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I'd like to know how to evaluate the integral

$$I=\int_0^\infty\frac{e^{-s^2}\sin(s)}{s}\,ds=\frac{\pi}{2}\text{erf}(1/2)$$

through the residue theorem. My first steps were to expand $\sin$ as exponentials, and note that the only pole is located at $s=0$ which is removable, so the residue is $0$. However, I can't seem to design a closed path in $\mathbb{C}$, enclosing the origin, that produces an error function. I would appreciate any help with determining a useful path to get the above result!

As a side note, the result can also be obtained by expanding $\sin(s)$ in a power series and integrating term-by-term. The result is an infinite sum of Gamma functions that ends up giving the power series for the error function. However, I am more interested in how this may be computed through the residue theorem, if at all.

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