Evaluating $\int^\infty_0e^{a\cos bx}\sin(a\sin bx)\frac{dx}{x}$

Question

Taken from Whittaker & Watson, chapter 6, problems on the theory of residues.

I want to show that

$$\int^\infty_0e^{a\cos bx}\sin(a\sin bx)\frac{dx}{x}=\tfrac{1}{2}\pi(e^a-1)$$

I can see that the function has a singularity at $z=0,$ but unfortunately because this is on the real axis, it seems that Jordan's lemma does not apply, and hence it is necessary to take the contour of two concentric semicircles, one of radius $\delta$ which is small, and one of radius $\rho$ which is large, centred at the origin and connected by the real axis, and then let $\delta\to0$ and $\rho\to\infty.$

However if I do this, then

$$\int^\infty_0e^{a\cos bx}\sin(a\sin bx)\frac{dx}{x}=\lim\limits_{\delta\to 0}\lim\limits_{\rho\to\infty}\int^\rho_{\delta}e^{a\cos bx}\sin(a\sin bx)\frac{dx}{x}$$

but it seems that no simplification has resulted.

What is the best method to evaluate this integral, and why?

Answer 1

Let me develop your idea of using two concentric contours:

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Let $0 < \delta < R$ and consider the $C = C(\delta, R)$ consisting of

• $\Gamma = \Gamma(R)$ : upper semi-circular arc of radius $R$ centered at $0$,
• $\gamma = \gamma_{\delta}$ : upper semi-circular arc of radius $\delta$ centered at $0$,
• $L = L(\delta, R) = [-R, -\delta] \cup [\delta, R]$.

Of course, we impose counter-clockwise orientation to $C$ as usual. Also we inherit orientation of $C$ to each components $\Gamma$, $\gamma$, $L$. Now let $b > 0$ and consider

$$ f(z) = \frac{\exp( a e^{ibz} )}{z}. $$

Since $f$ has no pole in the interior of $C$, we have

$$ 0 = \oint_{C} f(z) \, dz = \int_{\Gamma} f(z) \, dz + \int_{\gamma} f(z) \, dz + \int_{L} f(z) \, dz. $$

We will take $\delta \to 0^+$ and $R \to \infty$. To this end, we make the following observations:

1. Parametrize $\Gamma$ by $z = R e^{i\theta}$. Then

   $$ \int_{\Gamma} f(z) \, dz = i \int_{0}^{\pi} \exp\left( a e^{-bR\sin\theta}(\cos(bR\cos\theta)) + i\sin(bR\cos\theta)) \right) \, d\theta. $$

   Since the integrand of the RHS is bounded by $e^a$, by the dominated convergence theorem we can pass limit into the integral as $R\to\infty$. The result is

   $$ \lim_{R\to\infty} \int_{\Gamma(R)} f(z) \, dz = i\pi.$$

2. Similar computation for $\gamma$ shows that

   $$ \lim_{\delta\to0^+} \int_{\gamma(\delta)} f(z) \, dz = -i\pi e^a.$$

3. The integral of $f$ along $L$ is related to our integral by

   \begin{align*} \int_{\delta}^{R} e^{a\cos (bx)}\sin(a\sin (bx)) \, \frac{dx}{x} &= \frac{1}{2}\operatorname{Im} \int_{L(\delta, R)} f(z) \, dz \\ &= \frac{1}{2}\operatorname{Im} \left( -\int_{\Gamma(R)} f(z) \, dz - \int_{\gamma(\delta)} f(z) \, dz \right). \end{align*}

Therefore, taking $\delta \to 0$ and $R\to\infty$ we obtain

$$ \int_{0}^{\infty} e^{a\cos (bx)}\sin(a\sin (bx)) \, \frac{dx}{x} = \frac{\pi}{2}(e^a - 1) $$

as expected.

Answer 2

Not really in the spirit of residues, but it might still be interesting. We have:

$$ \int^N_0e^{a\cos bx}\sin(a\sin bx)\frac{dx}{x} = \int^N_0e^{a\cos bx}\Im \left(e^{ia\sin bx} \right)\frac{dx}{x} = \int^N_0\Im \left(e^{a e^{i bx}} \right)\frac{dx}{x} $$

for any $N>0$. Now we can use the infinite series for $\exp$: $$ \int^N_0\Im \left( \sum_{k \ge 0} \frac{a^k e^{ikbx}}{k!} \right)\frac{dx}{x} = \sum_{k \ge 1} \frac{a^k}{k!} \int^N_0\frac{\sin(kbx)}{x} dx = \sum_{k \ge 1} \frac{a^k}{k!} \int^{kbN}_0\frac{\sin(u)}{u} du $$ The term $k=0$ disappear because the imaginary part of a real constant is $0$ and you can do the sum/integral swap because the exponential series converge uniformely on any compact subset of the complex plane.

Now taking the limit on both sides as $N \to \infty $ and using the dominated convergence theorem leaves us with: $$ \int^\infty_0e^{a\cos bx}\sin(a\sin bx)\frac{dx}{x} = \sum_{k \ge 1} \frac{a^k}{k!} \lim_{N \to \infty} \left( \int^{kbN}_0\frac{\sin(u)}{u} du \right) = (e^a-1) \frac{\pi}{2}$$