Calculate $\int_{-\pi}^{\pi}\prod_{n=1}^\infty \left(1-\frac{t^2}{n^2}\right) e^{-izt}dt$

Question

Calculate $$\int_{-\pi}^{\pi}\prod_{n=1}^\infty \left(1-\frac{t^2}{n^2}\right) e^{-izt}dt$$ Any suggestions please?

Thank you very much.

Answer 1

Consider the integral \begin{align} I = \int_{-\pi}^{\pi} \left[ \prod_{k=1}^{\infty} \left( 1 - \frac{t^{2}}{k^{2}} \right) \right] e^{-i z t} \, dt \end{align} where \begin{align} \prod_{k=1}^{\infty} \left( 1 - \frac{t^{2}}{k^{2}} \right) = \frac{\sin(\pi t)}{\pi t} \end{align} for which \begin{align} I = \int_{-\pi}^{\pi} \frac{\sin(\pi t)}{\pi t} \, e^{-i z t} \, dt. \end{align} By direct integration this leads to \begin{align} I &= \frac{i}{2 \pi} \left[ Ei(-i(z+\pi)t) - Ei(i(z+\pi)t) \right]_{-\pi}^{\pi} \\ &= \frac{i}{2\pi} \left[ Ei(-i(\pi^{2}+ \pi z)) - Ei(i(\pi^{2} - \pi z)) + Ei(-i(\pi^{2} - \pi z)) - Ei(i(\pi^{2} - \pi z)) \right] \end{align} where $Ei(z)$ is the exponential integral defined by \begin{align} Ei(z) = - \int_{-z}^{\infty} \frac{e^{-t}}{t} \, dt. \end{align} Converting from the Exponential integrals to sine and cosine integrals, defined by \begin{align} Ci(z) &= - \int_{0}^{\infty} \frac{\cos(t)}{t} \, dt = \frac{1}{2}\left(Ei(iz)+ Ei(-iz) \right) \\ Si(z) &= - \int_{0}^{\infty} \frac{\sin(t)}{t} \, dt = \frac{1}{2i} \left( Ei(z)- Ei(-iz) \right) + \frac{\pi}{2}, \end{align} the integral is seen as \begin{align} I &= \frac{\sin(\pi^{2})}{2 \pi^{2}} \left[ Ci(\pi^{2} + \pi z) + Ci(\pi^{2} - \pi z) \right] \\ & \, \, \, + \frac{1}{\pi^{2}} \left( 2 \sin^{2}(\pi^{2}) - \cos^{2}(\pi^{2}) \right) \left[ Si(\pi^{2} + \pi z) + Si(\pi^{2} - \pi z) \right]. \end{align}