<p>Evaluate \(\displaystyle\int_0^1(-\ln x)^n\,dx\) for \(n\in\mathbb{N}\). [JEE Advanced 2009]</p>
Step-by-Step Solution
Key Concept: Let t = -ln x \to x = e^(-t), dx = -e^(-t)dt. \int_0^\infty tⁿ \cdot e^(-t) dt = \Gamma(n+1) = n\!
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<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>$$\int_0^1(-\ln x)^n dx=\int_\infty^0 t^n e^{-t}(-e^{-t})dt... $$</p>
<p>Wait: $dx = e^{-t}(-1)dt\cdot(-1) = e^{-t}dt$ (careful with signs).</p>
<p>$x=e^{-t}\Rightarrow dx=-e^{-t}dt$. When $x\to 0^+$, $t\to+\infty$; $x=1\Rightarrow t=0$.</p>
<p>$$\int_0^1(-\ln x)^n dx = \int_\infty^0 t^n(-e^{-t}dt)=\int_0^\infty t^n e^{-t}dt=\Gamma(n+1)=n\!$$</p>
Correct Answer: A