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let-put-f-t-0-e-ax-e-bx-x-2-e-tx-2-dx-with-t-0-and-a-gt-0-and-b-gt-0-find-a-integral-form-of-f-t-




Question Number 27974 by abdo imad last updated on 18/Jan/18
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).
$${let}\:{put}\:{f}\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{ax}} \:−\:{e}^{−{bx}} }{{x}^{\mathrm{2}} }\:{e}^{−{tx}^{\mathrm{2}} } \:{dx} \\ $$$${with}\:{t}\geqslant\mathrm{0}\:\:{and}\:{a}>\mathrm{0}\:{and}\:{b}>\mathrm{0} \\ $$$${find}\:{a}\:{integral}\:{form}\:{of}\:{f}\left({t}\right). \\ $$
Commented by abdo imad last updated on 20/Jan/18
after verifying that f is derivable on ]0,+∞[ we have   f^, (t)= −∫_0 ^∞  ( e^(−ax)  −e^(−bx) )e^(−tx^2 ) dx  =∫_0 ^∞  e^(−tx^2 −bx) dx −∫_0 ^∞  e^(−tx^2 −ax) dx   but  ∫_0 ^∞  e^(−tx^2 −ax) dx = ∫_0 ^∞   e^(−(  ((√t)x)^2  +2(a/(2(√t)))((√t)x) + (a^2 /(4t)) −(a^2 /(4t)))) dx  =  e^(a^2 /(4t))   ∫_0 ^∞   e^(−((√t)x +(a/( (√t))))^2 ) dx     the ch. (√t)x  +(a/( (√t)))=u give  ∫_0 ^∞   e^(−tx^2 −ax) dx = e^(a^2 /(4t))   ∫_(a/( (√t))) ^(+∞)   e^(−u^2 )  (du/( (√t)))  = (1/( (√t))) e^(a^2 /(4t))  (  ∫_0 ^∞  e^(−u^2 ) du − ∫_0 ^(a/( (√t)))   e^(−u^2 ) du)  = (1/( (√t))) e^(a^2 /(4t))   (    ((√π)/2)  − ∫_0 ^(a/( (√t)))   e^(−u^2 ) du) and by the same manner  we get  ∫_0 ^∞   e^(−tx^2 −bx) dx = (1/( (√t))) e^(b^2 /(4t)) (  ((√π)/2) − ∫_0 ^(b/( (√t))) e^(−u^2 ) du)  f^′ (t)= ((√π_ )/(2(√t)))(  e^(b^2 /(4t))  − e^(a^2 /(4t)) ) + ∫_0 ^(a/( (√t)))   e^(−u^2 ) du  −∫_0 ^(b/( (√t)))   e^(−u^2 ) du  = ((√π)/(2(√t))) (  e^(b^2 /(4t))   − e^(a^2 /(4t)) ) −∫_(a/( (√t))) ^(b/( (√t)))   e^(−u^2 )  du =  ψ(t) ⇒  f(t)= ∫_. ^t  ψ(u)du  +λ  .
