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Context: I saw this solution, and as far as I can tell it uses the following: $$\int_{\gamma-i\infty}^{\gamma+i\infty}Q(s)/x^sds=\sum_{s_k\text{ poles of } Q}\text{Res}_{s=s_k}Q(s)/x^s+\int_{\gamma'-i\infty}^{\gamma'+i\infty}Q(s)/x^sds,\tag1$$ where $\Re(s)=\gamma$ lies to the right of all the poles $s_k$ and $\Re(s)=\gamma'$ lies to the left of all the poles, and $$Q(s)=\int_0^\infty x^{s-1}f(x)dx,$$ so that $$f(x)=\int_{\gamma-i\infty}^{\gamma+i\infty}Q(s)/x^sds.$$

Question: When, if ever, is $(1)$ true? I have not seen it discussed anywhere other than MSE, where I have seen it a few times without explanation or justification.

Pardon me if this question is trivial, I don't have much experience with complex analysis.

clathratus
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    This is actually for the Mellin transform, not the Laplace transform, though they are related. In the case of the inverse Laplace transform with a $e^{st}$, for $t>0$ you close the contour out in the left half plane, and so the closed up vertical line contour encloses poles which are to its left. – Ian Apr 26 '21 at 18:43
  • @Ian ah yes forgive me for the mixup – clathratus Apr 26 '21 at 20:55

1 Answers1

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The residue theorem says that for $Q$ meromorphic with poles at $a_k$ and $C$ a rectangle $(\gamma',\gamma)+i(-T,T)$ with no poles on the boundary then $\int_{\partial C} Q(s)x^{-s}ds=2i\pi \sum_{a_k\in C} Res(Q(s)x^{-s})$ so it remains to check what happens as $|T|\to \infty$.

In your linked question the behavior as $|T|\to\infty$ is clear because $Q(s)$ has finitely many poles and it is rapidly decreasing on vertical strips, so $Q(x)x^{-s}$ is integrable on vertical lines and $\lim_{|T|\to \infty} \int_{\gamma'+iT}^{\gamma+iT} Q(s)x^{-s}ds= 0$.

Sometimes it works as well when $\int_{\gamma-i\infty}^{\gamma+i\infty}Q(s)/x^sds$ only converges conditionnally, sometimes it works when $Q(s)$ has infinitely many poles (or essential singularities or other kind of singularities), sometimes $\lim_{\gamma'\to -\infty}\int_{\gamma'-i\infty}^{\gamma'+i\infty}Q(s)x^{-s}ds=0$ so that $f(x) =2i\pi \sum Res(Q(s)x^{-s})$.

This is standard in the context of the residue theorem that every function is different and that we often need a bit of tricky investigation before finding the correct contours and limits.

reuns
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  • This makes sense, thank you. Could one apply the technique in your linked answer to the function $$F_k(x)=\sum_{n\ge1}\frac{1}{n^{2k+1}(e^{nx}-1)}?$$ – clathratus Apr 26 '21 at 21:01