Lebesgue theory and equivalence integral-series

Question

I have this exercise, but I really don't know how to start. Prove that: $$\sum_{n=1}^{\infty}{(-1)^n\frac{1}{(2n)(2n)!}}=\int_0^1{\sin{x}\log{x}dx}$$. How can I approach this kind of exercises in general?

Answer 1

Hint: use $$ \sin x=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)!}x^{2n-1}, \int_0^1x^n\ln x=-\frac{1}{(n+1)^2}.$$