# Gamma function convergence in $\mathbb{R}$

A few months ago I learned that Euler's gamma function: $$\int_0^{\infty} e^{-t}t^{x-1}dt$$ converges when $x \gt 0$. Then few weeks ago I find out from theory on power series that convergence for harmonic series $\sum {1 \over n^\alpha}$ can be found for $\alpha \in \mathbb{R}$ by using a integral criteria which says that series converges if and only if improper integral $\int_{1}^\infty {dx \over x^\alpha}$ converges, and that is right for $\alpha \gt 1$.

What I am trying to do here, is to implement integral criteria from power series to gamma function, but for ${1 \over t^{1 - x} }$ it doesn't seem to work since for $1-x \gt 1$ or $x \lt 0$, gamma function diverges, and i expect from integral criteria to converge. What am I doing wrong?

• You can check that $\int_{1}^{\infty} t^{x-1} e^{-t} \,dt$ converges for any $x$. This is because the exponential decay easily overcomes the polynomial growth (or decay) of $t^{x-1}$. So it is not the near-infinity behavior but the near-zero behavior that can go problematic. – Sangchul Lee Jan 2 '17 at 12:26

## 1 Answer

Let $u=t^{x/2}$ and assume $x>0$,

$$\int_1^\infty t^{x-1}e^{-t}\ dt=\frac2x\int_1^\infty e^{-u^{2/x}}\ du$$

As $x>0$, $2/x>0$, and so it is easy enough to check that $\int_1^\infty e^{-u^{2/x}}\ du$ will converge.

For $x<0$,

$$\int_1^\infty t^{x-1}e^{-t}\ dt=\frac2x\int_1^0e^{-u^{2/x}}\ du$$

And clearly this is finite.

For $x=0$, convergence is also trivial.

But for $\int_0^1t^{x-1}e^{-t}\ dt$, notice that if $x\ge1$, the integral is obviously finite.

For $1>x>0$, see that

$$\int_0^1t^{x-1}e^{-t}\ dt<\int_0^1t^{x-1}\ dt$$

And so this converges trivially. But for $x\le0$,

$$\int_0^1t^{x-1}e^{-t}\ dt>\int_0^1t^{x-1}e^{-1}\ dt$$

And so this diverges trivially.