Question

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

İntegralini çözün.

Answer 1

İntegralimiz :

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

İntegrali kısmi türev ile yazabiliriz.

$$\Xi(n,m)=\lim\limits_{s\to0}\frac{\partial^n}{\partial{s}^n}\int_0^1\:\frac{x^s}{1+x^m}\:dx$$

$\frac{1}{1+x^m}$ ifadesini açalım.

$$\Xi(n,m)=\lim\limits_{s\to0}\frac{\partial^n}{\partial{s}^n}\int_0^1\:x^s(1-x^m+x^{2m}-x^{3m}+\cdots)\:dx$$

İfadeyi sonsuz toplam ile yazalım.

$$\Xi(n,m)=\lim\limits_{s\to0}\frac{\partial^n}{\partial{s}^n}\int_0^1\:\sum_{k=0}^\infty\:(-1)^n\:x^{s+mk}\:dx$$

$$\Xi(n,m)=\lim\limits_{s\to0}\frac{\partial^n}{\partial{s}^n}\sum_{k=0}^\infty\:(-1)^n\int_0^1\:\:x^{s+mk}\:dx$$

İntegrali çözelim.

$$\Xi(n,m)=\lim\limits_{s\to0}\frac{\partial^n}{\partial{s}^n}\sum_{k=0}^\infty\:(-1)^n\:\frac{x^{s+mk+1}}{s+mk+1}\Bigg|_0^1$$

$$\Xi(n,m)=\lim\limits_{s\to0}\frac{\partial^n}{\partial{s}^n}\sum_{k=0}^\infty\:\frac{(-1)^n}{s+mk+1}$$

Sadeleştirelim.

$$\Xi(n,m)=\lim\limits_{s\to0}\frac{\partial^n}{\partial{s}^n}\frac{1}{m}\sum_{k=0}^\infty\:\frac{(-1)^n}{\frac{s}{m}+\frac{1}{m}+k}$$

Seriyi lerch zeta fonksiyonu ile yazabiliriz.Ayrıntılı bilgi için buraya bakılabilir.

$$\Xi(n,m)=\lim\limits_{s\to0}\frac{\partial^n}{\partial{s}^n}\frac{1}{m}\Phi\Big(-1,1,\frac{s}{m}+\frac{1}{2}\Big)$$

Şimdi sırayla $s$ ye göre $1.$ , $2.$ ve $n.$ türevleri alalım.

$$\Xi(n,m)=\lim\limits_{s\to0}\frac{\partial^{n-1}}{\partial{s}^{n-1}}\frac{1}{m^2}(-1)\Phi\Big(-1,2,\frac{s}{m}+\frac{1}{m}\Big)$$

$$\Xi(n,m)=\lim\limits_{s\to0}\frac{\partial^{n-2}}{\partial{s}^{n-2}}\frac{1}{m^3}(-1)(-2)\Phi\Big(-1,3,\frac{s}{m}+\frac{1}{m}\Big)$$

$$\Xi(n,m)=\lim\limits_{s\to0}\frac{(-1)^n}{m^{n+1}}\Gamma(n+1)\Phi\Big(-1,n+1,\frac{s}{m}+\frac{1}{m}\Big)$$

$s$ yerine $0$ yazalım.

$$\large\color{#A00000}{\boxed{\Xi(n,m)=\int_0^1\:\frac{\ln^n(x)}{1+x^m}\:dx=\frac{(-1)^n}{m^{n+1}}\Gamma(n+1)\Phi\Big(-1,n+1,\frac{1}{m}\Big)\\n\neq-1,-2,-3\cdots\:\:\:\:\:\:m\in\mathbb{R}}}$$