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1-ix-1-ix-n-1-ix-1-ix-m-1-1-x-2-dx-




Question Number 151983 by puissant last updated on 24/Aug/21
∫_(−∞) ^(+∞) (((1−ix)/(1+ix)))^n (((1+ix)/(1−ix)))^m (1/(1+x^2 ))dx
+(1ix1+ix)n(1+ix1ix)m11+x2dx
Answered by Olaf_Thorendsen last updated on 24/Aug/21
I = ∫_(−∞) ^(+∞) (((1−ix)/(1+ix)))^n (((1+ix)/(1−ix)))^m (1/(1+x^2 )) dx    • m = n, I = π (trivial)    • m ≠ n :  I = ∫_(−∞) ^(+∞) (e^(inarctan(−x)) /e^(inarctan(+x)) ).(e^(imarctan(+x)) /e^(imarctan(−x)) ).(1/(1+x^2 )) dx  I = ∫_(−∞) ^(+∞) (e^(2i(m−n)arctanx) /(1+x^2 )) dx  I = [(e^(2i(m−n)arctanx) /(2i(m−n)))]_(−∞) ^(+∞)   I = ((e^(iπ(m−n)) −e^(−iπ(m−n)) )/(2i(m−n)))  I = ((sin(π(m−n)))/(m−n)) = (0/(m−n)) = 0    My result is very strange.  You should verify the calculous.  I′m not sure it′s the good way.
I=+(1ix1+ix)n(1+ix1ix)m11+x2dxm=n,I=π(trivial)mn:I=+einarctan(x)einarctan(+x).eimarctan(+x)eimarctan(x).11+x2dxI=+e2i(mn)arctanx1+x2dxI=[e2i(mn)arctanx2i(mn)]+I=eiπ(mn)eiπ(mn)2i(mn)I=sin(π(mn))mn=0mn=0Myresultisverystrange.Youshouldverifythecalculous.Imnotsureitsthegoodway.
Commented by puissant last updated on 24/Aug/21
merci beaucoup Mr
mercibeaucoupMr
Commented by puissant last updated on 24/Aug/21
∫_(−∞) ^(+∞) (((1−ix)/(1+ix)))^n (((1+ix)/(1−ix)))^m (1/(1+x^2 ))dx  =∫_(−∞) ^(+∞) (((1+ix)/(1−ix)))^(m−n) (1/(1+x^2 ))dx  =∫_(−(π/2)) ^(+(π/2)) (((1+itant)/(1−itant)))^(m−n) dt =∫_(−(π/2)) ^(+(π/2)) (((cost+isint)/(cost−isint)))^(m−n) dt  =∫_(−(π/2)) ^(+(π/2)) e^(2it(m−n)) dt = [(e^(2it(m−n)) /(2i(m−n)))]_(−(π/2)) ^(+(π/2))   =((e^(iπ(m−n)) −e^(−iπ(m−n)) )/(2i(m−n)))  =((sin(π(m−n)))/((m−n))) = 0..  En fait voici ce que j′ai fait Mr mais  je voulais confirmer mon resultat..  Merci pour votre temps..  Cordialement..
+(1ix1+ix)n(1+ix1ix)m11+x2dx=+(1+ix1ix)mn11+x2dx=π2+π2(1+itant1itant)mndt=π2+π2(cost+isintcostisint)mndt=π2+π2e2it(mn)dt=[e2it(mn)2i(mn)]π2+π2=eiπ(mn)eiπ(mn)2i(mn)=sin(π(mn))(mn)=0..EnfaitvoicicequejaifaitMrmaisjevoulaisconfirmermonresultat..Mercipourvotretemps..Cordialement..
Commented by Olaf_Thorendsen last updated on 24/Aug/21
Excellent !  Cette integrale fait peur au depart mais  elle se calcule en 4 ou 5 lignes.  C′est souvent trompeur.
Excellent!Cetteintegralefaitpeuraudepartmaisellesecalculeen4ou5lignes.Cestsouventtrompeur.

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