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Question Number 42708 by prof Abdo imad last updated on 01/Sep/18

c a l c u l a t e f (α) = \int_{0}^{\infty} \frac{e^{- 2 x} - e^{- x}}{x^{2}} e^{- α x^{2}} d x w i t h α > 0

1) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{e^{- 2 x} - e^{- x}}{x^{2}} e^{- 2 x^{2}} d x

Commented by maxmathsup by imad last updated on 02/Sep/18

1) f^{^{'}} (α) = - \int_{0}^{+ \infty} (e^{- 2 x} - e^{- x}) e^{- α x^{2}} d x = \int_{0}^{\infty} e^{- α x^{2} - x} d x - \int_{0}^{\infty} e^{- α x^{2} - 2 x} d x

b u t \int_{0}^{\infty} e^{- α x^{2} - x} d x = \int_{0}^{\infty} e^{- {{(\sqrt{α} x)}^{2} + 2 \frac{\sqrt{α} x}{2 \sqrt{α}} + \frac{1}{4 α} - \frac{1}{4 α})} d x = \int_{0}^{\infty} e^{- {{(\sqrt{α} x + \frac{1}{2 \sqrt{α}})}^{2}} + \frac{1}{4 α}} d x

=_{\sqrt{α} x + \frac{1}{2 \sqrt{α}} = t} e^{\frac{1}{4 α}} \int_{0}^{\infty} e^{- t^{2}} \frac{d t}{\sqrt{α}} = \frac{\sqrt{π}}{2} \frac{e^{\frac{1}{4 α}}}{\sqrt{α}} \Rightarrow f (α) = \int_{.}^{α} \frac{\sqrt{π}}{2} \frac{e^{\frac{1}{4 x}}}{\sqrt{x}} d x + c

=_{\sqrt{x} = u} \frac{\sqrt{π}}{2} \int_{.}^{\sqrt{α}} \frac{e^{\frac{1}{4 u^{2}}}}{u} (2 u) d u + c = \sqrt{π} \int_{.}^{\sqrt{α}} e^{\frac{1}{4 u^{2}}} d u + c \Rightarrow f (α) = \sqrt{π} \int_{1}^{\sqrt{α}} e^{\frac{1}{4 u^{2}}} + c

c = f (1) = \int_{0}^{\infty} \frac{e^{- x^{2} - 2 x} - e^{- x^{2} - x}}{x^{2}} d x

2) \int_{0}^{\infty} \frac{e^{- 2 x} - e^{- x}}{x^{2}} e^{- 2 x^{2}} d x = f (2) = \sqrt{π} \int_{1}^{\sqrt{2}} e^{\frac{1}{4 u^{2}}} d u + f (1) \dots .