İntegralimiz :
Λ(n)=∫10Ei(x)lnn(x)dx
İntegrali kısmi türev ile yazalım.
Λ(n)=lim
Ei(x)=u ve x^s=dv olacak şekilde kısmi integral alalım.
\Lambda(n)=\lim\limits_{s\to0}\frac{\partial^n}{\partial{s}^n}\frac{Ei(x)x^{s+1}}{s+1}\Bigg|_0^1-\frac{1}{s+1}\int_0^1\:e^x\:x^s\:dx
\Lambda(n)=\lim\limits_{s\to0}\frac{\partial^n}{\partial{s}^n}\frac{Ei(1)}{s+1}-\frac{1}{s+1}\int_0^1\:e^x\:x^s\:dx
e^x ifadesini taylor ile açalım ve integrali alalım.
\Lambda(n)=\lim\limits_{s\to0}\frac{\partial^n}{\partial{s}^n}\frac{Ei(1)}{s+1}-\frac{1}{s+1}\int_0^1\:\sum_{k=0}^\infty\frac{x^{s+k}}{k!}\:dx
\Lambda(n)=\lim\limits_{s\to0}\frac{\partial^n}{\partial{s}^n}\frac{Ei(1)}{s+1}-\frac{1}{s+1}\sum_{k=0}^\infty\frac{1}{k!}\int_0^1\:x^{s+k}\:dx\Lambda(n)=\lim\limits_{s\to0}\frac{\partial^n}{\partial{s}^n}\frac{Ei(1)}{s+1}-\frac{1}{s+1}\sum_{k=0}^\infty\frac{1}{k!}\:\frac{x^{s+k+1}}{s+k+1}\Bigg|_0^1\Lambda(n)=\lim\limits_{s\to0}\frac{\partial^n}{\partial{s}^n}\frac{Ei(1)}{s+1}-\frac{1}{s+1}\sum_{k=0}^\infty\:\frac{1}{k!(s+k+1)}Türevi alalım.
\Lambda(n)=\lim\limits_{s\to0}\:(-1)^n\:n!(s+1)^{-n-1}Ei(1)\to\\-(-1)^n\:n\:(s+1)^{-n-1}\sum_{k=0}^\infty\frac{1}{k!(s+k+1)}-(-1)^n\:n!\:\sum_{k=0}^\infty\frac{1}{k!(s+k+1)^{n+1}}s yerine
0 verelim.
\Lambda(n)=(-1)^n!\:n!\:Ei(1)-(-1)^n\:n!\sum_{k=0}^\infty\frac{1}{(k+1)!}-(-1)^n\:n!\sum_{k=0}^\infty\frac{1}{k!(k+1)^{n+1}}Sadeleştirelim.
\Lambda(n)=(-1)^n!\:n!\:Ei(1)-(-1)^n\:n!\underbrace{\sum_{k=0}^\infty\frac{1}{(k+1)!}}_{\large\:e-1}-(-1)^n\:n!\sum_{k=0}^\infty\frac{1}{k!(k+1)^{n+1}}\color{#A00000}{\boxed{\Lambda(n)=\int_0^1\:Ei(x)\:\ln^n(x)\:dx=(-1)^n\:n!\bigg(Ei(1)-e+1-\sum_{k=0}^\infty\frac{1}{k!(k+1)^{n+1}}\bigg)}}Sondaki seri istenilirse hipergeometrik fonksiyon ilede yazılabilir.
\color{#A00000}{\boxed{\Lambda(n)=\int_0^1\:Ei(x)\:\ln^n(x)\:dx=(-1)^n\:n!\bigg(Ei(1)-e+1-{}_nF_n\Bigg(\begin{array}\,1,1,1\cdots1\\2,2,2\cdots2 \end{array}\Bigg|1\Bigg)\:\Bigg)}}