Prove $\int_0^{\infty } \left(\prod _{k=1}^n \frac{\sin \left(t/k^x\right)}{t/k^x}\right) \, dt = \frac{\pi}{2}$ for $x \ge 2$.

$I_n(x)$ is defined as the following.

$$I_n(x) := \int_0^{\infty } \left(\prod _{k=1}^n \frac{\sin \left(\displaystyle\frac{t}{ k^x}\right)}{\displaystyle\frac{t}{k^x}}\right) \, \mathbb{d}t$$

We know

$$I_1(1) = I_2(1) = I_3(1) = \frac{\pi}{2},$$ $$I_4(1) = \frac{1727 \pi}{3456}, I_5(1) = \frac{20652479 \pi}{41472000},$$ $$I_6(1) = \frac{2059268143 \pi}{4147200000}, I_7(1) = \frac{24860948333867803 \pi}{50185433088000000}, \cdots .$$

Now, prove

$$I_n(x) = \frac{\pi}{2}$$

for $x \ge 2$.

• any reason why should expect this to be true Commented Mar 20, 2016 at 16:03
• Aaaaaah now i understand!!!!! Commented Mar 20, 2016 at 16:08

Let's define $$\textrm{sinc}\left(x\right)=\begin{cases} \frac{\sin\left(x\right)}{x}, & x\neq0\\ 1, & x=0 \end{cases}$$ we have the following :
Theorem: Suppose that $\left\{ a_{n}\right\}$ is a sequence of positive numbers. Let $s_{n}=\sum_{k=1}^{n}a_{k}$ and $$\tau_{n}=\int_{0}^{\infty}\prod_{k=0}^{n}\textrm{sinc}\left(a_{k}t\right)dt$$ then $$0<\tau_{n}\leq\frac{\pi}{a_{0}n}$$ and the equality holds if $n=0$ or $a_{0}\geq s_{n}$ when $n\geq1$.
(See here for the proof) So if we define $$a_{k}=\frac{1}{\left(k+1\right)^{x}}$$ with $x\geq2$ we note that $$a_{0}=1\geq\sum_{k=1}^{\infty}\frac{1}{\left(k+1\right)^{x}}=\zeta\left(x\right)-1$$ and so we have $$I_{n}(x)=\int_{0}^{\infty}\prod_{k=1}^{n}\frac{\sin\left(t/k{}^{x}\right)}{t/k^{x}}dt=\int_{0}^{\infty}\prod_{k=0}^{n}\frac{\sin\left(t/\left(k+1\right)^{x}\right)}{t/\left(k+1\right)^{x}}dt=\frac{\pi}{2}.$$ Note: I think your product must start from $1$ since $$\frac{\sin\left(a/x\right)}{a/x}\overset{x\rightarrow0}{\rightarrow}0$$ and so all product becomes $0$.