Integral of lnt/(1+t) dt
The solution
The answer (Indefinite)
[src]
// -polylog(2, 1 + t) + pi*I*log(1 + t) for |1 + t| < 1\
/ || |
| || / 1 \ 1 |
| log(t) || -polylog(2, 1 + t) - pi*I*log|-----| for ------- < 1|
| ------ dt = C + |< \1 + t/ |1 + t| |
| 1 + t || |
| || __0, 2 /1, 1 | \ __2, 0 / 1, 1 | \ |
/ ||-polylog(2, 1 + t) + pi*I*/__ | | 1 + t| - pi*I*/__ | | 1 + t| otherwise |
\\ \_|2, 2 \ 0, 0 | / \_|2, 2 \0, 0 | / /
$$\int \frac{\log{\left(t \right)}}{t + 1}\, dt = C + \begin{cases} i \pi \log{\left(t + 1 \right)} - \operatorname{Li}_{2}\left(t + 1\right) & \text{for}\: \left|{t + 1}\right| < 1 \\- i \pi \log{\left(\frac{1}{t + 1} \right)} - \operatorname{Li}_{2}\left(t + 1\right) & \text{for}\: \frac{1}{\left|{t + 1}\right|} < 1 \\- i \pi {G_{2, 2}^{2, 0}\left(\begin{matrix} & 1, 1 \\0, 0 & \end{matrix} \middle| {t + 1} \right)} + i \pi {G_{2, 2}^{0, 2}\left(\begin{matrix} 1, 1 & \\ & 0, 0 \end{matrix} \middle| {t + 1} \right)} - \operatorname{Li}_{2}\left(t + 1\right) & \text{otherwise} \end{cases}$$
2
pi
- --- + 2*pi*I*log(2)
12
$$- \frac{\pi^{2}}{12} + 2 i \pi \log{\left(2 \right)}$$
=
2
pi
- --- + 2*pi*I*log(2)
12
$$- \frac{\pi^{2}}{12} + 2 i \pi \log{\left(2 \right)}$$
Use the examples entering the upper and lower limits of integration.