Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 33531 by Yozzzzy last updated on 18/Apr/18

∫_0 ^∞ (e^(−x^2 ) /(x^2 +1))dx=?

$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=? \\ $$

Commented by Rasheed.Sindhi last updated on 18/Apr/18

Ve....ry happy to see you back again,   if you are  old Yozzi!

$$\mathrm{Ve}....\mathrm{ry}\:\mathrm{happy}\:\mathrm{to}\:\mathrm{see}\:\mathrm{you}\:\mathrm{back}\:\mathrm{again},\: \\ $$$$\mathrm{if}\:\mathrm{you}\:\mathrm{are}\:\:\mathrm{old}\:\mathrm{Yozzi}! \\ $$

Commented by NECx last updated on 18/Apr/18

I′m most happy tosee you back

$${I}'{m}\:{most}\:{happy}\:{tosee}\:{you}\:{back} \\ $$

Commented by math khazana by abdo last updated on 18/Apr/18

let put I  = ∫_0 ^∞    (e^(−x^2 ) /(x^2  +1)) dx  we have   2I  = ∫_(−∞) ^(+∞)    (e^(−x^2 ) /(x^2  +1))dx .let consider the complex function  ϕ(z) = (e^(−z^2 ) /(1+z^2 ))  .the poles of ϕ are i and −i (simples)  Residus theorem give   ∫_(−∞) ^(+∞)  ϕ(z)dz =2i π Res(ϕ,i)   Res(ϕ,i) =lim_(z→i)  (z−i)ϕ(z)=lim_(z→i) (z−i) (e^(−z^2 ) /((z−i)(z+i)))  = (e^(−i^2 ) /(2i))  = (e/(2i)) ⇒ 2I  = 2iπ.(e/(2i)) = πe ⇒  ★  I = ((π e)/2) ★

$${let}\:{put}\:{I}\:\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} \:+\mathrm{1}}\:{dx}\:\:{we}\:{have}\: \\ $$$$\mathrm{2}{I}\:\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:.{let}\:{consider}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{−{z}^{\mathrm{2}} } }{\mathrm{1}+{z}^{\mathrm{2}} }\:\:.{the}\:{poles}\:{of}\:\varphi\:{are}\:{i}\:{and}\:−{i}\:\left({simples}\right) \\ $$$${Residus}\:{theorem}\:{give}\: \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\:\pi\:{Res}\left(\varphi,{i}\right)\: \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\left({z}−{i}\right)\varphi\left({z}\right)={lim}_{{z}\rightarrow{i}} \left({z}−{i}\right)\:\frac{{e}^{−{z}^{\mathrm{2}} } }{\left({z}−{i}\right)\left({z}+{i}\right)} \\ $$$$=\:\frac{{e}^{−{i}^{\mathrm{2}} } }{\mathrm{2}{i}}\:\:=\:\frac{{e}}{\mathrm{2}{i}}\:\Rightarrow\:\mathrm{2}{I}\:\:=\:\mathrm{2}{i}\pi.\frac{{e}}{\mathrm{2}{i}}\:=\:\pi{e}\:\Rightarrow \\ $$$$\bigstar\:\:{I}\:=\:\frac{\pi\:{e}}{\mathrm{2}}\:\bigstar \\ $$

Commented by MJS last updated on 18/Apr/18

but I get ∫_0 ^∞ (e^(−x^2 ) /(x^2 +1))dx≈.67165 by  computing with a math program

$$\mathrm{but}\:\mathrm{I}\:\mathrm{get}\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\approx.\mathrm{67165}\:\mathrm{by} \\ $$$$\mathrm{computing}\:\mathrm{with}\:\mathrm{a}\:\mathrm{math}\:\mathrm{program} \\ $$

Commented by math khazana by abdo last updated on 19/Apr/18

your math programe is not correct sir  the residus  theorem have fixed  the result ...

$${your}\:{math}\:{programe}\:{is}\:{not}\:{correct}\:{sir}\:\:{the}\:{residus} \\ $$$${theorem}\:{have}\:{fixed}\:\:{the}\:{result}\:... \\ $$

Commented by MJS last updated on 19/Apr/18

(∀x∈R: e^(−x^2 ) ≥(e^(−x^2 ) /(x^2 +1))) ∧ (∫_(−∞) ^(+∞) e^(−x^2 ) dx=(√π)) ⇒  ⇒ ∫_(−∞) ^(+∞) (e^(−x^2 ) /(x^2 +1))dx≤(√π) but eπ>(√π)

$$\left(\forall{x}\in\mathbb{R}:\:\mathrm{e}^{−{x}^{\mathrm{2}} } \geqslant\frac{\mathrm{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} +\mathrm{1}}\right)\:\wedge\:\left(\underset{−\infty} {\overset{+\infty} {\int}}\mathrm{e}^{−{x}^{\mathrm{2}} } {dx}=\sqrt{\pi}\right)\:\Rightarrow \\ $$$$\Rightarrow\:\underset{−\infty} {\overset{+\infty} {\int}}\frac{\mathrm{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\leqslant\sqrt{\pi}\:\mathrm{but}\:\mathrm{e}\pi>\sqrt{\pi} \\ $$

