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Find the value of $\int_{0}^{\pi}e^{\cos\theta}\cos(\sin\theta)d\theta$.


$e^{\cos\theta}\cos(\sin\theta)=Re(e^{e^{i\theta}})$,where $Re()$ represents real part of.
So our integral becomes $$ \int_{0}^{\pi}Re(e^{e^{i\theta}})d\theta $$
Now by Cauchy integral formula,
$$ \int_{0}^{2\pi}e^zdz=2\pi i $$

But in Cauchy formula, the limits of integration are from $0$ to $2\pi$ and in my question, limits of integration are from $0$ to $\pi$. So I think I cannot apply Cauchy formula as such. What should I do? Please help me, Thanks.

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    $\begingroup$ For $f(t) = e^{\cos(t)}\cos(\sin(t))$ we have $f(t) = f(2\pi -t)$ so $\int_0^{\pi} f(t){\rm d}t = \frac{1}{2}\int_0^{2\pi} f(t){\rm d}t$ $\endgroup$ – Winther Nov 24 '15 at 3:37
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$$I(b)=\int_0^\pi e^{b\cos x}\cos (b\sin x)dx\\ =\frac{1}{2}\int_{-\pi}^\pi e^{b\cos x}\cos (b\sin x)dx\\ =\frac{1}{2}\int_{0}^{2\pi} e^{b\cos x}\cos (b\sin x)dx\\ =Re\left(\frac{1}{2}\int_{0}^{2\pi} e^{be^{ix}}dx\right)\\ I'(b)=Re\left(\frac{1}{2}\int_{0}^{2\pi} \frac{\partial}{\partial b} e^{be^{ix}}dx\right)\\ =Re\left(\frac{1}{2}\int_{0}^{2\pi} ibe^{be^{ix}}e^{ix}dx\right)\\ =Re\left(\left.\frac{1}{2}e^{be^{ix}}\right]_0^{2\pi}\right)=0\\ \therefore I(b)=I(0)=\frac{1}{2}\int_0^{2\pi}dx=\pi\\ I(1)=\int_0^\pi e^{\cos x}\cos (\sin x)dx=\pi $$

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