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

$${\int }_0^{\mathrm{\infty }}\frac{e^{-x^2}}{x^2+1}dx=?$$

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

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

Commented by NECx last updated on 18/Apr/18

$$I^{\prime }mmosthappytoseeyouback$$

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

$$letputI={\int }_0^{\mathrm{\infty }}\frac{e^{-x^2}}{x^2+1}dxwehave$$$$2I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{-x^2}}{x^2+1}dx.letconsiderthecomplexfunction$$$$\phi \left(z\right)=\frac{e^{-z^2}}{1+z^2}.thepolesof\phi areiand-i\left(simples\right)$$$$Residustheoremgive$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,i)$$$$Res(\phi ,i)={lim}_{z\to i}(z-i)\phi \left(z\right)={lim}_{z\to i}(z-i)\frac{e^{-z^2}}{(z-i)(z+i)}$$$$=\frac{e^{-i^2}}{2i}=\frac{e}{2i}\Rightarrow 2I=2i\pi .\frac{e}{2i}=\pi e\Rightarrow$$$$★I=\frac{\pi e}{2}★$$

Commented by MJS last updated on 18/Apr/18

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

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

$$yourmathprogrameisnotcorrectsirtheresidus$$$$theoremhavefixedtheresult...$$

Commented by MJS last updated on 19/Apr/18

$$(\mathrm{\forall }x\in \mathbb{R}:{\mathrm{e}}^{-x^2}⩾\frac{{\mathrm{e}}^{-x^2}}{x^2+1})\wedge (\underset{-\mathrm{\infty }}{\stackrel{+\mathrm{\infty }}{\int }}{\mathrm{e}}^{-x^2}dx=\sqrt{\pi })\Rightarrow$$$$\Rightarrow \underset{-\mathrm{\infty }}{\stackrel{+\mathrm{\infty }}{\int }}\frac{{\mathrm{e}}^{-x^2}}{x^2+1}dx⩽\sqrt{\pi }\mathrm{but}\mathrm{e}\pi >\sqrt{\pi }$$

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

$$sirifyoufindsomethingwronginmymethod$$$$informme...$$

Commented by MJS last updated on 19/Apr/18

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

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

$$nevermindsirthekindofthisintegralismore$$$$difficulttotreat....$$

Commented by MJS last updated on 20/Apr/18

$$\mathrm{I}\mathrm{found}\mathrm{this}\mathrm{on}\mathrm{the}\mathrm{internet}:$$$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{{\mathrm{e}}^{-x^2}}{x^2+1}dx=\frac{\mathrm{e}\pi }{2}\times erfc\left(1\right)=\frac{\mathrm{e}\pi }{2}\times \frac{2}{\sqrt{\pi }}\underset{1}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-t^2}dt=$$$$=\mathrm{e}\sqrt{\pi }\times 0.139402792...=0.671646710...$$$$erfc\left(x\right)=\frac{2}{\sqrt{\pi }}\underset{x}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-t^2}dt$$$$\mathrm{so}\mathrm{we}\mathrm{would}\mathrm{need}\mathrm{to}\mathrm{find}$$$$r=\underset{1}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-t^2}dt=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-{(t+1)}^2}dt$$$$\mathrm{but}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{no}\mathrm{exact}\mathrm{solution}\mathrm{for}r$$

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

$$thankyousirforthisclarificationfor{\int }_1^{+\mathrm{\infty }}{e}^{-t^2}dxwecan$$$$giveaapproximatvalue...$$

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

$$r={\int }_1^0{e}^{-t^2}dt+{\int }_0^{+\mathrm{\infty }}{e}^{-x^2}dx=\frac{\sqrt{\pi }}{2}-{\int }_0^1{e}^{-t^2}dtbut$$$${\int }_0^1{e}^{-t^2}dt={\int }_0^1\left(\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{t}^{2n}}{n!}\right)dt$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n!}\frac{1}{2n+1}=1-\frac{1}{3}+\frac{1}{2.5}-\frac{1}{6.7}+\frac{1}{9.(4!)}-....$$$$and10termsforthisseriecangiveabettervalueforthisintegral...$$

Commented by MJS last updated on 20/Apr/18

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

Commented by Yozzzzy last updated on 21/Apr/18

$$Yes{it}^{\prime }sme.I^{\prime }vejustbeensobusywith$$$$universitythatIhavelittletimeduring$$$$theacademicyeartocontributetothis$$$$forum.I^{\prime }mgladithasstayedactivesince$$$$Istarteduni.I^{\prime }vegotananswerforthe$$$$integralwhichinvolvesintegratingunder$$$$theintegralsign.Onemayconsiderthe$$$$integral{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{-b(x^2+1)}}{x^2+1}dxwhereb⩾0$$$$andcreatedifferentialequationwithbas$$$$theindepedentvariable.$$$$InitiallyItriedusingcomplexanalysis$$$$butfoundmuchdifficultyinobtaininga$$$$suitablecontour.Nonetheless,Iwasableto$$$$getsomethingoutoftheattempts,theseintegrals:$$$$1){\int }_0^{\mathrm{\infty }}\frac{cos\left(x^2\right)-sin\left(x^2\right)}{1+x^4}dx=\frac{\pi }{4}{e}^{-1}\sqrt{2}$$$$2){\int }_0^{\mathrm{\infty }}\frac{x^2(cos\left(x^2\right)+sin\left(x^2\right))}{1+x^4}dx=\frac{\pi }{4}{e}^{-1}\sqrt{2}$$$$Hence{\int }_0^{\mathrm{\infty }}\frac{(x^2+1)sin\left(x^2\right)+(x^2-1)cos\left(x^2\right)}{1+x^4}dx=0.$$$$$$

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

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

Commented by Yozzzzy last updated on 24/Apr/18

$$Nonextyearismyfinalyear.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com