# The value of $\int_0^{\infty } \left(\prod _{i=1}^n \frac{\sin \left(\frac{x}{2 i-1}\right)}{\frac{x}{2 i-1}}\right) \, \mathbb{d}x$

I came across this problem on Facebook the other day, so probably you guys have already seen it.

Define the following integral: $$I(n):=\int_0^{\infty } \left(\prod _{i=1}^n \frac{\sin \left(\frac{x}{2 i-1}\right)}{\frac{x}{2 i-1}}\right) \, \mathbb{d}x$$

Using Mathematica, I found the symbolic solutions are:

$$I(1)=I(2)=\dots=I(7)=\frac\pi2$$

However,

$$I(8)=\frac{467807924713440738696537864469 }{935615849440640907310521750000}\pi\approx0.499999999993\pi$$ $$I(9)=\frac{17708695183056190642497315530628422295569865119}{35417390788301195294898352987527510935040000000}\pi\approx0.499999994\pi$$

I didn't go further. My question is, should $I(n)=\frac\pi2$ for all positive integers $n$, and the above results are just some roundoff errors in symbolic calculation?