Feynman's trick IX

Mellin transform of $\sin x$ and $\cos x$

We show the proof of the following result posted by @infseriesbot $$\int_{0}^{\infty} \frac{(\sin x - \cos x) \ln x}{\sqrt{x}} dx = \frac{\pi^{\frac{3}{2}}}{\sqrt{2}}$$

Proof

First we can start differentiating under the integral sign: $$\int_{0}^{\infty} \frac{(\sin x - \cos x) \ln x}{\sqrt{x}} dx = \int_{0}^{\infty} \frac{(\sin x - \cos x) \left(\frac{d}{dt}\Big|_{t=0+} x^t\right) }{\sqrt{x}} dx = \frac{d}{dt}\Big|_{t=0+} \int_{0}^{\infty} x^{t+\frac{1}{2}-1}(\sin x - \cos x) dx$$ The right hand side is the derivative of the Mellin transform of the function $\displaystyle (\sin x - \cos x)$ at $\displaystyle t+\frac{1}{2}$: We can split the Mellin into two, the Mellin of $\sin x$ and $\cos x$: $$\left\{ \mathscr{M} \sin x \right\} \left(t+\frac{1}{2}\right) = \int_{0}^{\infty} x^{t+\frac{1}{2}-1}\sin x dx$$ $$\left\{ \mathscr{M} \cos x \right\} \left(t+\frac{1}{2}\right) = \int_{0}^{\infty} x^{t+\frac{1}{2}-1}\cos x dx$$ Now we can expand $\sin x$ and $\cos x$ using the Euler's formula and the de Moivre theorem in order to apply the Ramanujan's master theorem: $$\sin(w) = \frac{e^{ix}-e^{-ix}}{2i} = \sum_{n=0}^{\infty} \frac{(ix)^n}{2in!} -\sum_{n=0}^{\infty} \frac{(-ix)^n}{2in!}=\sum_{n=0}^{\infty} \frac{(-x)^n\left[(-i)^n-(i)^n\right]}{2in!}$$ By de Moivre's theorem $$(-i)^{n} = -(e^{-\frac{i\pi}{2}})^{n} = \cos\left(-\frac{n\pi}{2}\right)+i\sin\left(-\frac{n\pi}{2}\right) = \cos\left(\frac{n\pi}{2}\right)-i\sin\left(\frac{n\pi}{2}\right)$$ $$i^{n} = (e^{\frac{i\pi}{2}})^{n} = \cos\left(\frac{n\pi}{2}\right)+i\sin\left(\frac{n\pi}{2}\right)$$ Then $$\frac{\left[(-i)^n-(i)^n\right]}{2i} = \frac{-2i\sin\left(\frac{n\pi}{2}\right)}{2i} = -\sin\left(\frac{n\pi}{2}\right)$$ Therefore $$\sin x = \sum_{n=0}^{\infty}\frac{(-x)^n[-\sin\left(\frac{n\pi}{2}\right)]}{n!}dw$$ In a similar way we can show: $$\cos x = \sum_{n=0}^{\infty}\frac{(-x)^n[\cos\left(\frac{n\pi}{2}\right)]}{n!}dw$$ Applying the Ramanujan's master theorem and using the fact that $\sin$ is odd and $\cos$ is even $$\int_{0}^{\infty} x^{t+\frac{1}{2}-1}\sin x dx = \Gamma\left(t+\frac{1}{2}\right)\sin\left(\frac{t\pi}{2}+\frac{\pi}{4}\right)$$ $$\int_{0}^{\infty} x^{t+\frac{1}{2}-1}\cos x dx = \Gamma\left(t+\frac{1}{2}\right)\cos\left(\frac{t\pi}{2}+\frac{\pi}{4}\right)$$ Therefore: $$\begin{align*} \frac{d}{dt}\Big|_{t=0+} \int_{0}^{\infty} x^{t+\frac{1}{2}-1}\sin x dx = & \frac{d}{dt}\Big|_{t=0+} \Gamma\left(t+\frac{1}{2}\right)\sin\left(\frac{t\pi}{2}+\frac{\pi}{4}\right) \\ =& \lim_{t \to 0+} \frac{1}{2} -\Gamma\left(t+\frac{1}{2}\right)\left[ \pi \sin \left(\frac{\pi t}{2} +\frac{\pi}{4}\right) +2\cos\left(\frac{\pi t}{2} + \frac{\pi}{4}\right) \psi^{(0)}\left(t+\frac{1}{2}\right)\right]\\ =& \frac{\pi^{\frac{3}{2}}}{2\sqrt{2}} + \sqrt{\frac{\pi}{2}} \psi^{(0)}\left(\frac{1}{2}\right) \end{align*}$$ $$\begin{align*} \frac{d}{dt}\Big|_{t=0+} \int_{0}^{\infty} x^{t+\frac{1}{2}-1}\cos x dx = & \frac{d}{dt}\Big|_{t=0+} \Gamma\left(t+\frac{1}{2}\right)\cos\left(\frac{t\pi}{2}-\frac{\pi}{4}\right) \\ =& \lim_{t \to 0+} \frac{1}{2} \Gamma\left(t+\frac{1}{2}\right)\left[2\cos\left(\frac{\pi t}{2} + \frac{\pi}{4}\right)\psi^{(0)}\left(t+\frac{1}{2}\right)- \pi \sin \left(\frac{\pi t}{2} +\frac{\pi}{4} \right)\right]\\ =& \sqrt{\frac{\pi}{2}} \psi^{(0)}\left(\frac{1}{2}\right) - \frac{\pi^{\frac{3}{2}}}{2\sqrt{2}} \end{align*}$$ Hence, we can conclude: $$\int_{0}^{\infty} \frac{(\sin x - \cos x) \ln x}{\sqrt{x}} dx = \frac{d}{dt}\Big|_{t=0+} \int_{0}^{\infty} x^{t+\frac{1}{2}-1}(\sin x - \cos x) dx = \frac{\pi^{\frac{3}{2}}}{\sqrt{2}}$$ $$\boxed{\int_{0}^{\infty} \frac{(\sin x - \cos x) \ln x}{\sqrt{x}} dx = \frac{\pi^{\frac{3}{2}}}{\sqrt{2}}}$$