Trouble calculating a integral

Question

I've difficulties calculating the following integral

$$\int_0^\infty y^k\frac{1}{\sqrt{2\pi}y}e^{\frac{-(\ln y)^2}{2}}[\sin (2\pi \ln y)]\,dy$$

This integral is a part of a density distribution of a random variable and I need to prove that this integral is equal to zero but I don't know how to do it. I've try to use the fact that $\sin$ is a even function but I don't know how to proceed any further than that. I appreciate any help. Thank you in advance.

Answer 1

You are considering $$I=\int_0^\infty y^k\frac{1}{\sqrt{2\pi}y}e^{\frac{-(\ln y)^2}{2}}[\sin (2\pi \ln y)]\,dy$$ Let $y=e^t$ to make $$I=\frac{1}{\sqrt{2\pi}}\int_{-\infty} ^\infty e^{k t-\frac{t^2}{2}} \sin (2 \pi t)\,dt=\frac{1}{\sqrt{2\pi}}\Im \int_{-\infty} ^\infty \exp\big[ (k+2 i \pi ) t-\frac{t^2}{2}\big]\,dt$$ Complete the square and you should arrive to the gaussian integral.

Edit

If you did what I suggested, you are suppose to have obtained

$$\int\exp\big[ (k+2 i \pi ) t-\frac{t^2}{2}\big]\,dt=-i \sqrt{\frac{\pi }{2}} e^{\frac{1}{2} (k+2 i \pi )^2} \text{erfi}\left(\frac{-i k+i t+2 \pi }{\sqrt{2}}\right)$$ $$\int_{-\infty} ^\infty \big[ (k+2 i \pi ) t-\frac{t^2}{2}\big]\,dt=\sqrt{2 \pi } e^{\frac{1}{2} (k+2 i \pi )^2}$$ $$\Im\Big[e^{\frac{1}{2} (k+2 i \pi )^2}\Big]=\sqrt{2 \pi } e^{\frac{k^2}{2}-2 \pi ^2} \sin (2 \pi k)$$ So, finally, the integral is just $$e^{\frac{k^2}{2}-2 \pi ^2} \sin (2 \pi k)$$ which is zero when $k$ is zero or a multiple of half an integer.

Answer 2

Using $y^k = e^{k\ln y} = \cosh(k\ln y) + \sinh(k\ln y)$ we have that the integral simplifies to

$$\int_0^{\infty}\sinh(k\ln y)\frac{1}{\sqrt{2\pi}y}e^{-\frac{(\ln y)^2}{2}}\sin(2\pi \ln y)\:dy$$

which is nonzero for $k\neq 0$