$ and by the substitution $s=\tan\theta$ and Euler's Beta function we have $$ S(\alpha) = \frac{\pi}{2\,\Gamma\left(\frac{1-\alpha}{2}\right)\cos\left(\frac{\pi}{4}(\alpha+1)\right)}.$$ Using differentiation under the integral sign, $$ \frac{d}{d\alpha} S(\alpha) = \lim_{T\to +\infty}\int_{0}^{T}x^\alpha \log(x)\sin(x^2)\,dx = \frac{\pi}{8}\cdot\frac{2\,\psi\left(\frac{1-\alpha}{2}\right)+\frac{\pi}{4}\tan\left(\frac{\pi}{4}(\alpha+1)\right)}{\Gamma\left(\frac{1-\alpha}{2}\right)\cos\left(\frac{\pi}{4}(\alpha+1)\right)}$$ such that $$ \lim_{T\to +\infty}\int_{0}^{T}\log(x)\sin(x^2)\,dx = \lim_{\alpha\to 0^-}S'(\alpha) = \frac{1}{8}\sqrt{\frac{\pi}{2}}\left[\pi+2\,\psi\left(\tfrac{1}{2}\right)\right]$$ where $\psi\left(\frac{1}{2}\right)=-\gamma-2\log(2)$ by Gauss Digamma Theorem. In the same way $$ \lim_{T\to +\infty}\int_{0}^{T}\log(x)\cos(x^2)\,dx = \lim_{\alpha\to 0^-}C'(\alpha) = -\frac{1}{8}\sqrt{\frac{\pi}{2}}\left[2\gamma+\pi+4\log 2\right].$$

The results can be combined in the non-trivial identity $$ \lim_{T\to +\infty}\int_{0}^{T}\log(x)\left[\sin(x^2)-\cos(x^2)\right]\,dx = \sqrt{\frac{\pi^3}{32}}.$$