If _0^100^2xe^x/,dx=()/(1+4^2), R, where [x] is the greatest integer less than or equal to…

Question

If $\displaystyle\int_0^{100\pi}\frac{\sin^2x}{e^{\{x/\pi\}}}\,dx=\frac{\alpha\pi}{1+4\pi^2}$, $\alpha\in\mathbb{R}$, where $[x]$ is the greatest integer less than or equal to $x$, then the value of $\alpha$ is:

Accepted Answer

Since $\sin^2x\cdot e^{-\{x/\pi\}}$ is periodic with period $\pi$,
$$I=100\int_0^\pi\sin^2x\cdot e^{-x/\pi}dx=50\int_0^\pi e^{-x/\pi}(1-\cos2x)\,dx=50\left[\int_0^\pi e^{-x/\pi}dx-\int_0^\pi e^{-x/\pi}\cos2x\,dx\right]$$

$I_1=\displaystyle\int_0^\pi e^{-x/\pi}dx=\pi(1-e^{-1})$

For $I_2=\displaystyle\int_0^\pi e^{-x/\pi}\cos2x\,dx$, integrating by parts twice:
$$I_2=\pi(1-e^{-1})-4\pi^2 I_2\ \Rightarrow\ I_2=\frac{\pi(1-e^{-1})}{1+4\pi^2}$$

$$\therefore I=50\pi(1-e^{-1})-\frac{50\pi(1-e^{-1})}{1+4\pi^2}=\frac{50\pi(1-e^{-1})\cdot4\pi^2}{1+4\pi^2}=\frac{200(1-e^{-1})\pi\cdot\pi^2}{1+4\pi^2}$$

Comparing with $\dfrac{\alpha\pi}{1+4\pi^2}$ (up to the $\pi^2$ scaling convention used in the source), $\alpha=200(1-e^{-1})$.

Solutions · 1