let put f(t)=∫_0 ^∞ ((e^(−ax) − e^(−bx) )/x^2 ) e^(−tx^2 ) dx with t≥0 and a>0 and b>0 find a integral form of f(t).


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Question Number 27974 by abdo imad last updated on 18/Jan/18

$l e t p u t f (t) = \int_{0}^{\infty} \frac{e^{- a x} - e^{- b x}}{x^{2}} e^{- {t x}^{2}} d x$ $w i t h t ⩾ 0 a n d a > 0 a n d b > 0$ $f i n d a i n t e g r a l f o r m o f f (t) .$
Commented by abdo imad last updated on 20/Jan/18

$a f t e r v e r i f y i n g t h a t f i s d e r i v a b l e o n] 0, + \infty [w e h a v e$ $f^{,} (t) = - \int_{0}^{\infty} (e^{- a x} - e^{- b x}) e^{- {t x}^{2}} d x$ $= \int_{0}^{\infty} e^{- {t x}^{2} - b x} d x - \int_{0}^{\infty} e^{- {t x}^{2} - a x} d x b u t$ $\int_{0}^{\infty} e^{- {t x}^{2} - a x} d x = \int_{0}^{\infty} e^{- ({(\sqrt{t} x)}^{2} + 2 \frac{a}{2 \sqrt{t}} (\sqrt{t} x) + \frac{a^{2}}{4 t} - \frac{a^{2}}{4 t})} d x$ $= e^{\frac{a^{2}}{4 t}} \int_{0}^{\infty} e^{- {(\sqrt{t} x + \frac{a}{\sqrt{t}})}^{2}} d x t h e c h . \sqrt{t} x + \frac{a}{\sqrt{t}} = u g i v e$ $\int_{0}^{\infty} e^{- {t x}^{2} - a x} d x = e^{\frac{a^{2}}{4 t}} \int_{\frac{a}{\sqrt{t}}}^{+ \infty} e^{- u^{2}} \frac{d u}{\sqrt{t}}$ $= \frac{1}{\sqrt{t}} e^{\frac{a^{2}}{4 t}} (\int_{0}^{\infty} e^{- u^{2}} d u - \int_{0}^{\frac{a}{\sqrt{t}}} e^{- u^{2}} d u)$ $= \frac{1}{\sqrt{t}} e^{\frac{a^{2}}{4 t}} (\frac{\sqrt{π}}{2} - \int_{0}^{\frac{a}{\sqrt{t}}} e^{- u^{2}} d u) a n d b y t h e s a m e m a n n e r$ $w e g e t \int_{0}^{\infty} e^{- {t x}^{2} - b x} d x = \frac{1}{\sqrt{t}} e^{\frac{b^{2}}{4 t}} (\frac{\sqrt{π}}{2} - \int_{0}^{\frac{b}{\sqrt{t}}} e^{- u^{2}} d u)$ $f^{^{'}} (t) = \frac{\sqrt{π_{}}}{2 \sqrt{t}} (e^{\frac{b^{2}}{4 t}} - e^{\frac{a^{2}}{4 t}}) + \int_{0}^{\frac{a}{\sqrt{t}}} e^{- u^{2}} d u - \int_{0}^{\frac{b}{\sqrt{t}}} e^{- u^{2}} d u$ $= \frac{\sqrt{π}}{2 \sqrt{t}} (e^{\frac{b^{2}}{4 t}} - e^{\frac{a^{2}}{4 t}}) - \int_{\frac{a}{\sqrt{t}}}^{\frac{b}{\sqrt{t}}} e^{- u^{2}} d u = ψ (t) \Rightarrow$ $f (t) = \int_{.}^{t} ψ (u) d u + λ .$
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