integral from 0 to 2 of 1/(ln(x))

解

\int_{0}^{2} \frac{1}{ln ( x )} d x

Ei (ln (1)) + Ei (ln (2)) - Ei (0)

解答ステップ

\int_{0}^{2} \frac{1}{ln ( x )} d x

未定義の (特異) 点を求める:

x = 1

存在する場合は

b, a < b < c, f (b) =

未定義

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{1} \frac{1}{ln ( x )} d x + \int_{1}^{2} \frac{1}{ln ( x )} d x

\int_{0}^{1} \frac{1}{ln ( x )} d x = Ei (ln (1))

\int_{1}^{2} \frac{1}{ln ( x )} d x = Ei (ln (2)) - Ei (0)

= Ei (ln (1)) + Ei (ln (2)) - Ei (0)