How to calculate this improper integral?

Solution 1:

1^st Solution. Define the sine integral by

$$ \operatorname{Si}(x) = \int_{0}^{x} \frac{\sin t}{t} \, \mathrm{d}t. $$

Using integartion by parts, it can be proved that

$$ \operatorname{Si}(x) = \frac{\pi}{2} + \mathcal{O}\left(\frac{1}{x}\right) \qquad\text{as } x \to \infty. $$

Now note that $e^{\cos\theta}\sin\sin\theta = \operatorname{Im}(e^{e^{i\theta}}-1) = \sum_{n=1}^{\infty} \frac{1}{n!}\sin(n\theta)$. Then by the Fubini's theorem, for $R > 0$,

\begin{align*} \int_{0}^{R} \frac{e^{\cos\theta}\sin\sin\theta}{\theta} \, \mathrm{d}\theta &= \int_{0}^{R} \frac{1}{\theta} \sum_{n=1}^{\infty} \frac{\sin(n\theta)}{n!} \, \mathrm{d}\theta \\ &= \sum_{n=1}^{\infty} \frac{1}{n!} \int_{0}^{R} \frac{\sin(n\theta)}{\theta} \, \mathrm{d}\theta \\ &= \sum_{n=1}^{\infty} \frac{1}{n!} \operatorname{Si}(nR) \\ &= \sum_{n=1}^{\infty} \frac{1}{n!} \left( \frac{\pi}{2} + \mathcal{O}\left( \frac{1}{nR} \right) \right) \\ &= \frac{\pi}{2}(e - 1) + \mathcal{O}\left(\frac{1}{R}\right). \end{align*}

So by letting $R \to \infty$, the integral converges to

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

2^nd Solution. It is well-known that

$$ \lim_{N\to\infty} \sum_{k=-N}^{N} \frac{1}{z + 2\pi k} = \frac{1}{2}\cot\left(\frac{z}{2}\right). $$

Moreover, this convergence is locally uniform (in the sense that the difference between the limit and the $N$-th partial sum, when understood as a meromorphic function on $\mathbb{C}$, converges to $0$ uniformly on any compact subsets of $\mathbb{C}$).

Using this and noting that $e^{\cos\theta}\sin\sin\theta = \operatorname{Im}(e^{e^{i\theta}} - e)$, we find

\begin{align*} \int_{-(2N+1)\pi}^{(2N+1)\pi} \frac{e^{\cos\theta}\sin\sin\theta}{\theta} \, \mathrm{d}\theta &= \operatorname{Im}\biggl( \int_{-(2N+1)\pi}^{(2N+1)\pi} \frac{e^{e^{i\theta}} - e}{\theta} \, \mathrm{d}\theta \biggr) \\ &= \operatorname{Im}\biggl( \int_{-\pi}^{\pi} (e^{e^{i\theta}} - e) \sum_{k=-N}^{N} \frac{1}{\theta + 2\pi k} \, \mathrm{d}\theta \biggr) \\ &\to \operatorname{Im}\biggl( \int_{-\pi}^{\pi} (e^{e^{i\theta}} - e) \frac{1}{2}\cot\left(\frac{\theta}{2}\right) \, \mathrm{d}\theta \biggr) \qquad\text{as } N \to \infty. \end{align*}

Now we substitute $z = e^{i\theta}$. Then using the identity $\cot(\theta/2) = i \frac{e^{i\theta} + 1}{e^{i\theta} - 1}$ and the residue theorem,

\begin{align*} \int_{-\infty}^{\infty} \frac{e^{\cos\theta}\sin\sin\theta}{\theta} \, \mathrm{d}\theta &= \operatorname{Im}\biggl( \frac{1}{2} \int_{|z|=1} \frac{(e^{z} - e)(z+1)}{z(z-1)} \, \mathrm{d}z \biggr) \\ &= \operatorname{Im}\biggl( \pi i \, \underset{z=0}{\operatorname{Res}} \frac{(e^{z} - e)(z+1)}{z(z-1)} \biggr) \\ &= \pi(e - 1). \end{align*}

Dividing both sides by $2$, we conclude that

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