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I need help to evaluate:$$I=\int^{\infty}_0 e^{\cos(x)}\sin(x+\sin(x))\frac{xdx}{1+x^2}$$

We have : \begin{align*}e^{\cos(x)}\sin(x+\sin(x))&=\text{Im}\left({e^{ix+e^{ix}}}\right)\\&= \text{Im}\left({\sum_{n=0}^{\infty}\frac{e^{i(n+1)x}}{n!}}\right)\\&= \sum_{n=0}^{\infty}\frac{\sin((n+1)x)}{n!}\end{align*}

Therfore: \begin{align*}I=\sum_{n=0}^{\infty}\frac{1}{n!}\int^{\infty}_0 \frac{x\sin((n+1)x)}{1+x^2}dx\end{align*}

How can we calculate this last integral? Is there another way to evaluate this integral?

Delta
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1 Answers1

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Utilize $\int_0^\infty \frac{x\sin ax}{1+x^2}dx=\frac\pi2 e^{-a}$ to continue \begin{align*}I=\sum_{n=0}^{\infty}\frac{1}{n!}\int^{\infty}_0 \frac{x\sin((n+1)x)}{1+x^2}dx = \frac\pi2 \sum_{n=0}^{\infty}\frac{e^{-(n+1)}}{n!}=\frac\pi2 e^{\frac1e-1} \end{align*}

Quanto
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