$\int_{0}^{\infty} \frac{1 - e^{-px} \sin(x)}{x} dx$ Evaluate Integral

Question

How do I solve the integral:

$$\int_{0}^{\infty} \frac{1 - e^{-px} \sin(x)}{x} dx$$

I know the answer is $ \arctan(p) $ but have no idea as to how to show that.

Any hints welcome!

Thanks

Answer 1

I assume that you are evaluating \begin{align} -\int_{0}^{\infty} \frac{e^{-px} \sin x}{x} dx,\tag1 \end{align} since \begin{align} \int_{0}^{\infty} \frac{1-e^{-px} \sin x}{x} dx \end{align} does not converge.

Therefore, $(1)$ can be evaluated using the technique of “Feynman Integration”. \begin{align} I(p)&=-\int_{0}^{\infty} \frac{e^{-px} \sin x}{x} dx\\ \frac{dI}{dp}&=-\int_{0}^{\infty}\frac{d}{dp}\left(\frac{ e^{-px} \sin x}{x}\right)\ dx\\ I'(p)&=\int_{0}^{\infty}e^{-px} \sin x\ dx\tag2\\ &=\frac{1}{p^2+1}, \end{align} where $(2)$ can be evaluated using IBP twice. See here. Hence \begin{align} \frac{dI}{dp}&=\frac{1}{p^2+1}\\ I&=\int\frac{1}{p^2+1}\ dp\\ &=\tan^{-1}(p)+C. \end{align} Now, we let $p \to\infty$ so that our integrand is $\lim\limits_{p\to\infty}I(p)=0$, implying that $C = -\dfrac{\pi}{2}$. Thus $$ \begin{align} -\int_{0}^{\infty} \frac{e^{-px} \sin x}{x} dx=\tan^{-1}(p)-\dfrac{\pi}{2}. \end{align} $$