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I understand the constant can have Mellin transformation $$ \int_0^\infty x^{s-1} dx = 2\pi \delta(i s) $$ This is related to integration $$ \int_{-\infty}^{\infty} e^{ikx} dx \sim \delta(k) $$

I also know Mellin transformation of $e^{i x}$, which gives $$ \int_0^\infty dx e^{ix} x^{s-1} = \exp\left[\frac{i\pi s}{2} \right] \Gamma(s) $$ which is also given in this answer: Mellin transform of $\sin x$ aka $\int^{\infty}_0 x^{s-1}\sin x dx $. This means trigonometric functions can be Mellin transformed.

Now my question is, can we Mellin transform $\cosh x$ or $\sinh x$? Obviously $$ \int_0^{\infty} x^{s-1} e^{-x} dx=\Gamma(s) $$ But the problem is for $e^x$ part. We understand it is divergent. But I wonder whether there exists any general function like Dirac delta to give an ANSWER to $$ \int_0^{\infty} x^{s-1} e^{x} dx =? $$ like Dirac delta function even though it is divergent

jtkw
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  • Have you tried to apply the same kind of contour deformation on the complex plane as in the linked answer ? Alternatively, you can work with a series representation by substituting the $e^x$ by the exponential series. – Abezhiko Jul 03 '23 at 18:38

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