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Question Number 73179 by mathmax by abdo last updated on 07/Nov/19
caoculate ∫_0 ^∞ ((arctan(x^2 −1))/(2x^2 +1))dx
$${caoculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$
Commented by mathmax by abdo last updated on 07/Nov/19
residus method but need more proof let A =∫_0 ^∞ ((arctan(x^2 −1))/(2x^2 +1))dx ⇒2A =∫_(−∞) ^(+∞) ((arctan(x^2 −1))/(2x^2 +1))dx let W(z)=((arctan(z^2 −1))/(2z^2 +1)) ⇒ W(z)=((arctan(z^2 −1))/(2(z^2 +(1/2)))) =((arctan(z^2 −1))/(2(z−(i/( (√2))))(z+(i/( (√2) ))))) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ Res(W,(i/( (√2)))) and Res(W,(i/( (√2))))=lim_(z→(i/( (√2)))) (z−(i/( (√2))))W(z) =((arctan(((i/( (√2))))^2 −1))/(2(((2i)/( (√2)))))) =(√2)((arctan(−(1/2)−1))/(4i)) =−((√2)/(4i)) arctan((3/2)) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ×(−((√2)/(4i)))arctan((3/2))=−((π(√2))/2) arctan((3/2))=2A ⇒ A =−((π(√2))/4) arctan((3/2)) and if the graph of the function x→((arctan(x^2 −1))/(2x^2 +1)) is on(x^′ ox) we take the absolute value...
$${residus}\:{method}\:{but}\:{need}\:{more}\:{proof}\:{let}\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{arctan}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$$$\Rightarrow\mathrm{2}{A}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{arctan}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:{let}\:{W}\left({z}\right)=\frac{{arctan}\left({z}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}{z}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$${W}\left({z}\right)=\frac{{arctan}\left({z}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}\left({z}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}\right)}\:=\frac{{arctan}\left({z}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}\left({z}−\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\left({z}+\frac{{i}}{\:\sqrt{\mathrm{2}}\:}\right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\:{and}\: \\ $$$${Res}\left({W},\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)={lim}_{{z}\rightarrow\frac{{i}}{\:\sqrt{\mathrm{2}}}} \:\left({z}−\frac{{i}}{\:\sqrt{\mathrm{2}}}\right){W}\left({z}\right)\:=\frac{{arctan}\left(\left(\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}\left(\frac{\mathrm{2}{i}}{\:\sqrt{\mathrm{2}}}\right)} \\ $$$$=\sqrt{\mathrm{2}}\frac{{arctan}\left(−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{4}{i}}\:=−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}{i}}\:{arctan}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi×\left(−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}{i}}\right){arctan}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=−\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:{arctan}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{2}{A} \\ $$$$\Rightarrow\:{A}\:=−\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\:{arctan}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:{and}\:{if}\:{the}\:{graph}\:{of}\:{the}\:{function} \\ $$$${x}\rightarrow\frac{{arctan}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}}\:{is}\:{on}\left({x}^{'} {ox}\right)\:{we}\:{take}\:{the}\:{absolute}\:{value}… \\ $$
Answered by mind is power last updated on 07/Nov/19
