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Prove that for all $a > 0$: $$\int\limits_0^{\pi/2}e^{-a\cos x}\cos(a\sin x)dx = \cfrac{\pi}{2} - \int\limits_0^a\cfrac{\sin x}{x}dx$$

I have no idea how to solve it. But the task looks very interesting. I understand some basics of complex analysis. But I know that it's not needed.

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\begin{align} \totald{}{a}\int_{0}^{\pi/2}\exp\pars{-a\expo{-\ic x}}\,\dd x & = -\int_{0}^{\pi/2}\exp\pars{-a\expo{-\ic x}}\expo{-ix}\,\dd x \\[3mm] & = -\ic\int_{x\ =\ 0}^{x\ =\ \pi/2}\exp\pars{-a\expo{-\ic x}}\,\dd\pars{\expo{-ix}} = -\ic\int_{1}^{-\ic}\expo{-at}\,\dd t \\[3mm] & = {i\over a}\pars{\expo{\ic a} - \expo{-a}} \end{align}


Then, \begin{align} &\int_{0}^{\pi/2}\exp\pars{-a\expo{-\ic x}}\,\dd x \\[3mm] = &\ {\pi \over 2} + \ic\int_{0}^{a}{\expo{\ic t} - \expo{-t} \over t}\,\dd t = {\pi \over 2} + \ic\int_{0}^{a}{\cos\pars{t} - \expo{-t} \over t}\,\dd t - \int_{0}^{a}{\sin\pars{t} \over t}\,\dd t \end{align}
$$ \color{#f00}{\Re\int_{0}^{\pi/2}\exp\pars{-a\expo{-\ic x}}\,\dd x} = \color{#f00}{{\pi \over 2} - \int_{0}^{a}{\sin\pars{t} \over t}\,\dd t} $$

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Hint. One may write, for $a>0$, $$ \begin{align} \int_0^{\pi/2}e^{-a\cos x}\cos(a\sin x)\,dx&=\Re \int_0^{\pi/2}e^{\large -ae^{ix}}\,dx \\\\&= \frac{\pi}2+\sum_{n=1}^\infty (-1)^n\frac{a^n}{n!}\int_0^{\pi/2}\cos(nx)\,dx \\\\&= \frac{\pi}2+\sum_{n=0}^\infty (-1)^n\frac{a^n}{n!}\frac{\sin(n\pi/2)}{n} \\\\&=\frac{\pi}2 -\sum_{p=0}^\infty \frac{(-1)^p}{(2p+1)!}\frac{a^{2p+1}}{(2p+1)} \\\\&= \cfrac{\pi}{2} - \int_0^a\cfrac{\sin x}{x}\:dx \end{align} $$ as announced.

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The LHS is the real part of: $$ J=\int_{0}^{\pi/2}\exp\left(-a\cos x+ia\sin x\right)\,dx =\int_{0}^{\pi/2}\exp\left(-a e^{-ix}\right)\,dx \tag{1}$$ that can be computed through a Taylor series expansion: $$ \text{Re}(J) = \sum_{n\geq 0}\frac{(-a)^n}{n!}\int_{0}^{\pi/2}\cos(nx)\,dx =\sum_{m\geq 0}\frac{(-a)^{2m+1}(-1)^m}{(2m+1)(2m+1)!}.\tag{2}$$ Now it is straightforward to check that: $$\frac{d}{da}\text{Re}(J) = \sum_{m\geq 0}\frac{(-1)^m a^{2m}}{(2m+1)!}=\frac{\sin(a)}{a}\tag{3}$$ hence the derivatives with respect to $a$ of the RHS and the LHS agree.
Since their value at $a=0$ is the same, RHS$\equiv$LHS.

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    $\begingroup$ Your proof is very similar to mine ;) $\endgroup$ – Olivier Oloa Jun 9 '16 at 17:13
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One may work this via complex analysis. Consider the contour integral

$$\oint_C dz \frac{e^{-a z}}{z} $$

where $C$ is a half-unit-circle in the right-half plane, plus the imaginary axis in between with a semicircular detour about the origin. Thus, the contour integral is equal to zero by Cauchy's theorem, and is also equal to

$$i \int_{-\pi/2}^{\pi/2} d\theta \, e^{-a e^{i \theta}} + PV \int_1^{-1} dy \frac{e^{-i a y}}{y} + i \epsilon \int_{\pi/2}^{-\pi/2} d\phi \, e^{i \phi} \frac{e^{-a \epsilon e^{i \phi}}}{\epsilon e^{i \phi}} $$

Thus, taking the imaginary part of the above equation and setting it equal to zero, we get

$$\int_{-\pi/2}^{\pi/2} d\theta \, e^{-a \cos{\theta}} \cos{(a \sin{\theta})} + \int_{-1}^1 dy \frac{\sin{a y}}{y} - \pi = 0$$

(Note that we do not need the $PV$ because the singularity at the origin is removable.) Halving and solving for the first integral (and substituting in the second integral), we get

$$\int_0^{\pi/2} d\theta \, e^{-a \cos{\theta}} \cos{(a \sin{\theta})} = \frac{\pi}{2} - \int_0^a dx \frac{\sin{x}}{x} $$

as asserted.

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