I would like to find the exact value of the following series:
$ \sum_{n=0}^{\infty} (-1)^{n+1}\int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t$
We can easily show that the series converges using the alternating series test:
$ 0 \leq\int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t \leq\frac{1}{3n+4}$
So $ \int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t \rightarrow_{n\rightarrow\infty} 0$
And $ \int_{0}^{1}\frac{t^{3n+6}}{1+t^3}\mathrm{d} t-\int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t=\int_{0}^{1}\frac{t^{3n+3}(t^3-1)}{1+t^3}\mathrm{d} t\leq0 $
So the series converges.
Do you have any idea to compute the sum?