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

Let

$$f(x)=\frac{\sin^2x}{e^{\left(\frac{x}{\pi}-\left[\frac{x}{\pi}\right]\right)}}$$

Since $$\sin^2x$$ has period $$\pi,$$ and $$\left\{\frac{x}{\pi}\right\}=\frac{x}{\pi}-\left[\frac{x}{\pi}\right]$$ also has period $$\pi,$$ the function $$f(x)$$ has period $$\pi.$$

Hence,

$$\int_0^{100\pi}f(x)\,dx=100\int_0^\pi f(x)\,dx$$

Therefore,

$$I=100\int_0^\pi e^{-x/\pi}\sin^2x\,dx$$

Using

$$\sin^2x=\frac{1-\cos2x}{2},$$

we get

$$I=50\left[\int_0^\pi e^{-x/\pi}\,dx-\int_0^\pi e^{-x/\pi}\cos2x\,dx\right]$$

Now,

$$\int_0^\pi e^{-x/\pi}\,dx=\left[-\pi e^{-x/\pi}\right]_0^\pi=\pi(1-e^{-1})$$

Also, using

$$\int e^{ax}\cos bx\,dx=\frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx),$$

with

$$a=-\frac1\pi,\qquad b=2,$$

we get

$$\int_0^\pi e^{-x/\pi}\cos2x\,dx=\frac{\pi(1-e^{-1})}{1+4\pi^2}$$

Hence,

$$I=50\left[\pi(1-e^{-1})-\frac{\pi(1-e^{-1})}{1+4\pi^2}\right]$$

$$=50\pi(1-e^{-1})\left[1-\frac1{1+4\pi^2}\right]$$

$$=50\pi(1-e^{-1})\left[\frac{4\pi^2}{1+4\pi^2}\right]$$

$$=\frac{200(1-e^{-1})\pi^3}{1+4\pi^2}$$

Comparing with

$$\frac{\alpha\pi^3}{1+4\pi^2},$$

we get

$$\alpha=200(1-e^{-1})$$