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.

[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]