$\int_0^1 \ln(x)^n \operatorname{Ei}(x) \, dx$ integralini her $n \geq 0$ tam sayisi icin hesaplayiniz.

Question

$Ei$'nin tanimi icin bu soruya bakabilirsiniz.

$\int_0^1 \ln(x)^n \operatorname{Ei}(x) \, dx$ integralini her $n \geq 0$ tam sayisi icin hesaplayiniz.

$n=0$ durumu icin  $$\begin{align} \int_{0}^{1} \text{Ei}(x) \, dx &= x \text{Ei}(x) \Big|^{1}_{0^{+}} - \int_{0}^{1}e^{x} \, dx \\ &= \text{Ei}(1)- e +1. \end{align}$$ $n=1$ durumu icin $$\begin{align}\int_{0}^{1} \ln(x) \text{Ei}(x) \, dx &= \text{Ei}(x) \left(x\ln (x) -x \right) \Bigg|^{1}_{0^{+}}-\int_{0}^{1} e^{x} \ln (x) \, dx + \int_{0}^{1} e^{x} \, dx \\ &= -\text{Ei}(1) - \Big(e^{x} \ln(x) - \text{Ei}(x) \Big) \Bigg|^{1}_{0^{+}} + e -1 \\ &= - \text{Ei}(1) + \text{Ei}(1) - \gamma + e -1 \\ &= -\gamma + e -1. \end{align}$$

Answer 1

İntegralimiz :

$$\Lambda(n)=\int_0^1\:Ei(x)\:\ln^n(x)\:dx$$

İntegrali kısmi türev ile yazalım.

$$\Lambda(n)=\lim\limits_{s\to0}\frac{\partial^n}{\partial{s}^n}\int_0^1\:Ei(x)\:x^s\:dx$$

$Ei(x)=u$ 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. $$\color{#A00000}{\boxed{\Lambda(n)=\int_0^1\:Ei(x)\:\ln^n(x)\:dx=(-1)^n\:n!\bigg(Ei(1)-e+1-{}_nF_n\Bigg(\begin{array}\,1,1,1\cdots1\\2,2,2\cdots2 \end{array}\Bigg|1\Bigg)\:\Bigg)}}$$

Comments

Toplam sembolünü hipergeometrik fonksiyon ile yazabiliriz.Ama bir işe yaramaz.

$$\sum_{k=0}^\infty\frac{1}{k!(k+1)^{n+1}}={}_nF_n \Bigg(\begin{array}\,1,1,1\cdots1\\2,2,2\cdots2 \end{array}\Bigg|1\Bigg)$$

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ise yaramaz derken? istedigim eklediginli haliydi.

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$\mbox{Ei}(x)$ yazınca işe yarıyor ya! O da yarar.

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Cevaba hipergeometrik fonksiyonlu halini de ekledim.İşe yaramıyor dememin nedeni ikisinin de kolayca hesaplanamıyor olması.