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 140977 by Mathspace last updated on 14/May/21

find ∫_0 ^∞ (e^(−t(1+x^2 )) /(1+x^2 ))dx with t≥0

find∫0∞e−t(1+x2)1+x2dxwitht⩾0

Answered by mathmax by abdo last updated on 14/May/21

let f(t)=∫_0 ^∞ (e^(−(1+x^2 )t) /(1+x^2 ))dx ⇒f^′ (t)=−∫_0 ^∞ e^(−(1+x^2 )t) dx =−e^(−t) ∫_0 ^∞ e^(−tx^2 ) dx =_(u=(√t)x) −e^(−t) ∫_0 ^∞ e^(−u^2 ) (du/( (√t))) =−(e^(−t) /( (√t))).((√π)/2) =−((√π)/2) (e^(−t) /( (√t))) ⇒f(t)=−((√π)/2)∫_0 ^t (e^(−u) /( (√u)))du +k f(0)=(π/2) =k ⇒f(t)=(π/2)−((√π)/2)∫_0 ^t (e^(−u) /( (√u)))du ((√u)=x) =(π/2)−((√π)/2)∫_0 ^(√t) (e^(−x^2 ) /x)(2x)dx =(π/2)−(√π)∫_0 ^(√t) e^(−x^2 ) dx ⇒ f(t)=(π/2)−(√π)∫_0 ^(√t) e^(−x^2 ) dx

letf(t)=∫0∞e−(1+x2)t1+x2dx⇒f′(t)=−∫0∞e−(1+x2)tdx=−e−t∫0∞e−tx2dx=u=tx−e−t∫0∞e−u2dut=−e−tt.π2=−π2e−tt⇒f(t)=−π2∫0te−uudu+kf(0)=π2=k⇒f(t)=π2−π2∫0te−uudu(u=x)=π2−π2∫0te−x2x(2x)dx=π2−π∫0te−x2dx⇒f(t)=π2−π∫0te−x2dx

Answered by qaz last updated on 14/May/21

∫_0 ^∞ (e^(−t(1+x^2 )) /(1+x^2 ))dx =−∫_0 ^t dt∫_0 ^∞ e^(−t(1+x^2 )) dx+(π/2) =−∫_0 ^t e^(−t) dt∫_0 ^∞ e^(−tx^2 ) dx+(π/2) =−∫_0 ^t e^(−t) ((Γ((1/2)))/(2t^(1/2) ))dt+(π/2) =−((√π)/2)∫_0 ^t t^(−(1/2)) e^(−t) dt+(π/2) =(π/2)[1−erf(t^2 )]

∫0∞e−t(1+x2)1+x2dx=−∫0tdt∫0∞e−t(1+x2)dx+π2=−∫0te−tdt∫0∞e−tx2dx+π2=−∫0te−tΓ(12)2t12dt+π2=−π2∫0tt−12e−tdt+π2=π2[1−erf(t2)]

Terms of Service

Privacy Policy

Contact: info@tinkutara.com