∫_0 ^(+∞) ((arctan(x^2 −1))/(2x^2 +1))dx=a ∫(dx/(2x^2 +1))=(1/( (√2)))arctan(x(√2)) a=[((arctan(x(√2)))/( (√2))).arctan(x^2 −1)]_0 ^(+∞) −(1/( (√2)))∫_0 ^(+∞) ((arctan(x(√2)).2xdx)/((x^2 −1)^2 +1)) a=(π^2 /(4(√2)))−(√2)∫_0 ^(+∞) ((arctan(x(√2))xdx)/((x^2 −1)^2 +1))=(π^2 /(4(√2)))−(1/( (√2)))∫_(−∞) ^(+∞) ((arctan(x(√2))xdx)/((x^2 −1)^2 +1)) on pose f(t)=∫_(−∞) ^(+∞) ((xarctan(tx))/((x^2 −1)^2 +1))dx t≥0 f′(t)=∫_(−∞) ^(+∞) ((x^2 dx)/((1+x^2 t^2 )(x^2 −1−i)(x^2 −1+i))) pols are x_0 =+(i/t),x^2 =1+i⇒x_2 =(2^(1/4) )e^((iπ)/8) x=−1+i⇒x_1 =(2^(1/4) )e^((i3π)/8) f′(t)=2iπ res(f,x_0 ,x_1 x_2 ) 2iπres(f,(i/t))=2iπ.(((−1)/t^2 )/((2i)t((((1+t^2 )/t^2 ))^2 +1)))=((−πt)/((2t^4 +2t^2 +1))) too be continued
$$\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{arctan}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx}=\mathrm{a} \\ $$$$\int\frac{\mathrm{dx}}{\mathrm{2x}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{arctan}\left(\mathrm{x}\sqrt{\mathrm{2}}\right) \\ $$$${a}=\left[\frac{{arctan}\left({x}\sqrt{\mathrm{2}}\right)}{\:\sqrt{\mathrm{2}}}.{arctan}\left({x}^{\mathrm{2}} −\mathrm{1}\right)\right]_{\mathrm{0}} ^{+\infty} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{arctan}\left(\mathrm{x}\sqrt{\mathrm{2}}\right).\mathrm{2xdx}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{a}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}\sqrt{\mathrm{2}}}−\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{arctan}\left(\mathrm{x}\sqrt{\mathrm{2}}\right)\mathrm{xdx}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{−\infty} ^{+\infty} \frac{\mathrm{arctan}\left(\mathrm{x}\sqrt{\mathrm{2}}\right)\mathrm{xdx}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{on}\:\mathrm{pose}\:\mathrm{f}\left(\mathrm{t}\right)=\int_{−\infty} ^{+\infty} \frac{\mathrm{xarctan}\left(\mathrm{tx}\right)}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\mathrm{dx} \\ $$$$\mathrm{t}\geqslant\mathrm{0} \\ $$$$\mathrm{f}'\left(\mathrm{t}\right)=\int_{−\infty} ^{+\infty} \frac{\mathrm{x}^{\mathrm{2}} \mathrm{dx}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}−\mathrm{i}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}+\mathrm{i}\right)} \\ $$$$\mathrm{pols}\:\mathrm{are}\:\mathrm{x}_{\mathrm{0}} =+\frac{\mathrm{i}}{\mathrm{t}},\mathrm{x}^{\mathrm{2}} =\mathrm{1}+\mathrm{i}\Rightarrow\mathrm{x}_{\mathrm{2}} =\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \right)\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{8}}} \\ $$$$\mathrm{x}=−\mathrm{1}+\mathrm{i}\Rightarrow\mathrm{x}_{\mathrm{1}} =\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \right)\mathrm{e}^{\frac{\mathrm{i3}\pi}{\mathrm{8}}} \\ $$$$\mathrm{f}'\left(\mathrm{t}\right)=\mathrm{2i}\pi\:\mathrm{res}\left(\mathrm{f},\mathrm{x}_{\mathrm{0}} ,\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} \right) \\ $$$$\mathrm{2i}\pi\mathrm{res}\left(\mathrm{f},\frac{\mathrm{i}}{\mathrm{t}}\right)=\mathrm{2i}\pi.\frac{\frac{−\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\left(\mathrm{2i}\right)\mathrm{t}\left(\left(\frac{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }{\mathrm{t}^{\mathrm{2}} }\right)^{\mathrm{2}} +\mathrm{1}\right)}=\frac{−\pi\mathrm{t}}{\left(\mathrm{2t}^{\mathrm{4}} +\mathrm{2t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\mathrm{too}\:\mathrm{be}\:\mathrm{continued} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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