Question

If \(\int_0^{100\pi} \frac{\sin^2 x}{e^{\left(\frac{x}{\pi} - \left[\frac{x}{\pi}\right]\right)}}dx = \frac{\alpha\pi^3}{1 + 4\pi^2},\ \alpha \in R\) where \([x]\) is the greatest integer less than or equal to x, then the value of \(\alpha\) is :

• A. \(200\,(1 - e^{-1})\)
• B. \(100\,(1 - e)\)
• C. \(50\,(e - 1)\)
• D. \(150\,(e^{-1} - 1)\)

Hint:

The integrand has period \(\pi\); reduce to 100 times the integral over \([0,\pi]\), where \([x/\pi] = 0\).

Answer 1

The Correct Option is A

The integrand \(\sin^2 x\, e^{-\left(\frac{x}{\pi} - \left[\frac{x}{\pi}\right]\right)}\) has period \(\pi\), so \(I = 100\int_0^{\pi} \sin^2 x\, e^{-x/\pi}dx\).
Let \(a = \frac{1}{\pi}\) and \(J = \int_0^{\pi}\sin^2 x\, e^{-ax}dx = \frac{1}{2}\int_0^{\pi}e^{-ax}dx - \frac{1}{2}\int_0^{\pi}e^{-ax}\cos 2x\, dx\).
\(\int_0^{\pi}e^{-ax}dx = \frac{1 - e^{-1}}{a} = \pi(1 - e^{-1})\).
Integrating by parts twice, \(\int e^{-ax}\cos 2x\, dx = \frac{e^{-ax}(2\sin 2x - a\cos 2x)}{a^2 + 4}\), so \(K = \int_0^{\pi}e^{-ax}\cos 2x\, dx = \frac{e^{-1}(-a) - (-a)}{a^2+4} = \frac{a(1 - e^{-1})}{a^2 + 4} = \frac{\pi(1 - e^{-1})}{1 + 4\pi^2}\).
\(J = \frac{\pi(1-e^{-1})}{2}\left[1 - \frac{1}{1+4\pi^2}\right] = \frac{\pi(1-e^{-1})}{2}\cdot\frac{4\pi^2}{1+4\pi^2} = \frac{2\pi^3(1-e^{-1})}{1+4\pi^2}\).
\(I = 100J = \frac{200\pi^3(1-e^{-1})}{1+4\pi^2}\), so \(\alpha = 200(1 - e^{-1})\).
Answer: (1) \(200\,(1 - e^{-1})\)