@user21820 Hi! I have a short question. I’m reading the book „The Gamma Function“ by J. Bonnar. On page ten he writes there ex. epsilon such that 0<=t^(x-1)e^(-t/2)<=1 for t=epsilon. Then he splits an integral at epsilon in two (1 to epsilon and epsilon to inf) and says the first is finite and for the second one for t >= epsilon he uses t^(x-1)e^(-t)=t^(x-1)e^(-t/2)e^(-t/2)<=e^(-t/2). This ist not really neat, or what do you think?

@DavidP I think something got mixed up in your description.

Can you just give the exact integral you want, and the intervals you want, using variables instead of "epsilon"?

If you want to prove convergence of ∫[0,∞] x^(c−1)·exp(−x/2) dx, it is indeed convenient to split it, because the tail is dominated by the exp part.

Yes - but isn't this estimation wrong. The first inequality - as he wrote - is for t=epsilon and the he uses this for every t>=epsilon.

Please phrase it using my variables, because I don't know what on earth is the point of saying "t=ε" here.

In particular, I would bound the tail for something like x∈[4c,∞).

(When c > 1.)

He splits \int_0^{\infty} t^(x-1)exp(-t) dt in \int_0^1 + \int_1^{\infty} and proves that the first one exists. That is clear. The his text is:

In the case of the second integral, first note that t^q exp(-t/2) \to 0 as t \to \infty for any q in \R. Hence for any x in \R there exits an \epsilon such that 0 \leq t^(x-1)e^(-t/2) \leq 1 for t = \epslion. So we further split the integral at \epsilon.

The first term is finite, being a finite integral. For the second term, when t \geq \epslion we have: t^(x-1)e^(-t)=(t^(x-1)e^(-t/2))e^(-t/2)<=e^(-t/2)

Ok I get the general idea the author is using, but I didn't read any of the formulae. I wouldn't do it that way only because it is obvious that my way works.

After all, he is still making an unjustified claim about a limit.

Why not just deal with everything at one go?

Thank you!

Ok sure!

Everything fine? Are you enjoying your weekend?

@DavidP Yea, going off soon though.

But everything is fine on my side. Hope you too are staying safe wherever you are!

Yes - thanks! Take care too. Bye!