Question

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

İntegralini çözün.

Answer 1

İntegralimiz :

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

İntegrali buradaki eşitliğin kısmi türevi olarak yazabiliriz.Eşitlik :

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

$$\lim\limits_{n\to1}\frac{\partial}{\partial{n}}\Xi(n,1)=\int_0^1\:\frac{\ln(x)\:\ln\big(\ln(x)\big)}{1+x}\:dx$$

Türevi alıp $n$ yerine $1$ verelim.($A\to$ Glaisher-Kinkelin sabiti)

$$\color{#A00000}{\boxed{\lim\limits_{n\to1}\frac{\partial}{\partial{n}}\Xi(n,1)=\int_0^1\:\frac{\ln(x)\:\ln\big(\ln(x)\big)}{1+x}\:dx=\frac{\pi^2}{12}\Big(12\ln(A)-1-i\pi-\ln(4\pi)\Big)\\\approx-0.449043-2.58386i}}$$