Commented by math khazana by abdo last updated on 19/Apr/18

sir if you find something wrong in my method  inform me ...

$${sir}\:{if}\:{you}\:{find}\:{something}\:{wrong}\:{in}\:{my}\:{method} \\ $$$${inform}\:{me}\:... \\ $$

Commented by MJS last updated on 19/Apr/18

the method seems ok, I′m not sure what′s  going on here... I will keep on studying this  problem and post the answer if I can find it

$$\mathrm{the}\:\mathrm{method}\:\mathrm{seems}\:\mathrm{ok},\:\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{what}'\mathrm{s} \\ $$$$\mathrm{going}\:\mathrm{on}\:\mathrm{here}...\:\mathrm{I}\:\mathrm{will}\:\mathrm{keep}\:\mathrm{on}\:\mathrm{studying}\:\mathrm{this} \\ $$$$\mathrm{problem}\:\mathrm{and}\:\mathrm{post}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{if}\:\mathrm{I}\:\mathrm{can}\:\mathrm{find}\:\mathrm{it} \\ $$

Commented by math khazana by abdo last updated on 19/Apr/18

nevermind sir  the kind of this integral is more  difficult to treat....

$${nevermind}\:{sir}\:\:{the}\:{kind}\:{of}\:{this}\:{integral}\:{is}\:{more} \\ $$$${difficult}\:{to}\:{treat}.... \\ $$

Commented by MJS last updated on 20/Apr/18

I found this on the internet:  ∫_0 ^∞ (e^(−x^2 ) /(x^2 +1))dx=((eπ)/2)×erfc(1)=((eπ)/2)×(2/(√π))∫_1 ^∞ e^(−t^2 ) dt=  =e(√π)×0.139402792...=0.671646710...  erfc(x)=(2/(√π))∫_x ^∞ e^(−t^2 ) dt  so we would need to find  r=∫_1 ^∞ e^(−t^2 ) dt=∫_0 ^∞ e^(−(t+1)^2 ) dt  but there′s no exact solution for r

$$\mathrm{I}\:\mathrm{found}\:\mathrm{this}\:\mathrm{on}\:\mathrm{the}\:\mathrm{internet}: \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\frac{\mathrm{e}\pi}{\mathrm{2}}×{erfc}\left(\mathrm{1}\right)=\frac{\mathrm{e}\pi}{\mathrm{2}}×\frac{\mathrm{2}}{\sqrt{\pi}}\underset{\mathrm{1}} {\overset{\infty} {\int}}\mathrm{e}^{−{t}^{\mathrm{2}} } {dt}= \\ $$$$=\mathrm{e}\sqrt{\pi}×\mathrm{0}.\mathrm{139402792}...=\mathrm{0}.\mathrm{671646710}... \\ $$$${erfc}\left({x}\right)=\frac{\mathrm{2}}{\sqrt{\pi}}\underset{{x}} {\overset{\infty} {\int}}\mathrm{e}^{−{t}^{\mathrm{2}} } {dt} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{would}\:\mathrm{need}\:\mathrm{to}\:\mathrm{find} \\ $$$${r}=\underset{\mathrm{1}} {\overset{\infty} {\int}}\mathrm{e}^{−{t}^{\mathrm{2}} } {dt}=\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−\left({t}+\mathrm{1}\right)^{\mathrm{2}} } {dt} \\ $$$$\mathrm{but}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{exact}\:\mathrm{solution}\:\mathrm{for}\:{r} \\ $$

Commented by abdo imad last updated on 20/Apr/18

thank you sir for this clarification for ∫_1 ^(+∞)  e^(−t^2 ) dx we can  give a approximat value...

$${thank}\:{you}\:{sir}\:{for}\:{this}\:{clarification}\:{for}\:\int_{\mathrm{1}} ^{+\infty} \:{e}^{−{t}^{\mathrm{2}} } {dx}\:{we}\:{can} \\ $$$${give}\:{a}\:{approximat}\:{value}... \\ $$

Commented by abdo imad last updated on 20/Apr/18

r = ∫_1 ^0  e^(−t^2 ) dt  +∫_0 ^(+∞)  e^(−x^2 ) dx=((√π)/2)  −∫_0 ^1  e^(−t^2 ) dt but  ∫_0 ^1   e^(−t^2 ) dt =∫_0 ^1  (Σ_(n=0) ^∞   (((−1)^n t^(2n) )/(n!)))dt  =Σ_(n=0) ^∞  (((−1)^n )/(n!))  (1/(2n+1)) = 1  −(1/3)  +(1/(2.5)) −(1/(6.7)) + (1/(9.(4!))) −....  and 10 termsfor this serie can give a better value for this integral...

