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Question Number 79107 by mathmax by abdo last updated on 22/Jan/20

calculate f(a) =∫_0 ^∞  e^(−(x^2  +(a/x^2 ))) dx with a>0

$${calculate}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left({x}^{\mathrm{2}} \:+\frac{{a}}{{x}^{\mathrm{2}} }\right)} {dx}\:{with}\:{a}>\mathrm{0} \\ $$

Commented bymathmax by abdo last updated on 23/Jan/20

f(a)=∫_0 ^∞  e^(−(x^2  +(a/x^2 ))) dx  changement x=(1/t) give  f(a)=−∫_0 ^∞   e^(−((1/t^2 )+at^2 )) (((−dt)/t^2 )) =∫_0 ^∞   (e^(−(at^2  +(1/t^2 ))) /t^2 )dt  ⇒f^′ (a) =−∫_0 ^∞   e^(−(at^2  +(1/t^2 ))) dt =_((√a)t =x)  −(1/(√a))∫_0 ^∞  e^(−{x^2 +(a/x^2 )})  dx  =−(1/(√a))f(a)⇒((f^′ (a))/(f(a))) =−(1/(√a)) ⇒ln∣f(a)∣=−2(√a) +c  f>0 ⇒f(a) =k e^(−2(√a))   k=f(0) =∫_0 ^∞  e^(−x^2 ) dx =((√π)/2) ⇒f(x) =((√π)/2) e^(−2(√a))

$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left({x}^{\mathrm{2}} \:+\frac{{a}}{{x}^{\mathrm{2}} }\right)} {dx}\:\:{changement}\:{x}=\frac{\mathrm{1}}{{t}}\:{give} \\ $$ $${f}\left({a}\right)=−\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+{at}^{\mathrm{2}} \right)} \left(\frac{−{dt}}{{t}^{\mathrm{2}} }\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−\left({at}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)} }{{t}^{\mathrm{2}} }{dt} \\ $$ $$\Rightarrow{f}^{'} \left({a}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left({at}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)} {dt}\:=_{\sqrt{{a}}{t}\:={x}} \:−\frac{\mathrm{1}}{\sqrt{{a}}}\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left\{{x}^{\mathrm{2}} +\frac{{a}}{{x}^{\mathrm{2}} }\right\}} \:{dx} \\ $$ $$=−\frac{\mathrm{1}}{\sqrt{{a}}}{f}\left({a}\right)\Rightarrow\frac{{f}^{'} \left({a}\right)}{{f}\left({a}\right)}\:=−\frac{\mathrm{1}}{\sqrt{{a}}}\:\Rightarrow{ln}\mid{f}\left({a}\right)\mid=−\mathrm{2}\sqrt{{a}}\:+{c} \\ $$ $${f}>\mathrm{0}\:\Rightarrow{f}\left({a}\right)\:={k}\:{e}^{−\mathrm{2}\sqrt{{a}}} \\ $$ $${k}={f}\left(\mathrm{0}\right)\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:\Rightarrow{f}\left({x}\right)\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:{e}^{−\mathrm{2}\sqrt{{a}}} \\ $$

Answered by ~blr237~ last updated on 22/Jan/20

let state u=(1/x)  then dx=−(du/u^2 )   f(a)=∫_0 ^∞ e^(−((1/u^2 )+au^2 )) (du/u^2 )   f∈C^(1 )  cause for all a>0 ∣e^(−(x^2 +(a/x^2 ))) ∣≤e^(−x^2 )   and ∫_(0 ) ^∞ e^(−x^2 ) dx=((√π)/2)  so converges  Now we can write   (df/da)=−∫_0 ^∞ e^(−(au^2 +(1/u^2 ))) du       =−(1/(√a))∫_0 ^∞ e^(−[(u(√a))^2 +(a/((u(√a))^2 ))]) d(u(√a))      =−(1/(√a)) ∫_(0 ) ^∞ e^(−(v^2 +(a/v^2 ))) dv     with v=u(√a)      So   (df/da)=−(1/(√a)) f(a)  Then  ln∣f(a)∣=−2(√a) +c ⇒f(a)=ke^(−2(√a))   with k>0  we know that lim_(a→0)  f(a)=k=((√π)/2)   Finaly    f(a)=((√π)/2) e^(−2(√a))

