Question Number 151983 by puissant last updated on 24/Aug/21

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.](https://www.tinkutara.com/question/Q151991.png)
Commented by puissant last updated on 24/Aug/21

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..](https://www.tinkutara.com/question/Q151994.png)
Commented by Olaf_Thorendsen last updated on 24/Aug/21
