Definite integral - possible evaluation using real methods?

Question

The book "inside interesting integrals" gives the following exercise for the chapter about contour integration and the residue theorem: $$\int_{0}^\infty \frac{e^{\cos x}\sin(\sin x)}{x}dx=\space\space ?$$ This can be solved using the function $$f(z)=\frac{\exp(e^{iz})}{z}$$ on a quarter-circular contour, and is pretty straightforward. The answer turns out to be $$\frac{\pi}{2}(e-1)$$ However, in the book, the author makes the following comment:

Edward Copson (1901-1980), who was professor of mathematics at the University of St. Andrews in Scotland, wrote "A definite integral which can be evaluated using Cauchy's method of residues can always be evaluated by other means, though generally not so simply." Here's an example of what Copson meant, an integral attributed to the great Cauchy himself. It is easily done with contour integration, but would (I think) otherwise be pretty darn tough.

Does anyone know how to evaluate this integral using real methods?

Answer 1

$$e^{\cos x}\sin(\sin x) = \text{Im}\, e^{\cos x+i\sin x} = \text{Im}\exp\left(e^{ix}\right) = \text{Im}\sum_{n\geq 0}\frac{e^{nix}}{n!}=\sum_{n\geq 1}\frac{\sin(nx)}{n!}$$ and since $\int_{0}^{+\infty}\frac{\sin(nx)}{x}\,dx = \frac{\pi}{2}$ for any $n>0$, we have $$ \int_{0}^{+\infty}\frac{e^{\cos x}\sin(\sin x)}{x}\,dx = \frac{\pi}{2}\sum_{n\geq 1}\frac{1}{n!}=\color{red}{\frac{\pi}{2}(e-1)}.$$

Answer 2

Perhaps surprisingly, a straightforward trick works. To this end we refer to the following easy-to-prove lemma.

Lemma. Define $\operatorname{Si}(x) = \int_{0}^{x} \frac{\sin t}{t} \, dt$. Then

1. $ \int_{0}^{x} \frac{\sin(yt)}{t} \, dt = \operatorname{Si}(xy)$, and
2. $ \operatorname{Si}(x) = \frac{\pi}{2} + \mathcal{O}(x^{-1})$ as $x \to \infty$.

Then for $R > 0$,

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

Letting $R \to \infty$ proves the claim.

Answer 3

More generally, if $f(z)$ is any function that has a Maclaurin series expansion that converges absolutely on the unit circle, then $$\begin{align} \int_{0}^{\infty} \frac{f(e^{ix}) - f(e^{-ix})}{x} \, \mathrm dx &= 2i \int_{0}^{\infty} \frac{1}{x} \sum_{n=0}^{\infty} \frac{f^{n}(0)}{n!} \, \sin(nx) \, \mathrm dx \\ &=2i \int_{0}^{\infty} \frac{1}{x} \sum_{n=1}^{\infty} \frac{f^{n}(0)}{n!} \, \sin(nx) \, \mathrm dx \\ &\overset{\spadesuit}{=} 2i \sum_{n=1}^{\infty} \frac{f^{(n)}(0)}{n!} \int_{0}^{\infty} \frac{\sin (nx)}{x} \, \mathrm dx \\ &= \pi i \sum_{n=1}^{\infty} \frac{f^{n}(0)}{n!} \\ &= \pi i \left(f(1) - f(0) \right). \end{align}$$

($\spadesuit$ See the addendum of Sangchul Lee's answer here.)

It's the case $\alpha =0$ of the formula I derived here.

If the Maclaurin series has a radius of converge greater than $1$, it will necessarily converge absolutely on the unit circle.

For $f(z) = e^{z}$, which is an entire function, we have $$ \int_{0}^{\infty} \frac{e^{e^{ix}}-e^{e^{ix}}}{x} \, \mathrm dx= 2i \int_{0}^{\infty} \frac{e^{\cos x}\sin(\sin x)}{x} \, \mathrm dx = \pi i \left(e-1 \right). $$

Some other examples:

$$ \int_{0}^{\infty} \frac{\sin(e^{ix})-\sin(e^{-ix})}{x} \, \mathrm dx = 2i \int_{0}^{\infty} \frac{\cos (\cos x) \sinh(\sin x)}{x} \, \mathrm dx = \pi i \sin(1) $$

$$\int_{0}^{\infty} \frac{\cos(e^{ix})-\cos(e^{-ix}) }{x} \, \mathrm dx = -2i\int_{0}^{\infty} \frac{\sin( \cos x) \sinh(\sin x)}{x} \, \mathrm dx = \pi i \left(\cos(1) - 1 \right) $$

$$\int_{0}^{\infty} \frac{\operatorname{Li}_{2}(e^{ix})-\operatorname{Li}_{2}(e^{-ix})}{x} \, \mathrm dx = 2i \int_{0}^{\infty} \frac{\operatorname{Cl}_{2}(x)}{x} \, \mathrm dx = \frac{\pi^{3}i }{6} $$

$$\int_{0}^{\infty} \frac{\frac{1}{\Gamma(e^{ix})}-\frac{1}{\Gamma(e^{-ix})}}{x} \, \mathrm dx = \pi i $$

Due to the fact that these integrals don't converge absolutely, they can be difficult to approximate numerically. Wolfram Alpha will sometimes return bizarre approximations.

To get a rough approximation, you can make the upper limit "large" and tell Wolfram Alpha to use the midpoint method/rule with lots of intervals.

See here , here and here.

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The same approach can be used to show that $$\int_{0}^{\infty} \frac{f(e^{ix})-f(e^{-ix})}{x} \, \cos(x) \, \mathrm dx = \pi i \left(f(1) -f(0) - \frac{f'(0)}{2} \right), $$ and, therefore, $$2i \int_{0}^{\infty} \frac{e^{\cos x} \sin(\sin x) \cos(x)}{x} \, \mathrm dx = \pi i \left(e- \frac{3}{2} \right). $$