# Show that $\int\limits_0^{\infty}\frac {dt}te^{\cos t}\sin\sin t=\frac {\pi}2(e-1)$

How do you show that

$$\int_{0}^{\infty}\frac{\mathrm{d}t}{t}\,\mathrm{e}^{\cos\left(t\right)}\, \sin\left(\sin\left(t\right)\right) = \frac{\pi}{2}\,\left(\,\mathrm{e} - 1\right)$$

I managed to get the left-hand side to equal the imaginary part of$$I=\int\limits_0^{\infty}\frac {dt}te^{e^{it}}$$But I’m not very sure what to do next. I’m thinking of a substitute $t\mapsto e^{it}$, but I’m not very sure how to evaluate the limit as $t\to\infty$. I also tried contour integration, but I’m not exactly sure what contour to draw.

• Seems like something you'd want to compute using residue theorem, I can't tell you the details right now though. – Wojowu Feb 26 '18 at 18:52
• The original integrand is an even function, and the second integrand, viewed as function is meromorphic with a pole at the origin, so you might be able to do with residues, using the boundary of a half-disk indented at the origin as the contour. – saulspatz Feb 26 '18 at 18:55
• Yes. Then you have to show that is goes to zero on the semi circle, and compute the residue at $z=0;$ the usual stuff. Oh, and you have to remember to adjust for the fact that you're using a half-circle near the origin. – saulspatz Feb 26 '18 at 19:04
• I haven't been able to make the contour I suggested work. I still think an indented contour is the way to go, but it takes a cleverer choice. – saulspatz Feb 26 '18 at 19:22
• @Crescendo How are you dealing with the integral on the positive imaginary axis? – saulspatz Mar 31 '18 at 17:24

$$e^{\cos t}\sin\sin t=\text{Im}\exp\left(e^{it}\right)=\text{Im}\sum_{n\geq 0}\frac{e^{nit}}{n!}=\sum_{n\geq 1}\frac{\sin(nt)}{n!}$$ and since for any $a>0$ we have $\int_{0}^{+\infty}\frac{\sin(at)}{t}\,dt=\frac{\pi}{2}$ it follows that $$\int_{0}^{+\infty}e^{\cos t}\sin\sin t\frac{dt}{t} = \frac{\pi}{2}\sum_{n\geq 1}\frac{1}{n!}=\frac{\pi}{2}(e-1),$$ pretty simple.

I have a counter-proposal: $$\begin{eqnarray*} \int_{0}^{+\infty}\left(e^{\cos t}\sin\sin t\right)^2\frac{dt}{t^2} &=&\frac{\pi}{2}\sum_{m,n\geq 1}\frac{\min(m,n)}{m!n!}\\&=&-\frac{\pi}{2}I_1(2)+\pi e(e-1)-2\pi e\int_{0}^{1}I_1(2x)e^{-x^2}\,dx. \end{eqnarray*}$$

Here's a way using contour integration. We first consider an indented semi-circular contour in the upper-half of the complex plane.

Since there are no residues inside the contour, we can write down

$$\int\limits_{\epsilon}^{R}dx\, f(x)+\int\limits_{\Gamma_{R}}dz\, f(z)-\int\limits_{\epsilon}^{R}dy\,\frac {e^{e^{-y}}}y+\int\limits_{\gamma_{\epsilon}}dz\, f(z)=0$$

where

$$f(z)=\frac {e^{e^{iz}}}z$$

It can be shown that as $\epsilon\to0$ and $R\to\infty$, the arc integrals become$$\int\limits_{\Gamma_{R}}dz\, f(z)=\frac {\pi i}2$$$$\int\limits_{\gamma_{\epsilon}}dz\, f(z)=-\frac {e\pi i}2$$

Putting everything together, we see that$$\int\limits_0^{\infty}dx\,\frac {e^{\cos x}}x(\cos\sin x+i\sin\sin x)x-\int\limits_0^{\infty}dy\,\frac {e^{e^{-y}}}y=\frac {\pi i}2(e-1)$$

Take the imaginary part of the first integral and the identity is established.