Show that if $a > 0$ and $b > 0$ that

\begin{align} \int_{0}^{\infty} {e}^{a \cos \left(b x\right)} \sin \left(a \sin \left(b x\right)\right) \frac{\mathrm{d}x}{x} = \frac{\pi}{2} \left({e}^{a} - 1\right) \\ \end{align}


I attempted to simplify the integral by putting it into a form of complex exponentials.

\begin{align} & \int_{0}^{\infty} {e}^{a {e}^{i b x}} \frac{\mathrm{d}x}{x} \\ = & \int_{0}^{\infty} {e}^{a \cos \left(b x\right) + i a \sin \left(b x\right)} \frac{\mathrm{d}x}{x} \\ = & \int_{0}^{\infty} {e}^{a \cos \left(b x\right)} \cos \left(\sin \left(b x\right)\right) \frac{\mathrm{d}x}{x} + i \int_{0}^{\infty} {e}^{a \cos \left(b x\right)} \sin \left(a \sin \left(b x\right)\right) \frac{\mathrm{d}x}{x} \\ \end{align}

We then have the result by linearity that

\begin{align} & \int_{- \infty}^{\infty} {e}^{a {e}^{i b x}} - {e}^{a {e}^{- i b x}} \frac{\mathrm{d}x}{x} \\ & = \int_{- \infty}^{\infty} {e}^{a \cos \left(b x\right)} \left({e}^{i a \sin \left(b x\right)} - {e}^{i a \sin \left(b x\right)}\right) \frac{\mathrm{d}x}{x} \\ & = 2 i \int_{- \infty}^{\infty} {e}^{a \cos \left(b x\right)} \sin \left(a \sin \left(b x\right)\right) \frac{\mathrm{d}x}{x} \\ & = 4 i I \\ \end{align}


\begin{align} I = \int_{0}^{\infty} {e}^{a \cos \left(b x\right)} \sin \left(a \sin \left(b x\right)\right) \frac{\mathrm{d}x}{x} \\ \end{align}

I am confused because when I take the residue at $0$, I find \begin{align} & {\text{Res}}_{z = 0} \frac{{e}^{a {e}^{i b z}} - {e}^{a {e}^{- i b z}}}{z} \\ & = {e}^{a} - {e}^{a} = 0 \\ \end{align}

  • 1
    $\begingroup$ A fairly similar integral was solved without the residue theorem by expanding into a series with the exponential integral, and simplifying. $\endgroup$ Aug 20, 2022 at 22:23
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    $\begingroup$ Thanks, I'll look into that method. The chapter I'm currently reading is about residues, but expanding the integrand into a Laurent series could give me an idea of the coefficient for $z^{-1}$. Edit: I just looked into your solution on the link. The use of the exponential integral was very clever. That is mentioned in about 200 pages, so I'll be sure to review your method when I get there. Thanks again. $\endgroup$
    – Talmsmen
    Aug 20, 2022 at 22:26
  • $\begingroup$ To see your error, take e.g. everyone's favourite semicircular contour: the arc's contribution doesn't vanish as $R\to\infty$. You get the same thing with $\int_{\Bbb R}\frac{\sin x}{x}\mathrm{d}x=\Im\int_{\Bbb R}\frac{\exp ix-1}{x}\mathrm{d}x$. $\endgroup$
    – J.G.
    Aug 21, 2022 at 7:36

2 Answers 2


Let $bx\to x$ to simplify the integral to\begin{align} & \int_{0}^{\infty} {e}^{a \cos x}\ \sin (a \sin x)\ \frac{dx}{x} \\ = &\ \Im \int_{0}^{\infty} {e}^{a e^{ix}}\ \frac{dx}{x} = \Im \int_{0}^{\infty} \sum_{k=0}^\infty\frac{(a e^{ix})^k}{k!}\frac{dx}x\\ = &\sum_{k=1}^\infty\frac{a^k}{k!} \int_{0}^{\infty} \frac{\sin(kx)}x\ dx = \sum_{k=1}^\infty\frac{a^k}{k!}\cdot \frac\pi2 = \frac{\pi}{2} \left({e}^{a} - 1\right) \end{align}

  • $\begingroup$ If $bx\to x$ why isn't there a $1/b$ factor, when changing the measure? $\endgroup$
    – Enrico M.
    Aug 21, 2022 at 10:04
  • $\begingroup$ @Laplacian - it cancels out in the ratio dx/x $\endgroup$
    – Quanto
    Aug 21, 2022 at 11:30
  • $\begingroup$ Right! Blindness :D $\endgroup$
    – Enrico M.
    Aug 21, 2022 at 11:34
  • 1
    $\begingroup$ The step $\int\sum=\sum\int$ is not trivial to justify. $\endgroup$
    – metamorphy
    Aug 22, 2022 at 4:44
  • 1
    $\begingroup$ @Quanto, What a onderful solution! How can you think about it? $\endgroup$
    – Lai
    Aug 24, 2022 at 10:34

$$\int_0^\infty e^{a e^{ib x}}\frac{dx}{x}=\int_0^\infty e^{a \cos(b x)+ia\sin(bx)}\frac{dx}{x}$$ $$=\int_0^\infty e^{a \cos (b x)}\Big(\cos(a\sin(bx))+i \sin (a \sin (b x))\Big) \frac{dx}{x}$$ Due to the fact that $$\int_{-\infty}^\infty e^{a \cos (b x)}\cos(a\sin(bx)) \frac{dx}{x}=0\,\,(\text{the integrand is odd})$$ and that $$\int_{-\infty}^0 e^{a \cos (b x)}\sin (a \sin (b x)) \frac{dx}{x}=\int_0^\infty e^{a \cos (b x)}\sin (a \sin (b x)) \frac{dx}{x}\,\,(\text{the integrand is even})$$ we can present the integrand in the form $$I=\int_0^\infty e^{a \cos (b x)}\sin (a \sin (b x)) \frac{dx}{x}=-\frac{i}{2}\int_{-\infty}^\infty e^{a e^{ib x}}\frac{dx}{x}=-\frac{i}{2}\int_{-\infty}^\infty e^{a e^{i x}}\frac{dx}{x}$$ where the integral is evaluated in the principal value sense. We go in the complex plane, adding a small arch (of the radius $r\to0$, around $x=0$) and a big arch (of the radius $R\to\infty$), closing the contour counter-clockwise, from $R$ to$-R$. There are no singular points inside the contour; therefore $$-\frac{i}{2}\oint e^{a e^{i z}}\frac{dz}{z}=0=I+I_r+I_R\qquad(1)$$ where $I_r, I_R$ are the integrals along the small and big arches correspondingly: $$I_r=-\pi i\underset{z=0}{\operatorname{Res}}\bigg(-\frac{i}{2} e^{a e^{i z}}\frac{1}{z}\bigg)=-\frac{\pi}{2}e^a\qquad(2)$$ (we integrate along the small arch clockwise, in the negative direction), and $$I_R=-\frac{i}{2}\int_0^\pi e^{ae^{iRe^{i\phi}}}id\phi=\frac{1}{2}\int_0^\pi e^{ae^{-R\sin\phi}e^{iR\cos\phi}}d\phi\to\frac{1}{2}\int_0^\pi d\phi=\frac{\pi}{2}\,\text{at}\,R\to\infty\quad(3)$$

Putting (2) and (3) into (1) $$I=-I_r-I_R=\frac{\pi}{2}\Big(e^a-1\Big)$$


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