If $I_{n} = \int_{0}^{1}x^n\cdot e^xdx,$ Then $\lim_{n\rightarrow \infty}\left(\sum_{k=1}^{n}\frac{I_{k+1}}{k}\right) =$

(1) If $\displaystyle I_{n} = \int_{0}^{1}x^n\cdot e^xdx$. Then $\displaystyle \lim_{n\rightarrow \infty}\left(\sum_{k=1}^{n}\frac{I_{k+1}}{k}\right) =$

$(2)$ Value of $\displaystyle \frac{\int_{0}^{\frac{\pi}{2}}e^{-x}\sin^{n}(x)dx}{\int_{0}^{\frac{\pi}{2}}e^{x}\sin^{n}(x)dx} = \;,$ Where $n\in \mathbb{N}$

$\bf{My\; Try}::$ for $(1)$ one:: Given $\displaystyle I_{n} = \int_{0}^{1}x^n\cdot e^xdx$

$\displaystyle \Rightarrow I_{n} = \left[x^n\cdot e^x\right]_{0}^{1}-n\int_{0}^{1}x^{n-1}\cdot e^xdx = e-nI_{n-1}\Rightarrow I_{n} = e-nI_{n-1}$

$\Rightarrow I_{n}+n\cdot I_{n-1} = e$

Now I did not understand how can i solve after that

Help Required.

Thanks

$(2)$ Let $\displaystyle I_{n} = \int_{0}^{\frac{\pi}{2}}e^{-x}\sin^{n}(x)dx$ and $\displaystyle J_{n} = \int_{0}^{\frac{\pi}{2}}e^{x}\sin^{n}(x)dx$

Now I did not understand how can i solve after that

Help Required

Thanks

• You have a recurrent formula, so you can use it up to the first term of the sequence. Have you tried to do it? – cool Dec 15 '13 at 12:10
• Two unrelated problems should be split up into two questions. – Ron Gordon Dec 15 '13 at 12:36
• Sorry Ron Gordon.......... – juantheron Dec 16 '13 at 3:33

Because $x \in (0,1)$, the sum

$$\sum_{k=1}^n \frac{x^k}{k}$$

converges as $n \to \infty$. Thus we can rearrange the integration and summation:

$$\sum_{k=1}^{\infty} \frac{I_{k+1}}{k} = \int_0^1 dx \, x \,e^x \sum_{k=1}^{\infty} \frac{x^k}{k} = -\int_0^1 dx \, x \,e^x \, \log{(1-x)}$$

The integral is easily evaluated using integration by parts:

$$-\int_0^1 dx \, x \,e^x \, \log{(1-x)} = \left [(1-x) e^x \log{(1-x)} \right ]_0^1+ \int_0^1 dx \frac{1-x}{1-x} e^x$$

which is equal to $e-1$. Thus, the limit is equal to $e-1$.