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

Compute the following integral:

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

Any hint, suggestion is welcome.

Solution 1:

Yet a different approach: parametric integration. Let $$ F(\lambda)=\int_{0}^{\infty} \frac{e^{-\lambda x} \sin(x)}{x}\,dx,\qquad\lambda>0. $$ Then $$ F'(\lambda)=-\int_{0}^{\infty} e^{-\lambda x} \sin(x)\,dx=-\frac{1}{1+\lambda^2}. $$ Integrating and taking into account that $\lim_{\lambda\to\infty}F(\lambda)=0$ we have $$ F(\lambda)=\frac\pi2-\arctan\lambda $$ and $$ \int_{0}^{\infty} \frac{e^{-x} \sin(x)}{x}\,dx=F(1)=\frac\pi4. $$

Solution 2:

Using Laplace Transform, $$\mathcal{L}(\sin(x)) = \frac{1}{s^2 + 1}$$ $$\mathcal{L}\left(\frac{\sin(x)}{x}\right) = \int_r^\infty \frac{1}{s^2 + 1} ds = \frac{\pi}{2} - \arctan(r)$$ Therefore, $$\int_0^\infty e^{-rx} \frac{\sin(x)}{x} dx = \frac{\pi}{2} - \arctan(r)$$ Substituting r = 1, $$\int_0^\infty e^{-x} \frac{\sin(x)}{x} dx = \frac{\pi}{4}$$

Solution 3:

Another approach: $$\begin{eqnarray*} \int_{0}^{\infty} dx\, \frac{e^{-x} \sin(x)}{x} &=& \int_{0}^{\infty}dx\, \frac{e^{-x}}{x} \sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1)!} \\ &=& \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} \int_{0}^{\infty}dx\, x^{2k} e^{-x} \\ &=& \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}(2k)! \\ &=& \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \hspace{5ex} \textrm{(Leibniz series for $\pi$)}\\ &=& \frac{\pi}{4}. \end{eqnarray*}$$