Q.Evaluate the following:

The Gamma integral is a special improper integral $\displaystyle\int_0^\infty e^{-x}x^{n-1}\,dx$, defined for every positive integer $n$ (and, more generally, every real $n>0$), and denoted $\Gamma(n)$ ("gamma of $n$").

Key facts leveraged before defining $\Gamma$: $e^\infty=\infty$, $e^{-\infty}=0$, and — by L'Hôpital's rule applied $m$ times — $\displaystyle\lim_{x\to\infty}x^m e^{-x}=0$ for every positive integer $m$, which is what makes the Gamma integral converge.

Derivation of $\Gamma(n)=(n-1)!$. Applying integration by parts to $I_n=\int_0^\infty e^{-x}x^n\,dx$ gives the recurrence $I_n=nI_{n-1}$; iterating down to $I_0=\int_0^\infty e^{-x}\,dx=1$ gives $I_n=n!$, i.e. $\displaystyle\int_0^\infty e^{-x}x^n\,dx=n!$. Re-indexing ($n\to n-1$) gives the Gamma-integral form

with the recurrence $\Gamma(n+1)=n\,\Gamma(n)$ and base value $\Gamma(1)=\int_0^\infty e^{-x}\,dx=1$.

Extending the technique by substitution. A substitution $t=ax$ (for $a>0$) converts a scaled exponential into the Gamma form:

Similarly $u=x^n$ (or writing $n=e^{\log n}$) converts other exponent forms, e.g. $\displaystyle\int_0^\infty \dfrac{x^n}{n^x}\,dx=\dfrac{n!}{(\log n)^{n+1}}$, and $x=u^2$ converts a Gaussian-type integral: $\displaystyle2\int_0^\infty e^{-x^2}x^{2n-1}\,dx=\Gamma(n)$.