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We must show that $$\int_1^{\infty}e^{-\pi x}(x^{\sigma/2} + x^{(1-\sigma)/2}) x^{-1} dx \ll (1 + |\sigma|)^{\pi |\sigma|}. $$ Here is my attempt, however I wondered if there is a one-trick wonder that makes the answer drop out. I'm especially curious about the $\pi$ in the numerator, which doesn't come naturally in my answer.

Notice that the integral is invariant under the transformation $\sigma \mapsto 1-\sigma $, so it is only necessary to consider the case where $\sigma \ge 1/2$. In this case, for $x\ge 1$, we have $x^{(1-\sigma)/2} \le x ^{1/4} \le x^{\sigma /2}$, so we can safely ignore the $x^{(1-\sigma)/2}$ term in the integral and absorb it into a constant: $$\int_1^{\infty}e^{-\pi x}(x^{\sigma/2} + x^{(1-\sigma)/2}) x^{-1} dx \ll \int_1^{\infty}e^{-\pi x}x^{\sigma/2} x^{-1} dx . $$ This looks a lot like the gamma function $\Gamma(s)$, so we extend the integral to $0$ (noting the integral still converges since $\sigma \ge 1/2$), and make the substitution $u = \pi x$: $$\int_1^{\infty}e^{-\pi x}x^{\sigma/2} x^{-1} dx \le \pi^{-\sigma/2} \int_0^{\infty}e^{-u}u^{\sigma/2} u^{-1} du = \pi^{-\sigma/2}\Gamma(\sigma /2). $$ From here we pick the integer $n$ such that $\sigma/2 \le n \le \sigma/2 + 1$, and we have: $$\Gamma(\sigma/2) \le \Gamma(n) = n! \le n^n \le (1 + \sigma/2)^{1 + \sigma/2}$$

and this implies the above result, however it seems a little bit stronger, so I wondered where the author of the current got the inequality from.

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There is probably something I am missing in this problem.

$$I(\sigma)=\int_1^{\infty}e^{-\pi x}\left(x^{\sigma/2} + x^{(1-\sigma)/2}\right) x^{-1}\, dx =E_{\frac{2-\sigma }{2}}(\pi )+E_{\frac{1+\sigma }{2}}(\pi )$$ where appear exponential integral functions (they are defined for $0 \leq \sigma \leq 1$).

In terms of the gamma function $$I(\sigma)=\frac{\Gamma \left(\frac{1-\sigma }{2},\pi \right) } {\pi ^{\frac{1-\sigma }{2}} }+\frac{\Gamma \left(\frac{\sigma }{2},\pi \right) }{\pi ^{\frac{\sigma }{2}} }$$

This makes $$ \frac{2 \,\Gamma \left(\frac{1}{4},\pi \right)}{\sqrt[4]{\pi }}\leq I(\sigma) \leq \Gamma (0,\pi )+\frac{\Gamma \left(\frac{1}{2},\pi \right)}{\sqrt{\pi }}$$ Numerically, these bounds are $\approx 0.0230332$ and $\approx 0.0230952$ (this is a very limited range).

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  • $\begingroup$ What’s the range of $\sigma$ in this answer? If we don’t discard the $(1-\sigma)/2$ term in the beginning we have to be careful because $\Gamma$ will become very large once we get close to $0$ (there is a pole there), this makes me dubious of your last displayed equation. This is why I looked only to the right of $1/2$. In any case the estimate needs to be valid for all real $\sigma$. $\endgroup$ – Quantaliinuxite Jun 11 at 9:52
  • $\begingroup$ @Quantaliinuxite. As I wrote it, it is for $0 \leq \sigma \leq 1$. I wrote There is probably something I am missing in this problem. $\endgroup$ – Claude Leibovici Jun 11 at 9:56

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