$$\left.{after}\:{verifying}\:{that}\:{f}\:{is}\:{derivable}\:{on}\:\right]\mathrm{0},+\infty\left[\:{we}\:{have}\:\right. \\ $$$${f}^{,} \left({t}\right)=\:−\int_{\mathrm{0}} ^{\infty} \:\left(\:{e}^{−{ax}} \:−{e}^{−{bx}} \right){e}^{−{tx}^{\mathrm{2}} } {dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{tx}^{\mathrm{2}} −{bx}} {dx}\:−\int_{\mathrm{0}} ^{\infty} \:{e}^{−{tx}^{\mathrm{2}} −{ax}} {dx}\:\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{tx}^{\mathrm{2}} −{ax}} {dx}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left(\:\:\left(\sqrt{{t}}{x}\right)^{\mathrm{2}} \:+\mathrm{2}\frac{{a}}{\mathrm{2}\sqrt{{t}}}\left(\sqrt{{t}}{x}\right)\:+\:\frac{{a}^{\mathrm{2}} }{\mathrm{4}{t}}\:−\frac{{a}^{\mathrm{2}} }{\mathrm{4}{t}}\right)} {dx} \\ $$$$=\:\:{e}^{\frac{{a}^{\mathrm{2}} }{\mathrm{4}{t}}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left(\sqrt{{t}}{x}\:+\frac{{a}}{\:\sqrt{{t}}}\right)^{\mathrm{2}} } {dx}\:\:\:\:\:{the}\:{ch}.\:\sqrt{{t}}{x}\:\:+\frac{{a}}{\:\sqrt{{t}}}={u}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{tx}^{\mathrm{2}} −{ax}} {dx}\:=\:{e}^{\frac{{a}^{\mathrm{2}} }{\mathrm{4}{t}}} \:\:\int_{\frac{{a}}{\:\sqrt{{t}}}} ^{+\infty} \:\:{e}^{−{u}^{\mathrm{2}} } \:\frac{{du}}{\:\sqrt{{t}}} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{{t}}}\:{e}^{\frac{{a}^{\mathrm{2}} }{\mathrm{4}{t}}} \:\left(\:\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}^{\mathrm{2}} } {du}\:−\:\int_{\mathrm{0}} ^{\frac{{a}}{\:\sqrt{{t}}}} \:\:{e}^{−{u}^{\mathrm{2}} } {du}\right) \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{{t}}}\:{e}^{\frac{{a}^{\mathrm{2}} }{\mathrm{4}{t}}} \:\:\left(\:\:\:\:\frac{\sqrt{\pi}}{\mathrm{2}}\:\:−\:\int_{\mathrm{0}} ^{\frac{{a}}{\:\sqrt{{t}}}} \:\:{e}^{−{u}^{\mathrm{2}} } {du}\right)\:{and}\:{by}\:{the}\:{same}\:{manner} \\ $$$${we}\:{get}\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{tx}^{\mathrm{2}} −{bx}} {dx}\:=\:\frac{\mathrm{1}}{\:\sqrt{{t}}}\:{e}^{\frac{{b}^{\mathrm{2}} }{\mathrm{4}{t}}} \left(\:\:\frac{\sqrt{\pi}}{\mathrm{2}}\:−\:\int_{\mathrm{0}} ^{\frac{{b}}{\:\sqrt{{t}}}} {e}^{−{u}^{\mathrm{2}} } {du}\right) \\ $$$${f}^{'} \left({t}\right)=\:\frac{\sqrt{\pi_{} }}{\mathrm{2}\sqrt{{t}}}\left(\:\:{e}^{\frac{{b}^{\mathrm{2}} }{\mathrm{4}{t}}} \:−\:{e}^{\frac{{a}^{\mathrm{2}} }{\mathrm{4}{t}}} \right)\:+\:\int_{\mathrm{0}} ^{\frac{{a}}{\:\sqrt{{t}}}} \:\:{e}^{−{u}^{\mathrm{2}} } {du}\:\:−\int_{\mathrm{0}} ^{\frac{{b}}{\:\sqrt{{t}}}} \:\:{e}^{−{u}^{\mathrm{2}} } {du} \\ $$$$=\:\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{{t}}}\:\left(\:\:{e}^{\frac{{b}^{\mathrm{2}} }{\mathrm{4}{t}}} \:\:−\:{e}^{\frac{{a}^{\mathrm{2}} }{\mathrm{4}{t}}} \right)\:−\int_{\frac{{a}}{\:\sqrt{{t}}}} ^{\frac{{b}}{\:\sqrt{{t}}}} \:\:{e}^{−{u}^{\mathrm{2}} } \:{du}\:=\:\:\psi\left({t}\right)\:\Rightarrow \\ $$$${f}\left({t}\right)=\:\int_{.} ^{{t}} \:\psi\left({u}\right){du}\:\:+\lambda\:\:.\: \\ $$

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