Computing $ \sum_{n=0}^{\infty} (-1)^{n+1}\int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t$ 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?
 A: $$\sum_{n=0}^{m-1} (-1)^{n}\int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t = \int^1_0 \frac{t^3}{1+t^3} \sum_{n=0}^{m-1} (-t^3)^{n}\mathrm{d} t$$
$$ =  \int^1_0 \frac{t^3}{1+t^3} \frac{1 - (-t^3)^m }{1+t^3} \mathrm{d} t.$$
Thus by the Dominated Convergence Theorem, $$ \sum_{n=0}^{\infty} (-1)^n \int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t = \lim_{m\to\infty} \int^1_0 \frac{t^3}{1+t^3} \frac{1 - (-t^3)^m }{1+t^3} \mathrm{d} t$$ 
$$ = \int^1_0 \frac{t^3}{(1+t^3)^2} \mathrm{d} t = \frac{2\sqrt{3}\pi +\log 64 -9}{54}.$$
A: We have
$$ \begin{align*}
\sum_{n=0}^{\infty} (-1)^{n+1} \int_{0}^{1} \frac{t^{3(n+1)}}{t^3 + 1} \; dt
& = - \int_{0}^{1} \frac{t^3}{(t^3 + 1)^2} \; dt \\
&= \left[ \frac{t}{3} \frac{1}{t^3 + 1}\right]_{0}^{1} - \frac{1}{3} \int_{0}^{1} \frac{dt}{t^3 + 1} \\
&= \frac{1}{6} - \frac{1}{3} \int_{0}^{1} \left( \frac{1}{3(t+1)} -\frac{2 t-1}{6(t^2-t+1)}+\frac{1}{2(t^2-t+1)} \right) \; dt \\
&= \frac{1}{6} - \frac{1}{3} \left( \frac{1}{3} \log 2 + 0 + \frac{\pi}{3\sqrt{3}} \right) \\
&= \frac{1}{6} - \frac{1}{9} \left( \log 2 + \frac{\pi}{\sqrt{3}} \right).
\end{align*}$$
