UPSC 2016 Maths Optional Paper 1 Q1c — Step-by-Step Solution

10 marks · Section A

Indefinite integrals · Calculus · asked 7× in 13 yrs · Read the full method →

Question

Evaluate $I=\displaystyle\int_0^1\sqrt[3]{x\log\!\left(\frac1x\right)}\,dx$ .

Substitution $x=e^{-t}$ turns $\int_0^1 x^{m}(-\log x)^{n}\,dx$ into a Gamma integral $\Gamma(n+1)/(m+1)^{n+1}$ .

I=\int_0^1\Big(x\log\tfrac1x\Big)^{1/3}\,dx=\int_0^1 x^{1/3}\big(-\log x\big)^{1/3}\,dx,

using $\log\tfrac1x=-\log x\ge0$ on $(0,1)$ .

Let $x=e^{-t}$ , so $\log x=-t$ , $-\log x=t$ , and $dx=-e^{-t}\,dt$ . As $x:0\to1$ , $t:\infty\to0$ :

I=\int_{\infty}^{0}\big(e^{-t}\big)^{1/3}\,t^{1/3}\,\big(-e^{-t}\big)\,dt=\int_0^{\infty} e^{-t/3}\,t^{1/3}\,e^{-t}\,dt=\int_0^{\infty}t^{1/3}e^{-4t/3}\,dt.

Here $s-1=\tfrac13\Rightarrow s=\tfrac43$ and $p=\tfrac43$ . Hence

I=\frac{\Gamma\!\left(\tfrac43\right)}{\left(\tfrac43\right)^{4/3}}.

Using $\Gamma\!\left(\tfrac43\right)=\tfrac13\Gamma\!\left(\tfrac13\right)$ and $\left(\tfrac43\right)^{4/3}=\dfrac{4^{4/3}}{3^{4/3}}$ :

I=\Gamma\!\left(\tfrac43\right)\left(\tfrac34\right)^{4/3}=\frac{3^{4/3}}{4^{4/3}}\,\Gamma\!\left(\tfrac43\right)=\frac{3\sqrt[3]{6}}{8}\,\Gamma\!\left(\tfrac43\right).

\boxed{\;I=\frac{\Gamma\!\left(\tfrac43\right)}{\left(\tfrac43\right)^{4/3}}=\left(\frac34\right)^{4/3}\Gamma\!\left(\frac43\right)\approx0.6085.\;}