Compute $\int_0^\infty \frac{e^{-tx}\sin(x)}{x}dx$

I have to compute$$\int_0^\infty \dfrac{e^{-tx}\cdot \sin(x)}{x}dx$$

This is following a helping problem $$\int_0^\infty e^{-tx}\cdot \sin(x)dx$$ which using IPB two times turned out to be $$\dfrac{1}{1+t^2}$$

I think there must be a substitution to get to the first problem, but I just cannot see it. Any hint appreciated.

Define $$I = \int \limits_0 ^\infty e^{-tx}\frac{\sin(x)}{x}dx$$ then $$\dfrac{dI}{dt}=-\int \limits_0 ^\infty e^{-tx}\sin(x)dx = -\frac{1}{1+t^2}$$ hence $$I = -\arctan(t) + D$$

We fix $D$ using $I(0)=D=\dfrac\pi 2$, easily obtainable through complex analysis.

• You don't need to use complex analysis. Simply note that $I(\infty)=0$ to find $D$. (Now you can plug in $t=0$, giving $I(0)=\frac\pi2$ without complex analysis.) – Akiva Weinberger Sep 28 '14 at 23:41
• @columbus8myhw That's true, easily obtained from $0\leq I \leq \int\limits_0^{+\infty}e^{-tx}dx$ – Frédéric Sep 29 '14 at 7:50

We can compute $\int_0^{\infty}e^{-tx}\sin x/x dx$ for a positive $t$. Indeed, define $$F(t,x):=\frac{\sin x}xe^{-tx}.$$ If $\delta$ is a positive number, then for any non-negative $x$, $$\sup_{s\geqslant \delta}\partial_tF(s,x)|\leqslant e^{-\delta x}.$$ Since the map $x\mapsto e^{-\delta x}$ is integrable on $[0,\infty)$, we can take the derivative under the integral.


\begin{align} &\color{#00f}{\large\int_{0}^{\infty}{\expo{-tx}\sin\pars{x} \over x}\,\dd x} =\int_{0}^{\infty}\expo{-tx}\pars{\half\int_{-1}^{1}\expo{\ic kx}\,\dd k} =\half\int_{-1}^{1}\int_{0}^{\infty}\expo{\pars{\ic k - t}x}\,\dd k \\[3mm]&=\half\int_{-1}^{1}{1 \over t - \ic k}\,\dd k =t\int_{0}^{1}{\dd k \over t^{2} + k^{2}} =\int_{0}^{1/t}{\dd k \over 1 + k^{2}}= \arctan\pars{1 \over t} \\[3mm]&=\color{#00f}{\Large{\pi \over 2} - \arctan\pars{t}} \end{align}

\begin{aligned} \int_{0}^{\infty} e^{-tx} \frac{\sin x}{x} \, \mathrm{d}x &= \int_{0}^{\infty} e^{-tx} \sin x \int_{0}^{\infty} e^{-zx} \,\mathrm{d}z\,\mathrm{d}x\\ &= \int_{0}^{\infty}\!\!\int_{0}^{\infty} e^{-x\left(z+t\right)} \sin x\,\mathrm{d}x\,\mathrm{d}z\\ &= \int_{0}^{\infty} \frac{\mathrm{d}z}{\left(z+t\right)^2 +1}\\ &= \left[\lim_{N\to\infty}\arctan(N+t)\right] - \arctan t \\ &= \frac{\pi}{2} - \arctan t\\ &= \operatorname{arccot} t \end{aligned}

• Note: We can say that $\displaystyle{\frac{1}{x} = \int_{0}^{\infty} e^{-zx}\,\mathrm{d}z}$ because we are integrating over the interval $(0, \infty)$, i.e., $x>0$, so the integral converges. – user149844 Sep 28 '14 at 21:24