Question Number 151983 by puissant last updated on 24/Aug/21
$$\int_{−\infty} ^{+\infty} \left(\frac{\mathrm{1}−{ix}}{\mathrm{1}+{ix}}\right)^{{n}} \left(\frac{\mathrm{1}+{ix}}{\mathrm{1}−{ix}}\right)^{{m}} \frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$
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)
$$\mathrm{I}\:=\:\int_{−\infty} ^{+\infty} \left(\frac{\mathrm{1}−{ix}}{\mathrm{1}+{ix}}\right)^{{n}} \left(\frac{\mathrm{1}+{ix}}{\mathrm{1}−{ix}}\right)^{{m}} \frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$$$ \\ $$$$\bullet\:{m}\:=\:{n},\:\mathrm{I}\:=\:\pi\:\left(\mathrm{trivial}\right) \\ $$$$ \\ $$$$\bullet\:{m}\:\neq\:{n}\:: \\ $$$$\mathrm{I}\:=\:\int_{−\infty} ^{+\infty} \frac{{e}^{{in}\mathrm{arctan}\left(−{x}\right)} }{{e}^{{in}\mathrm{arctan}\left(+{x}\right)} }.\frac{{e}^{{im}\mathrm{arctan}\left(+{x}\right)} }{{e}^{{im}\mathrm{arctan}\left(−{x}\right)} }.\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$$$\mathrm{I}\:=\:\int_{−\infty} ^{+\infty} \frac{{e}^{\mathrm{2}{i}\left({m}−{n}\right)\mathrm{arctan}{x}} }{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$$$\mathrm{I}\:=\:\left[\frac{{e}^{\mathrm{2}{i}\left({m}−{n}\right)\mathrm{arctan}{x}} }{\mathrm{2}{i}\left({m}−{n}\right)}\right]_{−\infty} ^{+\infty} \\ $$$$\mathrm{I}\:=\:\frac{{e}^{{i}\pi\left({m}−{n}\right)} −{e}^{−{i}\pi\left({m}−{n}\right)} }{\mathrm{2}{i}\left({m}−{n}\right)} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{sin}\left(\pi\left({m}−{n}\right)\right)}{{m}−{n}}\:=\:\frac{\mathrm{0}}{{m}−{n}}\:=\:\mathrm{0} \\ $$$$ \\ $$$$\mathrm{My}\:\mathrm{result}\:\mathrm{is}\:\mathrm{very}\:\mathrm{strange}. \\ $$$$\mathrm{You}\:\mathrm{should}\:\mathrm{verify}\:\mathrm{the}\:\mathrm{calculous}. \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{it}'\mathrm{s}\:\mathrm{the}\:\mathrm{good}\:\mathrm{way}. \\ $$
Commented by puissant last updated on 24/Aug/21
$${merci}\:{beaucoup}\:{Mr} \\ $$
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)
$$\int_{−\infty} ^{+\infty} \left(\frac{\mathrm{1}−{ix}}{\mathrm{1}+{ix}}\right)^{{n}} \left(\frac{\mathrm{1}+{ix}}{\mathrm{1}−{ix}}\right)^{{m}} \frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\int_{−\infty} ^{+\infty} \left(\frac{\mathrm{1}+{ix}}{\mathrm{1}−{ix}}\right)^{{m}−{n}} \frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\int_{−\frac{\pi}{\mathrm{2}}} ^{+\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{1}+{itant}}{\mathrm{1}−{itant}}\right)^{{m}−{n}} {dt}\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{+\frac{\pi}{\mathrm{2}}} \left(\frac{{cost}+{isint}}{{cost}−{isint}}\right)^{{m}−{n}} {dt} \\ $$$$=\int_{−\frac{\pi}{\mathrm{2}}} ^{+\frac{\pi}{\mathrm{2}}} {e}^{\mathrm{2}{it}\left({m}−{n}\right)} {dt}\:=\:\left[\frac{{e}^{\mathrm{2}{it}\left({m}−{n}\right)} }{\mathrm{2}{i}\left({m}−{n}\right)}\right]_{−\frac{\pi}{\mathrm{2}}} ^{+\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{{e}^{{i}\pi\left({m}−{n}\right)} −{e}^{−{i}\pi\left({m}−{n}\right)} }{\mathrm{2}{i}\left({m}−{n}\right)} \\ $$$$=\frac{{sin}\left(\pi\left({m}−{n}\right)\right)}{\left({m}−{n}\right)}\:=\:\mathrm{0}.. \\ $$$${En}\:{fait}\:{voici}\:{ce}\:{que}\:{j}'{ai}\:{fait}\:{Mr}\:{mais} \\ $$$${je}\:{voulais}\:{confirmer}\:{mon}\:{resultat}.. \\ $$$${Merci}\:{pour}\:{votre}\:{temps}.. \\ $$$${Cordialement}.. \\ $$
Commented by Olaf_Thorendsen last updated on 24/Aug/21
$$\mathrm{Excellent}\:! \\ $$$$\mathrm{Cette}\:\mathrm{integrale}\:\mathrm{fait}\:\mathrm{peur}\:\mathrm{au}\:\mathrm{depart}\:\mathrm{mais} \\ $$$$\mathrm{elle}\:\mathrm{se}\:\mathrm{calcule}\:\mathrm{en}\:\mathrm{4}\:\mathrm{ou}\:\mathrm{5}\:\mathrm{lignes}. \\ $$$$\mathrm{C}'\mathrm{est}\:\mathrm{souvent}\:\mathrm{trompeur}.\: \\ $$