1-calculate-0-e-xt-2-dt-with-x-gt-0-2-find-the-value-of-0-e-t-2-e-2t-2-t-2-dt-by-using-fubinni-theorem-

Question Number 53271 by Abdo msup. last updated on 19/Jan/19

1) c a l c u l a t e \int_{0}^{\infty} e^{- {x t}^{2}} d t w i t h x > 0

2) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{e^{- t^{2}} - e^{- 2 t^{2}}}{t^{2}} d t b y u s i n g

f u b i n n i t h e o r e m .

Commented by maxmathsup by imad last updated on 22/Jan/19

1) w e h a v e \int_{0}^{\infty} e^{- {x t}^{2}} d t =_{t \sqrt{x} = u} \int_{0}^{\infty} e^{- u^{2}} \frac{d u}{\sqrt{x}} = \frac{1}{\sqrt{x}} \int_{0}^{\infty} e^{- u^{2}} d u

= \frac{1}{\sqrt{x}} \frac{\sqrt{π}}{2} = \frac{\sqrt{π}}{2 \sqrt{x}}

2) w e h a v e \int_{1}^{2} \frac{\sqrt{π}}{2 \sqrt{x}} d x = \sqrt{π} {[\sqrt{x}]}_{1}^{2} = \sqrt{π} (\sqrt{2} - 1) b u t

\int_{1}^{2} \frac{\sqrt{π}}{2 \sqrt{x}} d x = \int_{1}^{2} (\int_{0}^{\infty} e^{- {x t}^{2}} d t) d x = \int_{0}^{\infty} (\int_{1}^{2} e^{- {x t}^{2}} d x) d t (b y f u b i n i)

= \int_{0}^{\infty} ({[- \frac{1}{t^{2}} e^{- {x t}^{2}}]}_{x = 1}^{x = 2}) d t = \int_{0}^{\infty} \frac{e^{- t^{2}} - e^{- 2 t^{2}}}{t^{2}} d t \Rightarrow

\int_{0}^{\infty} \frac{e^{- t^{2}} - e^{- 2 t^{2}}}{t^{2}} d t = \sqrt{π} (\sqrt{2} - 1)

Commented by maxmathsup by imad last updated on 22/Jan/19

a n t h e r m e t h o d l e t a > 0 a n d b > 0 a n d f (a, b) = \int_{0}^{\infty} \frac{e^{- {a x}^{2}} - e^{- {b x}^{2}}}{x^{2}} d x \Rightarrow \frac{\partial f}{\partial a} (a, b) = - \int_{0}^{\infty} e^{- {a x}^{2}} d x

= - \int_{0}^{\infty} e^{- {(\sqrt{a} x)}^{2}} d x =_{\sqrt{a} x = t} - \int_{0}^{\infty} e^{- t^{2}} \frac{d t}{\sqrt{a}}

= - \frac{1}{\sqrt{a}} \frac{\sqrt{π}}{2} = - \frac{\sqrt{π}}{2 \sqrt{a}} \Rightarrow f (a, b) = - \sqrt{π} \int \frac{d a}{2 \sqrt{a}} = - \sqrt{π} \sqrt{a} + c

c = f (0, b) = \int_{0}^{\infty} \frac{1 - e^{- {b x}^{2}}}{x^{2}} d x = φ (b) \Rightarrow φ^{^{'}} (b) = \int_{0}^{\infty} e^{- {b x}^{2}} d x = \frac{\sqrt{π}}{2 \sqrt{b}} \Rightarrow

φ (b) = \sqrt{π} \sqrt{b} + c_{0} b u t φ (0) = c_{0} \Rightarrow φ (b) = \sqrt{π} \sqrt{b} \Rightarrow f (a, b) = - \sqrt{π} \sqrt{a} + \sqrt{π} \sqrt{b}

= \sqrt{π} (\sqrt{b} - \sqrt{a}) f i n a l l y \int_{0}^{\infty} \frac{e^{- t^{2}} - e^{- 2 t^{2}}}{t^{2}} d t = f (1, 2) = \sqrt{π} (\sqrt{2} - 1) .