I wanna evaluate this integral $$I=\int_{0}^{+\infty}{\exp(\cos(x))\sin(\sin(x))\over x}dx$$ so i wrote $g(x)=\exp(\cos(x))\sin(\sin(x))$ as a Fourier series and I found out that $$g(x)=\sum_{i=0}^{+\infty} {\sin(nx)\over n!}$$ and then dividing by $x$ we have $$\frac{g(x)}{x} = \sum_{i=0}^{+\infty} {\sin(nx)\over xn!}$$ and the integrating from $0$ to infinity we have $$\int_{0}^{+\infty}{\exp(\cos(x))\sin(\sin(x))\over x}\,dx = \int_{0}^{+\infty}\sum_{i=0}^{+\infty} {\sin(nx)\over xn!}=\sum_{i=0}^{+\infty} \int_{0}^{\infty}{\sin(nx)\over xn!}$$ because of the uniform convergence of the series then we found out that that integral is equal to $$\frac{\pi\exp(1)}{2}$$ while when I look in this integral in wolframalpha I found out that this integral from 0 to 200 is 2.69...which does not equal the previous result so what is the problem=
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The $n=0$ term is zero, so the answer should be $\frac{\pi (e-1)}{2}$ – Conrad Jun 27 '19 at 17:25
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The numerator is the imaginary part of $\exp(e^{ix})$. – Angina Seng Jun 27 '19 at 17:27
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thank you conrad – Mossaab Sama Jun 27 '19 at 18:59