Question-65581

Question Number 65581 by aliesam last updated on 31/Jul/19

Commented by mathmax by abdo last updated on 31/Jul/19

l e t A = \int e^{(\frac{1}{x} - x)} d x w e h a v e e^{u} = \sum_{n = 0}^{\infty} \frac{u^{n}}{n!} w i t h r a d i u s i n f i n i t e \Rightarrow

e^{(\frac{1}{x} - x)} = \sum_{n = 0}^{\infty} \frac{1}{n!} {(\frac{1}{x} - x)}^{n} = \sum_{n = 0}^{\infty} \frac{1}{n!} (\sum_{k = 0}^{n} C_{n}^{k} x^{k} {(\frac{1}{x})}^{n - k})

= \sum_{n = 0}^{\infty} \frac{1}{n!} \sum_{k = 0}^{n} C_{n}^{k} x^{k} . x^{k - n} = \sum_{n = 0}^{\infty} \frac{1}{n!} \sum_{k = 0}^{n} C_{n}^{k} x^{2 k - n} \Rightarrow

A = \sum_{n = 0}^{\infty} \frac{1}{n!} (\sum_{k = 0}^{n} C_{n}^{k} \int x^{2 k - n} d x)

= \sum_{n = 0}^{\infty} \frac{1}{n!} (\sum_{k = 0}^{n} \frac{C_{n}^{k}}{2 k - n + 1} x^{2 k - n + 1} + λ)

= \sum_{n = 0}^{\infty} \sum_{k = 0}^{n} \frac{C_{n}^{k}}{n! (2 k - n + 1)} x^{2 k - n + 1} + λ e