I came up with this problem while messing around in Desmos and was very surprised by the solution I got from Wolfram Alpha. The integral calculator is unable to solve it and WolframAlpha does not provide an indefinite integral that could hint at a possible solution.
My intuition tells me to try integration by parts with $u=\ln(x)$ but I don't know how to take the integral of $\sin\left(\frac{1}{x^{2}}\right)$.
Best Answer
Letting $x\mapsto \frac{1}{\sqrt x}$ yields $$ I=-\frac{1}{4} \int_0^{\infty} \frac{\sin x \ln x}{x^{\frac{3}{2}}} d =-\frac{1}{4} I^{\prime}\left(-\frac{1}{2}\right) $$ where $$ \begin{aligned} I(a) & =\int_0^{\infty} x^{a-1} \sin x d x \\ & =-\Im \int_0^{\infty} x^{a-1} e^{-i x} d x \\ & =-\Im\left(\frac{\Gamma(a)}{i^a}\right) \\ & =-\Gamma(a) \Im\left(e^{-\frac{\pi}{2} i a}\right) \\ & =\Gamma(a) \sin \left(\frac{\pi}{2} a\right) \end{aligned} $$ By logarithmic differentiation, we have $$ \begin{aligned} I(a) & =\Gamma(a) \sin \left(\frac{\pi}{2} a\right) \\ \ln I(a) & =\ln \Gamma(a)+\ln \sin \left(\frac{\pi}{2} a\right) \\ I^{\prime}(a) & =I(a)\left[\psi(a)+\frac{\pi}{2} \cot \left(\frac{\pi}{2} a\right)\right] \\ I^{\prime}\left(-\frac{1}{2}\right) & =-\Gamma\left(-\frac{1}{2}\right) \sin \frac{\pi}{4}\left[\psi\left(-\frac{1}{2}\right)-\frac{\pi}{2}\right] \\ & =2 \sqrt{\pi} \cdot \frac{1}{\sqrt{2}}\left(2-\gamma-\ln 4-\frac{\pi}{2}\right) \end{aligned} $$ now we can conclude that $$ \boxed{I=\sqrt{\frac{\pi}{2}}\left(\frac{\gamma}{2}+\frac{\pi}{4}+\ln 2-1\right)} $$