Question

Express each of the following integrals as a $\mathrm{\Gamma }$ function. By computer, evaluate numerically both the $\mathrm{\Gamma }$ function and the original integral. ${\int }_0^1{x}^2{(\ln \frac{1}{x})}^{3}dx$ Hint: Put $x=e^{-u}$

Short answer

$\frac{2}{27}$

Step-by-step solution

- Substitution

Use the substitution given in the hint: Let $x=e^{-u}$. Then, $dx=-e^{-u}du$.

- Change of limits

Determine the new limits of integration for the variable $u$. When $x=0$, $u=\mathrm{\infty }$. When $x=1$, $u=0$. Thus, the integral is now ${\int }_{\mathrm{\infty }}^0{x}^2{(\ln \frac{1}{x})}^3(-e^{-u}du)$.

- Simplify the integrand

Substitute $x=e^{-u}$ into the integrand: $x^2=(e^{-u}{)}^2=e^{-2u}$ and $\ln \frac{1}{x}=\ln e^u=u$. Now the integral becomes ${\int }_{\mathrm{\infty }}^0{e}^{-2u}{u}^3(-e^{-u})du$.

- Integrate

Simplify the integral: ${\int }_{\mathrm{\infty }}^0{e}^{-3u}{u}^3(-du)={\int }_0^{\mathrm{\infty }}{e}^{-3u}{u}^{3}du$. This integral is in the form of a Gamma function: $\mathrm{\Gamma }(n)={\int }_0^{\mathrm{\infty }}{t}^{n-1}{e}^{-t}dt$. Rewrite the integral as ${\int }_0^{\mathrm{\infty }}{u}^3{e}^{-3u}du=\frac{1}{3^4}{\int }_0^{\mathrm{\infty }}(3u{)}^3{e}^{-(3u)}d(3u)$.

- Express in Gamma function

Recognize that the expression is now: $\frac{1}{3^4}{\int }_0^{\mathrm{\infty }}{t}^3{e}^{-t}dt$, where $t=3u$. According to the definition of the Gamma function, this is $\frac{1}{3^4}\mathrm{\Gamma }(4)$ as $\mathrm{\Gamma }(n)=(n-1)!$.

- Evaluate the Gamma function

Calculate $\mathrm{\Gamma }(4)$: $\mathrm{\Gamma }(4)=3!=6$. Thus, the integral becomes $\frac{1}{3^4}\ast 6=\frac{6}{81}=\frac{2}{27}$. Verify this using a computational tool to ensure accuracy.

Key concepts

Gamma function

The Gamma function, denoted as $\mathrm{\Gamma }(n)$, is a continuous extension of the factorial function to the real and complex plane. Specifically, for any positive integer $n$, $\mathrm{\Gamma }(n)=(n-1)!$. It is defined via an improper integral: \