I understand the recurrence relation of the gamma function and why it is well defined for $x\ge 1$. But doesn't $\lim_{t \to 0} \frac{e^{-t}} {\sqrt{t} } = \infty$?
I get how you can define $Γ(0.5) = Γ(1.5)\cdot 2$, but to prove the recurrence relation you need to assume the integral converges. So how is the function well defined for $0<x<1$?

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    $\begingroup$ The improper integral still converges (look at the integral of $1/\sqrt{t}$ for example). $\endgroup$ Jan 27, 2022 at 11:17

1 Answer 1


It is well defined for $0\lt{x}\lt1$ because if $x=\frac1{y},$ then $$\int_0^{\infty}t^{\frac1{y}-1}e^{-t}\,\mathrm{d}t=y\int_0^{\infty}\frac1{y}t^{\frac1{y}-1}e^{-t}\,\mathrm{d}t=y\int_0^{\infty}e^{-t^y}\,\mathrm{d}y,$$ and the latter can easily be shown to converge. For the record, the fact that $$\lim_{t\to0}\frac1{t^{1-x}e^t}=\infty$$ for $0\lt{x}\lt1$ does not imply the integral diverges. The integral $$\int_0^1\frac1{x^p}\,\mathrm{d}x$$ converges for $0\lt{p}\lt1.$


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