Sr Examen
Integral de lnt/(1+t) dt
Solución
Ha introducido
[src]
1 / | | log(t) | ------ dt | 1 + t | / 0
$$\int\limits_{0}^{1} \frac{\log{\left(t \right)}}{t + 1}\, dt$$
Integral(log(t)/(1 + t), (t, 0, 1))
Respuesta (Indefinida)
[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 | / /
$$\int \frac{\log{\left(t \right)}}{t + 1}\, dt = C + \begin{cases} i \pi \log{\left(t + 1 \right)} - \operatorname{Li}_{2}\left(t + 1\right) & \text{for}\: \left|{t + 1}\right| < 1 \\- i \pi \log{\left(\frac{1}{t + 1} \right)} - \operatorname{Li}_{2}\left(t + 1\right) & \text{for}\: \frac{1}{\left|{t + 1}\right|} < 1 \\- i \pi {G_{2, 2}^{2, 0}\left(\begin{matrix} & 1, 1 \\0, 0 & \end{matrix} \middle| {t + 1} \right)} + i \pi {G_{2, 2}^{0, 2}\left(\begin{matrix} 1, 1 & \\ & 0, 0 \end{matrix} \middle| {t + 1} \right)} - \operatorname{Li}_{2}\left(t + 1\right) & \text{otherwise} \end{cases}$$
Respuesta
[src]
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
$$- \frac{\pi^{2}}{12} + 2 i \pi \log{\left(2 \right)}$$
-pi^2/12 + 2*pi*i*log(2)
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.