Feynman's trick

Integral involving the Feynman's integral trick

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— 級数bot (@infseriesbot) June 16, 2021

Today we show the proof of this result which involves the famous Feynman's integral trick twice. The post of @infseriesbot has a typo (the sign is wrong). First, we differentiate the parameter $a$ under the integral sign: \begin{align*} \frac{d}{da} \int_{0}^{\infty} \frac{(1-\cos a x)\ln(x)}{x^2}dx =& \int_{0}^{\infty} \frac{d}{da} \frac{(1-\cos a x)\ln(x)}{x^2}dx\\ =&\int_{0}^{\infty} \frac{ \sin a x \ln (x)}{x} dx\\ =&\int_{0}^{\infty} \frac{ \sin w \ln (\frac{w}{a})}{w} dw \quad (w \mapsto ax)\\ =&\int_{0}^{\infty} \frac{ \sin w \ln (w)}{w} dw-\ln (a)\int_{0}^{\infty} \frac{ \sin w }{w} dw \\ =&\int_{0}^{\infty} \frac{\sin w \ln (w)}{w} dw-\frac{\pi \ln(a)}{2}\\\ =& \int_{0}^{\infty} \frac{ \sin w \frac{d}{dt} w^{t} \big|_{t=0+}}{w} dw-\frac{\pi \ln(a)}{2}\\ =& \frac{d}{dt}\Big|_{t=0+}\int_{0}^{\infty} w^{t-1} \sin w dw -\frac{\pi \ln(a)}{2} \end{align*} Here we are using the fact that $\displaystyle \frac{d}{dt} w^{t} \big|_{t=0+} =\ln(w)$ and the fact that $\displaystyle \int_{0}^{\infty} \frac{\sin(w)}{w}dw = \frac{\pi}{2}$ (a famous result obtained through complex analysis). Note that the first integral is the Mellin transform of $\sin w$, so we could use the following expansion series of $\sin$ (a consequence of the Euler's formula and the de Moivre theorem) in oder to apply the Ramanujan's master theorem: \[ \sin w = \sum_{n=0}^{\infty}\frac{(-w)^n[-\sin\left(\frac{n\pi}{2}\right)]}{n!} \] Then, by Ramanujan's master theorem: \begin{align*} \frac{d}{dt}\Big|_{t=0+}\int_{0}^{\infty} w^{t-1} \sin w dw= &\frac{d}{dt}\Big|_{t=0+} \left[\Gamma(t) \sin\left(\frac{t\pi}{2}\right)\right] \\ =& \lim_{t \to 0+}\Gamma(t)\left[\frac{\pi}{2}\cos\left(\frac{t\pi}{2}\right) +\psi^0(t)\sin\left(\frac{\pi t}{2} \right) \right]\\ =& \lim_{t \to 0+}\Gamma(t+1)\left[\frac{\pi}{2}\frac{\cos\left(\frac{t\pi}{2}\right)}{t} +\psi^0(t)\frac{\sin\left(\frac{\pi t}{2} \right)}{t} \right] \end{align*} Update: I found a cool way to calculate this limit:
Note that \begin{align*} \frac{\pi}{2}\frac{\cos\left(\frac{t\pi}{2}\right)}{t} = &\frac{\pi}{2}\frac{1}{t}-\frac{\pi^3}{2^3}\frac{t}{2!}+\frac{\pi^5}{2^5}\frac{t^3}{4!}-\frac{\pi^7}{2^7}\frac{t^5}{6!}+... \\ =& \frac{\pi}{2}\frac{1}{t}+\mathcal{O}(t) \end{align*} Also note that \begin{align*}\psi^0(t)\frac{\sin\left(\frac{\pi t}{2} \right)}{t} =&\psi^0(t)\frac{\pi}{2}-\psi^0(t)\frac{\pi^3}{2^3}\frac{t^2}{3!}+\psi^0(t)\frac{\pi^5}{2^5}\frac{t^4}{5!}-\psi^0(t)\frac{\pi^7}{2^7}\frac{t^6}{7!}+...\\ =& \psi^0(t)\frac{\pi}{2}-t\psi^0(t)\left(\frac{\pi^3}{2^3}\frac{t}{3!}-\frac{\pi^5}{2^5}\frac{t^3}{5!}+\frac{\pi^7}{2^7}\frac{t^5}{7!}-...\right)\\ =& \psi^0(t)\frac{\pi}{2}-t\psi^0(t)\mathcal{O}(t) \end{align*} Therefore \begin{align*}\frac{\pi}{2}\frac{\cos\left(\frac{t\pi}{2}\right)}{t} +\psi^0(t)\frac{\sin\left(\frac{\pi t}{2} \right)}{t} =& \frac{\pi}{2}\frac{1}{t}+\psi^0(t)\frac{\pi}{2}+\mathcal{O}(t)-t\psi^0(t)\mathcal{O}(t) \\ =& \frac{\pi}{2}\left(\frac{1}{t}+ \psi^0(t)\right)+\mathcal{O}(t)-t\psi^0(t)\mathcal{O}(t)\\ =&\frac{\pi}{2}\psi^{0}(t+1)+\mathcal{O}(t)-t\psi^0(t)\mathcal{O}(t)\\ \end{align*} \[\lim_{t \to 0+}\Gamma(t+1)\left[\frac{\pi}{2}\frac{\cos\left(\frac{t\pi}{2}\right)}{t} +\psi^0(t)\frac{\sin\left(\frac{\pi t}{2} \right)}{t} \right] =\lim_{t \to 0+}\Gamma(t+1)\left[\frac{\pi}{2}\psi^{0}(t+1)+\mathcal{O}(t)-t\psi^0(t)\mathcal{O}(t) \right]=-\frac{\pi}{2}\gamma\]
where $\displaystyle \lim_{t \to 0+} \psi^0(t+1) =-\gamma$ and $\displaystyle \lim_{t \to 0+} t\psi^0(t) =-1$

Therefore
\begin{align*} \frac{d}{da} \int_{0}^{\infty} \frac{(1-\cos a x)\ln(x)}{x^2}dx =& -\frac{\pi}{2} \gamma -\frac{\pi \ln(a)}{2}\\ \Longrightarrow \int_{0}^{\infty} \frac{(1-\cos a x)\ln(x)}{x^2}dx = &-\frac{\pi}{2} \int a\gamma +a\ln(a) da + C\\ =& -\frac{a\pi\gamma}{2} - \frac{a\pi \ln(a)}{2} + \frac{\pi a}{2} + C \end{align*} Note that if $a=0$ then $C=0$ therefore \[\boxed{ \int_{0}^{\infty} \frac{(1-\cos a x)\ln(x)}{x^2}dx = -\frac{\pi a}{2}(\gamma +\ln(a)-1)}\]