$${let}\:{state}\:{u}=\frac{\mathrm{1}}{{x}}\:\:{then}\:{dx}=−\frac{{du}}{{u}^{\mathrm{2}} }\: \\ $$ $${f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−\left(\frac{\mathrm{1}}{{u}^{\mathrm{2}} }+{au}^{\mathrm{2}} \right)} \frac{{du}}{{u}^{\mathrm{2}} }\: \\ $$ $${f}\in{C}^{\mathrm{1}\:} \:{cause}\:{for}\:{all}\:{a}>\mathrm{0}\:\mid{e}^{−\left({x}^{\mathrm{2}} +\frac{{a}}{{x}^{\mathrm{2}} }\right)} \mid\leqslant{e}^{−{x}^{\mathrm{2}} } \:\:{and}\:\int_{\mathrm{0}\:} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\sqrt{\pi}}{\mathrm{2}}\:\:{so}\:{converges} \\ $$ $${Now}\:{we}\:{can}\:{write}\: \\ $$ $$\frac{{df}}{{da}}=−\int_{\mathrm{0}} ^{\infty} {e}^{−\left({au}^{\mathrm{2}} +\frac{\mathrm{1}}{{u}^{\mathrm{2}} }\right)} {du} \\ $$ $$\:\:\:\:\:=−\frac{\mathrm{1}}{\sqrt{{a}}}\int_{\mathrm{0}} ^{\infty} {e}^{−\left[\left({u}\sqrt{{a}}\right)^{\mathrm{2}} +\frac{{a}}{\left({u}\sqrt{{a}}\right)^{\mathrm{2}} }\right]} {d}\left({u}\sqrt{{a}}\right) \\ $$ $$\:\:\:\:=−\frac{\mathrm{1}}{\sqrt{{a}}}\:\int_{\mathrm{0}\:} ^{\infty} {e}^{−\left({v}^{\mathrm{2}} +\frac{{a}}{{v}^{\mathrm{2}} }\right)} {dv}\:\:\:\:\:{with}\:{v}={u}\sqrt{{a}}\: \\ $$ $$\:\:\:{So}\:\:\:\frac{{df}}{{da}}=−\frac{\mathrm{1}}{\sqrt{{a}}}\:{f}\left({a}\right) \\ $$ $${Then}\:\:{ln}\mid{f}\left({a}\right)\mid=−\mathrm{2}\sqrt{{a}}\:+{c}\:\Rightarrow{f}\left({a}\right)={ke}^{−\mathrm{2}\sqrt{{a}}} \:\:{with}\:{k}>\mathrm{0} \\ $$ $${we}\:{know}\:{that}\:\underset{{a}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{f}\left({a}\right)={k}=\frac{\sqrt{\pi}}{\mathrm{2}}\: \\ $$ $${Finaly}\:\:\:\:{f}\left({a}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}\:{e}^{−\mathrm{2}\sqrt{{a}}} \: \\ $$ $$ \\ $$ $$ \\ $$

Commented bymsup trace by abdo last updated on 22/Jan/20

thanks sir

$${thanks}\:{sir} \\ $$

Answered by mind is power last updated on 22/Jan/20

on pose x=a^(1/4) u  ⇒f(a)=∫_0 ^(+∞) a^(1/4) e^(−(√a)(x^2 +(1/x^2 ))) dx  x=(1/y)⇒  =a^(1/4) ∫_0 ^(+∞) e^(−(√a)(y^2 +(1/y^2 ))) (dy/y^2 )  ⇒2f(a)a^(−(1/4)) =∫_0 ^(+∞) (1+(1/y^2 ))e^(−(√a)(y−(1/y))^2 −2(√a))   ⇒2f(a)a^(−(1/4)) e^(2(√a)) =∫_0 ^(+∞) (1+(1/y^2 ))e^(−(√a)(y−(1/y))^2 ) dy  =∫_0 ^1 (1+(1/y^2 ))e^(−(√a)(y−(1/y))^2 ) +∫_1 ^(+∞) (1+(1/y^2 ))e^(−(√a)(y−(1/y))^2 ) dy  u=y−(1/y),in first t=(1/y)in 2nd⇒  ⇒∫_0 ^(+∞)  e^(−(√a)u^2 ) du +∫_0 ^1 (1+(1/t^2 ))e^(−(√a)(t−(1/t))^2 ) dt  =2∫_0 ^(+∞) e^(−(√a)(u^2 )) du,w=a^(1/4) u⇒  =2∫_0 ^(+∞) e^(−w^2 ) (dw/a^(1/4) )=((√π)/a^(1/4) )  ⇒2f(a)a^(−(1/4)) e^(2(√a)) =((√π)/a^(1/4) )⇒f(a)=((√π)/(2e^(2(√a)) ))

