1)calculate∫_0 ^∞ e^(−xt^2 ) dt with x>0 2) find the value of ∫_0 ^∞ ((e^(−t^2 ) −e^(−2t^2 ) )/t^2 ) dt by using fubinni theorem .


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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 bymaxmathsup 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 bymaxmathsup by imad last updated on 22/Jan/19

$a n t h e r m e t h o d l e t a > 0 a n d b > 0 a n d f (a, b) = \int_{0}^{\infty} \frac{e^{- {a x}^{2}} - e^{- {b x}^{2}}}{x^{2}} d x \Rightarrow \frac{\partial f}{\partial a} (a, b) = - \int_{0}^{\infty} e^{- {a x}^{2}} d x$ $= - \int_{0}^{\infty} e^{- {(\sqrt{a} x)}^{2}} d x =_{\sqrt{a} x = t} - \int_{0}^{\infty} e^{- t^{2}} \frac{d t}{\sqrt{a}}$ $= - \frac{1}{\sqrt{a}} \frac{\sqrt{π}}{2} = - \frac{\sqrt{π}}{2 \sqrt{a}} \Rightarrow f (a, b) = - \sqrt{π} \int \frac{d a}{2 \sqrt{a}} = - \sqrt{π} \sqrt{a} + c$ $c = f (0, b) = \int_{0}^{\infty} \frac{1 - e^{- {b x}^{2}}}{x^{2}} d x = φ (b) \Rightarrow φ^{^{'}} (b) = \int_{0}^{\infty} e^{- {b x}^{2}} d x = \frac{\sqrt{π}}{2 \sqrt{b}} \Rightarrow$ $φ (b) = \sqrt{π} \sqrt{b} + c_{0} b u t φ (0) = c_{0} \Rightarrow φ (b) = \sqrt{π} \sqrt{b} \Rightarrow f (a, b) = - \sqrt{π} \sqrt{a} + \sqrt{π} \sqrt{b}$ $= \sqrt{π} (\sqrt{b} - \sqrt{a}) f i n a l l y \int_{0}^{\infty} \frac{e^{- t^{2}} - e^{- 2 t^{2}}}{t^{2}} d t = f (1, 2) = \sqrt{π} (\sqrt{2} - 1) .$
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