- #1

- 33

- 0

## Main Question or Discussion Point

I attempted to prove the following equality, but to no avail. Anyone is willing to lend a hand?

[tex]\int_0^{\infty} s^{2t-2v} e^{i w s} ds + \int_0^{\infty} s^{2t-2v} e^{-i w s} ds = \left[ \left( \frac{1}{-iw}\right)^{2t-2v+1} + \left( \frac{1}{iw}\right)^{2t-2v+1} \right] \Gamma(2t-2v + 1)[/tex],

where [tex]i = \sqrt{-1}[/tex], [tex]s > 0[/tex], [tex]\Gamma(\cdot)[/tex] is gamma function, [tex]-\pi \leq w \leq \pi[/tex], [tex]0 < t <1[/tex], and [tex]1 \leq v \leq \infty[/tex] is integer.

I got

[tex]\int_0^{\infty} s^{2t-2v} e^{i w s} ds + \int_0^{\infty} s^{2t-2v} e^{-i w s} ds = \left[ \left( \frac{1}{-iw}\right)^{2t-2v+1} + \left( \frac{1}{iw}\right)^{2t-2v+1} \right] \Gamma(2t-2v + 1)[/tex],

where [tex]i = \sqrt{-1}[/tex], [tex]s > 0[/tex], [tex]\Gamma(\cdot)[/tex] is gamma function, [tex]-\pi \leq w \leq \pi[/tex], [tex]0 < t <1[/tex], and [tex]1 \leq v \leq \infty[/tex] is integer.

I got

**almost**all the RHS, except the power terms. It seems strange as IMHO it is only true when [tex]t[/tex] is integer and that [tex]2t-2v+1 \geq 0[/tex].
Last edited: