Question

$$\int_0^1\:\frac{\ln{x}}{1+x}\:dx$$

İntegralini çözün.

Answer 1

İntegralimiz :

$$\Xi(1,1)=\int_0^1\:\frac{\ln(x)}{1+x}\:dx$$

Buradaki eşitlikte $n$ yerine $1$ verelim.Eşitlik :

$$\Xi(n,1)=\int_0^1\:\frac{\ln^n(x)}{1+x}\:dx=(-1)^n\:(1-2^{-n})\:\Gamma(n+1)\zeta(n+1)$$

$$\Xi(1,1)=\int_0^1\:\frac{\ln(x)}{1+x}\:dx=-2^{-1}\:\Gamma(2)\zeta(2)$$

$$\large\color{#A00000}{\boxed{\Xi(1,1)=\int_0^1\:\frac{\ln(x)}{1+x}\:dx=-\frac{\pi^2}{12}}}$$