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Question Number 42504 by maxmathsup by imad last updated on 26/Aug/18
let x>0 prove that ∫_0 ^∞ ((e^(−t^2 ) ln(1+xt^2 ))/t^2 ) dt =π ∫_0 ^(√x) e^(1/u^2 ) du .
$${let}\:{x}>\mathrm{0}\:{prove}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{e}^{−{t}^{\mathrm{2}} } {ln}\left(\mathrm{1}+{xt}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }\:{dt}\:=\pi\:\int_{\mathrm{0}} ^{\sqrt{{x}}} \:\:{e}^{\frac{\mathrm{1}}{{u}^{\mathrm{2}} }} \:\:{du}\:. \\ $$
Commented by maxmathsup by imad last updated on 28/Aug/18
let f(x) = ∫_0 ^∞ ((e^(−t^2 ) ln(1+xt^2 ))/t^2 )dt ⇒f^′ (x)= ∫_0 ^∞ (e^(−t^2 ) /(1+xt^2 ))dt ⇒ 2f^′ (x) = ∫_(−∞) ^(+∞) (e^(−t^2 ) /(1+xt^2 ))dt let consider the complex function ϕ(z) = (e^(−z^2 ) /(1+xz^2 )) we have ϕ(z) = (e^(−z^2 ) /(((√x)z−i)((√x)z +i))) =(e^(−z^2 ) /(x(z−(i/( (√x))))(z+(i/( (√x))))))) the poles of ϕ are +^− (i/( (√x))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,(i/( (√x)))) but Res(ϕ,(i/( (√x))))= (e^(1/x) /(x(((2i)/( (√x)))))) = (e^(1/x) /(2i(√x))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (e^(1/x) /(2i(√x))) =π (e^(1/x) /( (√x))) ⇒ f^′ (x) =(π/2) (e^(1/x) /( (√x))) ⇒ f(x) =(π/2) ∫_0 ^x (e^(1/t) /( (√t)))dt +c but c=f(0) =0 ⇒f(x) =(π/2) ∫_0 ^x (e^(1/t) /( (√t))) dt =_((√t)=u) (π/2) ∫_0 ^(√x) (e^(1/u^2 ) /u) (2u)du = π ∫_0 ^(√x) e^(1/u^2 ) du .
$${let}\:{f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{t}^{\mathrm{2}} } {ln}\left(\mathrm{1}+{xt}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }{dt}\:\Rightarrow{f}^{'} \left({x}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{e}^{−{t}^{\mathrm{2}} } }{\mathrm{1}+{xt}^{\mathrm{2}} }{dt}\:\Rightarrow \\ $$$$\mathrm{2}{f}^{'} \left({x}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{e}^{−{t}^{\mathrm{2}} } }{\mathrm{1}+{xt}^{\mathrm{2}} }{dt}\:\:{let}\:{consider}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{−{z}^{\mathrm{2}} } }{\mathrm{1}+{xz}^{\mathrm{2}} }\:\:{we}\:{have}\:\varphi\left({z}\right)\:=\:\frac{{e}^{−{z}^{\mathrm{2}} } }{\left(\sqrt{{x}}{z}−{i}\right)\left(\sqrt{{x}}{z}\:+{i}\right)}\:=\frac{{e}^{−{z}^{\mathrm{2}} } }{{x}\left({z}−\frac{{i}}{\:\sqrt{{x}}}\right)\left({z}+\frac{{i}}{\left.\:\sqrt{{x}}\right)}\right)} \\ $$$${the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}\:\frac{{i}}{\:\sqrt{{x}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\frac{{i}}{\:\sqrt{{x}}}\right)\:\:{but}\:\:\:{Res}\left(\varphi,\frac{{i}}{\:\sqrt{{x}}}\right)=\:\frac{{e}^{\frac{\mathrm{1}}{{x}}} }{{x}\left(\frac{\mathrm{2}{i}}{\:\sqrt{{x}}}\right)}\:=\:\frac{{e}^{\frac{\mathrm{1}}{{x}}} }{\mathrm{2}{i}\sqrt{{x}}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\:\frac{{e}^{\frac{\mathrm{1}}{{x}}} }{\mathrm{2}{i}\sqrt{{x}}}\:=\pi\:\frac{{e}^{\frac{\mathrm{1}}{{x}}} }{\:\sqrt{{x}}}\:\Rightarrow\:{f}^{'} \left({x}\right)\:=\frac{\pi}{\mathrm{2}}\:\frac{{e}^{\frac{\mathrm{1}}{{x}}} }{\:\sqrt{{x}}}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{e}^{\frac{\mathrm{1}}{{t}}} }{\:\sqrt{{t}}}{dt}\:\:+{c}\:\:\:{but}\:\:{c}={f}\left(\mathrm{0}\right)\:=\mathrm{0}\:\Rightarrow{f}\left({x}\right)\:=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{x}} \:\:\frac{{e}^{\frac{\mathrm{1}}{{t}}} }{\:\sqrt{{t}}}\:{dt} \\ $$$$=_{\sqrt{{t}}={u}} \:\:\:\:\frac{\pi}{\mathrm{2}}\:\:\int_{\mathrm{0}} ^{\sqrt{{x}}} \:\:\frac{{e}^{\frac{\mathrm{1}}{{u}^{\mathrm{2}} }} }{{u}}\:\left(\mathrm{2}{u}\right){du}\:\:=\:\pi\:\:\:\int_{\mathrm{0}} ^{\sqrt{{x}}} \:\:\:\:{e}^{\frac{\mathrm{1}}{{u}^{\mathrm{2}} }} \:{du}\:. \\ $$

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