4
$\begingroup$

(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

$\endgroup$
  • $\begingroup$ You have a recurrent formula, so you can use it up to the first term of the sequence. Have you tried to do it? $\endgroup$ – cool Dec 15 '13 at 12:10
  • 1
    $\begingroup$ Two unrelated problems should be split up into two questions. $\endgroup$ – Ron Gordon Dec 15 '13 at 12:36
  • $\begingroup$ Sorry Ron Gordon.......... $\endgroup$ – juantheron Dec 16 '13 at 3:33
3
$\begingroup$

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 $e-1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.