Integral of the day IV

Integral related to the Mellin transform of $e^{ax}$

Today we show the proof of this integral posted by @mvs_rpi \[\int_{0}^{\infty} \frac{\ln x}{\sqrt{x}} e^{-x}\sin x dx = \frac{\sqrt{\pi}}{2i}\left[ \frac{-\gamma -\ln(4-4i)}{\sqrt{1-i}} + \frac{( \gamma+ \ln(4+4i))}{\sqrt{1+i}} \right] \]

Proof

\begin{align*} \int_{0}^{\infty} \frac{\ln x}{\sqrt{x}} e^{-x}\sin x dx =& \int_{0}^{\infty} \frac{\left(\frac{d}{dt}\Big|_{t=0+} x^t\right) }{\sqrt{x}} e^{-x}\sin x dx\\ =&\frac{d}{dt}\Big|_{t=0+} \int_{0}^{\infty} x^{t-\frac{1}{2}} e^{-x}\sin x dx\\ =&\frac{d}{dt}\Big|_{t=0+} \int_{0}^{\infty} x^{t-\frac{1}{2}} e^{-x}\left[\frac{e^{ix}-e^{-ix}}{2i}\right] dx\\ =& \frac{d}{dt}\Big|_{t=0+} \underbrace{\frac{1}{2i}\int_{0}^{\infty} x^{t-\frac{1}{2}} e^{x(-1+i)} dx}_{I_{1}} + \frac{d}{dt}\Big|_{t=0+} \underbrace{\frac{1}{2i}\int_{0}^{\infty} x^{t-\frac{1}{2}} e^{x(-1-i)} dx}_{I_{2}} \end{align*} Recall the Mellin transform: \[\int_{0}^{\infty} x^{s-1}e^{ax} dx \quad a\in \mathbb{C}\] \[ e^{ax} = \sum_{n=0}^{\infty} \frac{a^nx^n}{n!} = \sum_{n=0}^{\infty} \frac{(-a)^n(-x)^n}{n!} = \sum_{n=0}^{\infty} \frac{\phi(n)(-x)^n}{n!}\] where $\displaystyle \phi(n) = (-a)^n$ By Ramanujan's master theorem \[\int_{0}^{\infty} x^{s-1}e^{ax} dx = \Gamma(s)(-a)^{-s} \quad \operatorname{Re}(s)>0, \; \operatorname{Re}(a) <0\] If $\displaystyle s = t+\frac{1}{2}, \; \displaystyle a = -1+i$ \[\int_{0}^{\infty} x^{t-\frac{1}{2}}e^{x(-1+i)} dx = \Gamma\left(t+\frac{1}{2}\right)(1-i)^{-t-\frac{1}{2}} \] \begin{align*} I_{1} = \frac{d}{dt}\Big|_{t=0+} \frac{1}{2i}\int_{0}^{\infty} x^{t-\frac{1}{2}}e^{x(-1+i)} dx = & \frac{d}{dt}\Big|_{t=0+} \frac{1}{2i}\Gamma\left(t+\frac{1}{2}\right)(1-i)^{-t-\frac{1}{2}}\\ =& \frac{\sqrt{\pi}( -\gamma -\ln(4-4i))}{2i\sqrt{1-i}} \end{align*} If $\displaystyle s = t+\frac{1}{2}, \; \displaystyle a = -1-i$ \begin{align*} I_{2} = \frac{d}{dt}\Big|_{t=0+} \frac{1}{2i}\int_{0}^{\infty} x^{t-\frac{1}{2}}e^{x(-1-i)} dx =& I_{1} = \frac{d}{dt}\Big|_{t=0+} \frac{1}{2i}\Gamma\left(t+\frac{1}{2}\right)(1+i)^{-t-\frac{1}{2}} \\ =& \frac{\sqrt{\pi}( -\gamma -\ln(4+4i))}{2i\sqrt{1+i}} \end{align*} Therefore \[ I = I_{1}-I_{2} =+ \frac{\sqrt{\pi}}{2i}\left[ \frac{-\gamma -\ln(4-4i)}{\sqrt{1-i}} + \frac{( \gamma+ \ln(4+4i))}{\sqrt{1+i}} \right] \approx -.0.23611\]
\[\boxed{ \int_{0}^{\infty} \frac{\ln x}{\sqrt{x}} e^{-x}\sin x dx = \frac{\sqrt{\pi}}{2i}\left[ \frac{-\gamma -\ln(4-4i)}{\sqrt{1-i}} + \frac{( \gamma+ \ln(4+4i))}{\sqrt{1+i}} \right] }\]