# Show that $\int_1^{\infty}e^{-\pi x}(x^{\sigma/2} + x^{(1-\sigma)/2}) x^{-1} dx \ll (1 + |\sigma|)^{\pi |\sigma|}$

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.

$$I(\sigma)=\int_1^{\infty}e^{-\pi x}\left(x^{\sigma/2} + x^{(1-\sigma)/2}\right) x^{-1}\, dx =E_{\frac{2-\sigma }{2}}(\pi )+E_{\frac{1+\sigma }{2}}(\pi )$$ 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 $$\frac{2 \,\Gamma \left(\frac{1}{4},\pi \right)}{\sqrt{\pi }}\leq I(\sigma) \leq \Gamma (0,\pi )+\frac{\Gamma \left(\frac{1}{2},\pi \right)}{\sqrt{\pi }}$$ Numerically, these bounds are $$\approx 0.0230332$$ and $$\approx 0.0230952$$ (this is a very limited range).
• 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 $\sigma$. – Quantaliinuxite Jun 11 at 9:52
• @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. – Claude Leibovici Jun 11 at 9:56