<p>Evaluate \(I=\displaystyle\int_0^\infty\frac{e^{-x}\sin x}{x}\,dx\) [JEE Advanced 2008]</p>
Step-by-Step Solution
Key Concept: Define I(a) = \int_0^\infty e^(-ax)sinx/x dx. I'(a) = -\int_0^\infty e^(-ax)sinx dx = -1/(a^2+1). Integrate: I(a) = -arctan(a)+C. I(\infty)=0 \to C=\pi/2. I(1)=\pi/2-\pi/4=\pi/4.
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<p>Let $I(a)=\int_0^\infty\frac{e^{-ax}\sin x}{x}dx$. Differentiate:</p>
<p>$I'(a)=-\int_0^\infty e^{-ax}\sin x\,dx=-\frac{1}{a^2+1}$</p>
<p>Integrate: $I(a)=-\arctan a+C$. As $a\to\infty$: $I(\infty)=0=-\frac{\pi}{2}+C\Rightarrow C=\frac{\pi}{2}$.</p>
<p>$I(a)=\frac{\pi}{2}-\arctan a$. At $a=1$: $I(1)=\frac{\pi}{2}-\frac{\pi}{4}=\boxed{\frac{\pi}{4}}$.</p>
Correct Answer: A