Question

a) Γ(n) =∫[infinity]0tn−1e−tdta)Show that Γ(n+ 1) =nΓ(n) using integration by parts.b) Show that Γ(n+ 1) =n! , wherenis a positive integer, by repeatedly applying the result in parta)c) Compute Γ(5).d) Prove what the value of Γ(12) should be by evaluating the gamma function forn= 1/2 and usingthe fact that∫[infinity]0e−u2du=√π2.

Answer 1

$\Gamma(n)=\displaystyle\int_0^\infty t^{n-1}e^{-t}\,\mathrm dt$

a. Integrate by parts by taking

$u=t^{n-1}\implies\mathrm du=(n-1)t^{n-2}\,\mathrm dt$

$\mathrm dv=e^{-t}\,\mathrm dt\implies v=-e^{-t}$

Then

$\displaystyle\Gamma(n)=-t^{n-1}e^{-t}\bigg|_0^\infty+(n-1)\int_0^\infty t^{n-2}e^{-t}\,\mathrm dt$

We have

$\displaystyle\lim_{t\to\infty}\frac{t^{n-1}}{e^t}=0$

and so

$\Gamma(n)=(n-1)\displaystyle\int_0^\infty t^{n-2}e^{-t}\,\mathrm dt=(n-1)\Gamma(n-1)$

or, replacing $n\to n+1$, $\Gamma(n+1)=n\Gamma(n)$.

b. From the above recursive relation, we find

$\Gamma(n+1)=n\Gamma(n)=n(n-1)\Gamma(n-1)=n(n-1)(n-2)\Gamma(n-2)=\cdots=n(n-1)(n-2)\cdots2\cdot1\Gamma(1)$

Now,

$\Gamma(1)=\displaystyle\int_0^\infty e^{-t}\,\mathrm dt=1$

and so we're left with $\Gamma(n+1)=n!$.

c. Using the previous result, we find $\Gamma(5)=4!=24$.

d. If the question is asking to find $\Gamma(12)$, then you can just use the same approach as in (c).

But if you're supposed to find $\Gamma\left(\frac12\right)$, we have

$\displaystyle\Gamma\left(\frac12\right)=\int_0^\infty t^{-1/2}e^{-t}\,\mathrm dt$

Substitute

$u=t^{1/2}\implies u^2=t\implies 2u\,\mathrm du=\mathrm dt$

Then

$\displaystyle\Gamma\left(\frac12\right)=\int_0^\infty\frac1ue^{-u^2}(2u\,\mathrm du)=2\int_0^\infty e^{-u^2}\,\mathrm du=\frac{2\sqrt\pi}2=\sqrt\pi$