matematikk.net

[tex]\large I=\int_{0}^{1}\frac{x^{100}-1}{\ln(x)}\,dx[/tex]

La [tex]I(\alpha) = \int_0^1 \dfrac{x^\alpha-1}{\ln(x)}dx[/tex]. Da har man at $$\dfrac{dI}{d\alpha} = \int_0^1 \dfrac{d}{d\alpha}\left(\dfrac{x^\alpha-1}{\ln(x)}\right)dx = \int_0^1 x^\alpha dx = \dfrac{1}{\alpha+1}$$
slik at [tex]I(\alpha) = \ln|\alpha+1|+C[/tex]. Ettersom $I(0) = 0$ må $C = 0$, og dermed følger [tex]I(100) = \ln(101)[/tex].

MatIsa skrev:La [tex]I(\alpha) = \int_0^1 \dfrac{x^\alpha-1}{\ln(x)}dx[/tex]. Da har man at $$\dfrac{dI}{d\alpha} = \int_0^1 \dfrac{d}{d\alpha}\left(\dfrac{x^\alpha-1}{\ln(x)}\right)dx = \int_0^1 x^\alpha dx = \dfrac{1}{\alpha+1}$$
slik at [tex]I(\alpha) = \ln|\alpha+1|+C[/tex]. Ettersom $I(0) = 0$ må $C = 0$, og dermed følger [tex]I(100) = \ln(101)[/tex].

bra,
"differentiate under the integral sign" funker...