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let-x-gt-0-and-F-x-0-arctan-xt-2-1-t-2-dt-1-find-a-simple-form-of-F-x-2-find-the-value-of-0-arctan-2t-2-1-t-2-dt-3-find-the-value-of-0-arctan-3t-2-1-t-2

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Question Number 38211 by prof Abdo imad last updated on 22/Jun/18
let x>0 and F(x)= ∫_0 ^(+∞) ((arctan(xt^2 ))/(1+t^2 ))dt 1) find a simple form of F(x) 2)find the value of ∫_0 ^∞ ((arctan(2t^2 ))/(1+t^2 ))dt 3)find the value of ∫_0 ^∞ ((arctan(3t^2 ))/(1+t^2 ))dt.
$${let}\:{x}>\mathrm{0}\:{and}\:{F}\left({x}\right)=\:\int_{\mathrm{0}} ^{+\infty} \:\frac{{arctan}\left({xt}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{F}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left(\mathrm{2}{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \frac{{arctan}\left(\mathrm{3}{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}. \\ $$
Commented by math khazana by abdo last updated on 26/Jun/18
1) we have F(x)=(1/2) ∫_(−∞) ^(+∞) ((arctan(xt^2 ))/(1+t^2 ))dt but ∫_(−∞) ^(+∞) ((arctan(xt^2 ))/(1+t^2 ))dt= π{ arctan((√(2x)) +1) +arctan((√(2x)) −1) −arctanx} ⇒ F(x)=(π/2){ arctan((√(2x)) +1)+arctan((√(2x)) −1)−arctan(x)} 2) ∫_0 ^∞ ((arctan(2t^2 ))/(1+t^2 ))dt =F(2) =(π/2){arctan(3) +arctan(1) −arctan(2)} =(π/2){ arctan(3)−arctan(2)}+(π^2 /8) 3) ∫_0 ^∞ ((arctan(3t^2 ))/(1+t^2 ))dt =F(3) =(π/2){ arctan((√6) +1)+arctan((√6)−1)arctan(3)}.
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left({xt}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:{but} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\frac{{arctan}\left({xt}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}= \\ $$$$\pi\left\{\:{arctan}\left(\sqrt{\mathrm{2}{x}}\:+\mathrm{1}\right)\:+{arctan}\left(\sqrt{\mathrm{2}{x}}\:−\mathrm{1}\right)\:−{arctanx}\right\} \\ $$$$\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\pi}{\mathrm{2}}\left\{\:{arctan}\left(\sqrt{\mathrm{2}{x}}\:+\mathrm{1}\right)+{arctan}\left(\sqrt{\mathrm{2}{x}}\:−\mathrm{1}\right)−{arctan}\left({x}\right)\right\} \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left(\mathrm{2}{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:={F}\left(\mathrm{2}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}\left\{{arctan}\left(\mathrm{3}\right)\:+{arctan}\left(\mathrm{1}\right)\:−{arctan}\left(\mathrm{2}\right)\right\} \\ $$$$=\frac{\pi}{\mathrm{2}}\left\{\:{arctan}\left(\mathrm{3}\right)−{arctan}\left(\mathrm{2}\right)\right\}+\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\left.\mathrm{3}\right)\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{3}{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:={F}\left(\mathrm{3}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}\left\{\:{arctan}\left(\sqrt{\mathrm{6}}\:+\mathrm{1}\right)+{arctan}\left(\sqrt{\mathrm{6}}−\mathrm{1}\right){arctan}\left(\mathrm{3}\right)\right\}. \\ $$$$ \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 26/Jun/18
F(3)=(π/2){ arctan((√6) +1)+arctan((√6) −1)−arctan(3)}
$${F}\left(\mathrm{3}\right)=\frac{\pi}{\mathrm{2}}\left\{\:{arctan}\left(\sqrt{\mathrm{6}}\:+\mathrm{1}\right)+{arctan}\left(\sqrt{\mathrm{6}}\:−\mathrm{1}\right)−{arctan}\left(\mathrm{3}\right)\right\} \\ $$

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