Definite Integration
Gamma + Special Values
Grade None
Question:
<p>Evaluate \(\displaystyle\int_0^1\sqrt{-\ln x}\,dx\). [JEE Advanced 2009]</p>
<li>\(\dfrac{\sqrt{\pi}}{2}\)</li>
<li>\(\sqrt{\pi}\)</li>
<li>\(\dfrac{\pi}{2}\)</li>
<li>\(1\)</li>
Step-by-Step Solution
Key Concept: Let t=-ln x \to x=e^(-t), dx=-e^(-t)dt. \int_0^\infty \sqrt{t} \cdot e^(-t)dt = \Gamma(3/2) = (1/2)\Gamma(1/2) = \sqrt\pi/2.
<div class='solution'>
<p>Let $t=-\ln x\Rightarrow x=e^{-t}$, $dx=-e^{-t}dt$. Limits: $x=0\to t=\infty$; $x=1\to t=0$.</p>
<p>$$\int_0^1\sqrt{-\ln x}\,dx=\int_\infty^0\sqrt{t}(-e^{-t})dt=\int_0^\infty t^{1/2}e^{-t}dt=\Gamma\left(\frac{3}{2}\right)=\frac{1}{2}\Gamma\left(\frac{1}{2}\right)=\frac{\sqrt{\pi}}{2}$$</p>
</div>
Correct Answer: A