How to solve the following result in integration?

Question

I want to prove the following inequality:

$$\int_{a}^{b}{|\cosh(\sqrt{iw})|^2dw} \geq \int_{a}^{b} \sinh^2\big(\sqrt{\frac{w}{2}}\big) dw$$

How to solve above integral? I am unable to keep $\iota (i)$ out of the first integral.

Answer 1

We have: $$ \sinh\sqrt{\frac{w}{2}}=\sqrt{\frac{w}{2}}\sum_{n\geq 0}\frac{w^n}{2^n(2n+1)!} \tag{1} $$ and: $$ \cosh(\sqrt{iw})=\sum_{n\geq 0}\frac{i^n w^n}{(2n)!} \tag{2}$$ so: $$ \sinh^2\sqrt{\frac{w}{2}}=\frac{1}{2}\sum_{n\geq 1}\frac{2^n w^n}{(2n)!}\tag{3}$$ and: $$ \left\|\cosh\sqrt{iw}\right\|^2 = \left(\sum_{n\geq 0}\frac{i^n w^n}{(2n)!}\right)\cdot\left(\sum_{n\geq 0}\frac{(-1)^n i^n w^n}{(2n)!}\right)\\=\cosh^2\sqrt{\frac{w}{2}}\cos^2\sqrt{\frac{w}{2}}+\sinh^2\sqrt{\frac{w}{2}}\sin^2\sqrt{\frac{w}{2}}\tag{4}$$ is easy to compare with $(3)$, since $\cosh^2\geq \sinh^2$ and $\sin^2+\cos^2=1$.