# Evaluating $\int_0^\infty \frac{e^{-kx}\sin x}x\,\mathrm dx$

How to evaluate the following integral?

$$\int_0^\infty \frac{e^{-kx}\sin x}x\,\mathrm dx$$

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Differentiating with respect to $k$, one finds an easily computable integral $$-\int_0^{\infty}e^{-kx}\sin x\,dx=-\frac{1}{1+k^2}$$ Integrating back with respect to $k$ and using that for $k\rightarrow\infty$ the integral is $0$, we obtain $$\int_0^{\infty}\frac{e^{-kx}\sin x}{x}\,dx=\arctan k^{-1}.$$

You can use Parseval's theorem, i.e.

$$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dw \, F(w) G^*(w)$$

where $F$ and $G$ are the Fourier transforms of $f$ and $g$, respectively. In this case:

$$f(x) = e^{-k x} \theta(x) \implies F(w) = \frac{1}{k-i w}$$

$$g(x) = \frac{\sin{x}}{x} \implies G(w) = \begin{cases}\pi & |w| < 1 \\ 0 & |w| > 1 \end{cases}$$

Then the integral is equal to

$$\frac{\pi}{2 \pi} \int_{-1}^1 \frac{dw}{k-i w} = \frac{i}{2} \log{\left( \frac{i k + 1}{i k -1}\right)} = \arctan{\frac{1}{k}}$$


Note that the integral is

$$\mathcal{L}\bigg(\frac {\sin(t)}t\bigg)=F(k)$$ Where $$\mathcal{L}$$ is the Laplace transform operator

You can find the transformation using power series, which ultimately gives $$F(k)=\arctan(\frac 1k)$$

One more option:\begin{align}\int_0^\infty\int_0^\infty e^{-(k+y)x}\sin x\mathrm{d}x\mathrm{d}y&=\Im\int_0^\infty\int_0^\infty e^{-(k+y-i)x}\mathrm{d}x\mathrm{d}y\\&=\int_0^\infty\tfrac{1}{(k+y)^2+1}\mathrm{d}y\\&=[\arctan(k+y)]_0^\infty\\&=\tfrac{\pi}{2}-\arctan k.\end{align}