[tex]\Gamma(n)=\displaystyle\int_0^\infty t^{n-1}e^{-t}\,\mathrm dt[/tex]
a. Integrate by parts by taking
[tex]u=t^{n-1}\implies\mathrm du=(n-1)t^{n-2}\,\mathrm dt[/tex]
[tex]\mathrm dv=e^{-t}\,\mathrm dt\implies v=-e^{-t}[/tex]
Then
[tex]\displaystyle\Gamma(n)=-t^{n-1}e^{-t}\bigg|_0^\infty+(n-1)\int_0^\infty t^{n-2}e^{-t}\,\mathrm dt[/tex]
We have
[tex]\displaystyle\lim_{t\to\infty}\frac{t^{n-1}}{e^t}=0[/tex]
and so
[tex]\Gamma(n)=(n-1)\displaystyle\int_0^\infty t^{n-2}e^{-t}\,\mathrm dt=(n-1)\Gamma(n-1)[/tex]
or, replacing [tex]n\to n+1[/tex], [tex]\Gamma(n+1)=n\Gamma(n)[/tex].
b. From the above recursive relation, we find
[tex]\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)[/tex]
Now,
[tex]\Gamma(1)=\displaystyle\int_0^\infty e^{-t}\,\mathrm dt=1[/tex]
and so we're left with [tex]\Gamma(n+1)=n![/tex].
c. Using the previous result, we find [tex]\Gamma(5)=4!=24[/tex].
d. If the question is asking to find [tex]\Gamma(12)[/tex], then you can just use the same approach as in (c).
But if you're supposed to find [tex]\Gamma\left(\frac12\right)[/tex], we have
[tex]\displaystyle\Gamma\left(\frac12\right)=\int_0^\infty t^{-1/2}e^{-t}\,\mathrm dt[/tex]
Substitute
[tex]u=t^{1/2}\implies u^2=t\implies 2u\,\mathrm du=\mathrm dt[/tex]
Then
[tex]\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[/tex]
