find-the-value-of-0-1-x-E-1-x-dx-

Question Number 26569 by abdo imad last updated on 26/Dec/17

f i n d t h e v a l u e o f \int_{0}^{1} x E (\frac{1}{x}) d x

Commented by abdo imad last updated on 30/Dec/17

l e t p u t I = \int_{0}^{1} x E (\frac{1}{x}) d x a n d b y t h e c h a n g e m e n t \frac{1}{x} = t

I = \int_{1}^{\propto} \frac{E (t)}{t^{3}} d t = {l i m}_{n - >\propto} S_{n} w i t h S_{n} = \sum_{k = 1}^{k = n - 1} \int_{k}^{k + 1} \frac{E (t)}{t^{3}} d t

S_{n} = \sum_{k = 1}^{k = n - 1} k . \int_{k}^{k + 1} \frac{d t}{t^{3}} = \frac{1}{2} \sum_{k = 1}^{k = n - 1} k (\frac{1}{k^{2}} - \frac{1}{{(k + 1)}^{2}})

= \frac{1}{2} \sum_{k = 1}^{k = n - 1} \frac{2 k + 1}{k {(k + 1)}^{2}}

= \sum_{k = 1}^{k = n - 1} \frac{1}{{(k + 1)}^{2}} + \frac{1}{2} \sum_{k = 1}^{k = n - 1} \frac{1}{k {(k + 1)}^{2}} b u t

\sum_{k = 1}^{k = n - 1} \frac{1}{k {(k + 1)}^{2}} = \sum_{k = 1}^{k = n - 1} \frac{1}{k} - \sum_{k = 1}^{n - 1} \frac{1}{k + 1} - \sum_{k = 1}^{k = n - 1} \frac{1}{{(k + 1)}^{2}}

a f t e r a l l c a l c u l u s w e f i n d

S_{n} = \frac{1}{2} \sum_{k = 1}^{k = n} \frac{1}{k^{2}} - \frac{1}{2} (H_{n} - H_{n - 1}) b u t {l i m}_{n - >\propto} H_{n} - H_{n - 1} = 0

a n d {l i m}_{n - >\propto} \sum_{k = 1}^{k = n} \frac{1}{k^{2}} = \frac{π^{2}}{6} \Rightarrow {l i m}_{n - >\propto} S_{n} = I = \frac{π^{2}}{12} .