# Evaluating $\int_{0}^{\infty}\frac{1-e^{-t}}{t}\sin{t}\operatorname d\!t$

Find this integral $$I=\int_{0}^{\infty}\dfrac{1-e^{-t}}{t}\sin{t}\operatorname d\!t$$

I know this $$\int_{0}^{\infty}\dfrac{\sin{t}}{t}\operatorname d\!t=\dfrac{\pi}{2}$$But I can't find this value,Thank you

Hints:

• One has $\displaystyle\frac{1-\mathrm e^{-t}}t=\int_0^1\mathrm e^{-xt}\mathrm dx$
• For every real numbers $x$ and $t$, one has $\mathrm e^{-xt}\sin t=\Im(\mathrm e^{-(x-\mathrm i)t})$
• For every complex number $z$ such that $\Re z\gt0$, one has $\displaystyle\int_0^\infty\mathrm e^{-zt}\mathrm dt=\frac1z$
• For every real number $x$, one has $\displaystyle\frac1{x-\mathrm i}=\frac{x+\mathrm i}{x^2+1}$ hence $\displaystyle\Im\left(\frac1{x-\mathrm i}\right)=\frac1{x^2+1}$
• And finally, one has $\displaystyle\int_0^1\frac1{x^2+1}\mathrm dx=\frac\pi4$

Extension/Consequence: for every nonnegative $a$, $$\int_{0}^{\infty}\dfrac{1-\mathrm e^{-at}}{t}\sin{t}\,\mathrm dt=\arctan a$$

• Do we always have to consider the imaginary part, i mean $e^{-xt}sint=Im(e^{-(x-1)t})$ is clear, but $\dfrac{1}{x-i}$, has nothing to do with $\sin$. Is it a general rule ?(I've never integrated using real and imaginary parts) – OBDA Mar 10 '14 at 12:48
• @O.B.D.A. Dunno what you call a "general" rule but it is definitely profitable to view sine and cosine as the imaginary and real parts of a complex exponential. One can argue that exponentials are simpler than sines and cosines. – Did Mar 10 '14 at 13:18

Since you know that $$\int_0^\infty \frac{\sin t}tdt=\frac\pi2$$ so it suffices to find $$\int_0^\infty\frac{e^{-t}}t\sin tdt$$ so let $$f(x)=\int_0^\infty\frac{e^{-t}}t\sin (xt)dt=\int_0^\infty h(x,t)dt$$ so using Leibniz theorem and since $$\left|\frac{\partial h}{\partial x}(x,t)\right|\le e^{-t}\in L^1((0,\infty))$$ so we have $$f'(x)=\int_0^\infty\cos(xt)e^{-t}dt=\frac1{1+x^2}$$ Now since $f(0)=0$ then we find $$f(x)=\arctan x$$ and we have the desired result by taking $x=1$.

Integration by parts twice yields \begin{align} \frac{\mathrm{d}}{\mathrm{d}a}\int_0^\infty\frac{1-e^{-at}}{t}\sin(t)\,\mathrm{d}t &=\int_0^\infty e^{-at}\sin(t)\,\mathrm{d}t\\ &=\frac1a\int_0^\infty e^{-at}\cos(t)\,\mathrm{d}t\\ &=\frac1{a^2}-\frac1{a^2}\int_0^\infty e^{at}\sin(t)\,\mathrm{d}t\\ &=\frac1{1+a^2} \end{align} Noting that when $a=0$ the original integral is $0$, we get $$\int_0^\infty\frac{1-e^{-at}}{t}\sin(t)\,\mathrm{d}t=\arctan(a)$$

By Frullani's theorem $$\int_{0}^{+\infty}\frac{e^{it}-e^{(i-a)t}}{t}\,dt =\log(1+ia)$$ hence by taking the imaginary part:

$$\int_{0}^{+\infty}\frac{1-e^{-at}}{t}\,\sin(t)\,dt = \arctan(a).$$

As an alternative, by using the Laplace transform we have: $$\mathcal{L}(\sin x)=\frac{1}{1+s^2},\qquad\mathcal{L}^{-1}\left(\frac{1-e^{-at}}{t}\right)= I_{s\leq a}(s)$$ hence: $$I = \int_{0}^{a}\frac{ds}{1+s^2}=\arctan(a).$$