$${r}\:=\:\int_{\mathrm{1}} ^{\mathrm{0}} \:{e}^{−{t}^{\mathrm{2}} } {dt}\:\:+\int_{\mathrm{0}} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\sqrt{\pi}}{\mathrm{2}}\:\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{t}^{\mathrm{2}} } {dt}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{e}^{−{t}^{\mathrm{2}} } {dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} {t}^{\mathrm{2}{n}} }{{n}!}\right){dt} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:=\:\mathrm{1}\:\:−\frac{\mathrm{1}}{\mathrm{3}}\:\:+\frac{\mathrm{1}}{\mathrm{2}.\mathrm{5}}\:−\frac{\mathrm{1}}{\mathrm{6}.\mathrm{7}}\:+\:\frac{\mathrm{1}}{\mathrm{9}.\left(\mathrm{4}!\right)}\:−.... \\ $$$${and}\:\mathrm{10}\:{termsfor}\:{this}\:{serie}\:{can}\:{give}\:{a}\:{better}\:{value}\:{for}\:{this}\:{integral}... \\ $$

Commented by MJS last updated on 20/Apr/18

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Commented by Yozzzzy last updated on 21/Apr/18

Yes it′s me. I′ve just been so busy with   university that I have little time during   the academic year to contribute to this   forum. I′m glad it has stayed active since  I started uni. I′ve got an answer for the   integral which involves integrating under  the integral sign. One may consider the   integral ∫_(−∞) ^∞ (e^(−b(x^2 +1)) /(x^2 +1))dx where b≥0  and create differential equation with b as  the indepedent variable.  Initially I tried using complex analysis  but found much difficulty in obtaining a   suitable contour. Nonetheless, I was able to   get something out of the attempts, these integrals:  1)∫_0 ^∞ ((cos(x^2 )−sin(x^2 ))/(1+x^4 ))dx=(π/4)e^(−1) (√2)  2)∫_0 ^∞ ((x^2 (cos(x^2 )+sin(x^2 )))/(1+x^4 ))dx=(π/4)e^(−1) (√2)  Hence ∫_0 ^∞ (((x^2 +1)sin(x^2 )+(x^2 −1)cos(x^2 ))/(1+x^4 ))dx=0.

$${Yes}\:{it}'{s}\:{me}.\:{I}'{ve}\:{just}\:{been}\:{so}\:{busy}\:{with}\: \\ $$$${university}\:{that}\:{I}\:{have}\:{little}\:{time}\:{during}\: \\ $$$${the}\:{academic}\:{year}\:{to}\:{contribute}\:{to}\:{this}\: \\ $$$${forum}.\:{I}'{m}\:{glad}\:{it}\:{has}\:{stayed}\:{active}\:{since} \\ $$$${I}\:{started}\:{uni}.\:{I}'{ve}\:{got}\:{an}\:{answer}\:{for}\:{the}\: \\ $$$${integral}\:{which}\:{involves}\:{integrating}\:{under} \\ $$$${the}\:{integral}\:{sign}.\:{One}\:{may}\:{consider}\:{the}\: \\ $$$${integral}\:\int_{−\infty} ^{\infty} \frac{{e}^{−{b}\left({x}^{\mathrm{2}} +\mathrm{1}\right)} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\:{where}\:{b}\geqslant\mathrm{0} \\ $$$${and}\:{create}\:{differential}\:{equation}\:{with}\:{b}\:{as} \\ $$$${the}\:{indepedent}\:{variable}. \\ $$$${Initially}\:{I}\:{tried}\:{using}\:{complex}\:{analysis} \\ $$$${but}\:{found}\:{much}\:{difficulty}\:{in}\:{obtaining}\:{a}\: \\ $$$${suitable}\:{contour}.\:{Nonetheless},\:{I}\:{was}\:{able}\:{to}\: \\ $$$${get}\:{something}\:{out}\:{of}\:{the}\:{attempts},\:{these}\:{integrals}: \\ $$$$\left.\mathrm{1}\right)\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left({x}^{\mathrm{2}} \right)−{sin}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{4}} }{dx}=\frac{\pi}{\mathrm{4}}{e}^{−\mathrm{1}} \sqrt{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right)\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} \left({cos}\left({x}^{\mathrm{2}} \right)+{sin}\left({x}^{\mathrm{2}} \right)\right)}{\mathrm{1}+{x}^{\mathrm{4}} }{dx}=\frac{\pi}{\mathrm{4}}{e}^{−\mathrm{1}} \sqrt{\mathrm{2}} \\ $$$${Hence}\:\int_{\mathrm{0}} ^{\infty} \frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right){sin}\left({x}^{\mathrm{2}} \right)+\left({x}^{\mathrm{2}} −\mathrm{1}\right){cos}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{4}} }{dx}=\mathrm{0}. \\ $$$$ \\ $$

Commented by Rasheed.Sindhi last updated on 23/Apr/18

Did you complete your graduation?

$$\mathrm{Did}\:\mathrm{you}\:\mathrm{complete}\:\mathrm{your}\:\mathrm{graduation}? \\ $$

Commented by Yozzzzy last updated on 24/Apr/18

No next year is my final year.

$${No}\:{next}\:{year}\:{is}\:{my}\:{final}\:{year}.\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com