$${on}\:{pose}\:{x}={a}^{\frac{\mathrm{1}}{\mathrm{4}}} {u} \\ $$ $$\Rightarrow{f}\left({a}\right)=\int_{\mathrm{0}} ^{+\infty} {a}^{\frac{\mathrm{1}}{\mathrm{4}}} {e}^{−\sqrt{{a}}\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)} {dx} \\ $$ $${x}=\frac{\mathrm{1}}{{y}}\Rightarrow \\ $$ $$={a}^{\frac{\mathrm{1}}{\mathrm{4}}} \int_{\mathrm{0}} ^{+\infty} {e}^{−\sqrt{{a}}\left({y}^{\mathrm{2}} +\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right)} \frac{{dy}}{{y}^{\mathrm{2}} } \\ $$ $$\Rightarrow\mathrm{2}{f}\left({a}\right){a}^{−\frac{\mathrm{1}}{\mathrm{4}}} =\int_{\mathrm{0}} ^{+\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right){e}^{−\sqrt{{a}}\left({y}−\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} −\mathrm{2}\sqrt{{a}}} \\ $$ $$\Rightarrow\mathrm{2}{f}\left({a}\right){a}^{−\frac{\mathrm{1}}{\mathrm{4}}} {e}^{\mathrm{2}\sqrt{{a}}} =\int_{\mathrm{0}} ^{+\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right){e}^{−\sqrt{{a}}\left({y}−\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} } {dy} \\ $$ $$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right){e}^{−\sqrt{{a}}\left({y}−\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} } +\int_{\mathrm{1}} ^{+\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right){e}^{−\sqrt{{a}}\left({y}−\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} } {dy} \\ $$ $${u}={y}−\frac{\mathrm{1}}{{y}},{in}\:{first}\:{t}=\frac{\mathrm{1}}{{y}}{in}\:\mathrm{2}{nd}\Rightarrow \\ $$ $$\Rightarrow\int_{\mathrm{0}} ^{+\infty} \:{e}^{−\sqrt{{a}}{u}^{\mathrm{2}} } {du}\:+\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){e}^{−\sqrt{{a}}\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} } {dt} \\ $$ $$=\mathrm{2}\int_{\mathrm{0}} ^{+\infty} {e}^{−\sqrt{{a}}\left({u}^{\mathrm{2}} \right)} {du},{w}={a}^{\frac{\mathrm{1}}{\mathrm{4}}} {u}\Rightarrow \\ $$ $$=\mathrm{2}\int_{\mathrm{0}} ^{+\infty} {e}^{−{w}^{\mathrm{2}} } \frac{{dw}}{{a}^{\frac{\mathrm{1}}{\mathrm{4}}} }=\frac{\sqrt{\pi}}{{a}^{\frac{\mathrm{1}}{\mathrm{4}}} } \\ $$ $$\Rightarrow\mathrm{2}{f}\left({a}\right){a}^{−\frac{\mathrm{1}}{\mathrm{4}}} {e}^{\mathrm{2}\sqrt{{a}}} =\frac{\sqrt{\pi}}{{a}^{\frac{\mathrm{1}}{\mathrm{4}}} }\Rightarrow{f}\left({a}\right)=\frac{\sqrt{\pi}}{\mathrm{2}{e}^{\mathrm{2}\sqrt{{a}}} } \\ $$

Commented bymsup trace by abdo last updated on 22/Jan/20

thanks sir.

$${thanks}\:{sir}. \\ $$

Commented bymind is power last updated on 22/Jan/20

y′re welcom

$${y}'{re}\:{welcom} \\ $$

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