Question Number 68038 by mathmax by abdo last updated on 03/Sep/19 | ||
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx} \\ $$ | ||
Commented by mathmax by abdo last updated on 07/Sep/19 | ||
$${let}\:{f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{arctan}\left({x}^{\mathrm{2}} −{t}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:{dx}\Rightarrow{f}\left(\mathrm{1}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx} \\ $$$${f}^{'} \left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)\left(\mathrm{1}+\left({x}^{\mathrm{2}} −{t}\right)^{\mathrm{2}} \right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)\left({x}^{\mathrm{4}} −\mathrm{2}{tx}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow\mathrm{2}{f}^{'} \left({t}\right)\:=\int_{−\infty} ^{+\infty\:} \:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)\left({x}^{\mathrm{4}} −\mathrm{2}{tx}^{\mathrm{2}\:} +{t}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$${let}\:{W}\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} \:+\mathrm{4}\right)\left({z}^{\mathrm{4}} −\mathrm{2}{tz}^{\mathrm{2}} \:+{t}^{\mathrm{2}\:} +\mathrm{1}\right)}\:\:{poles}\:{of}\:{W}? \\ $$$${z}^{\mathrm{2}} \:+\mathrm{4}\:=\mathrm{0}\:\Rightarrow{z}\:=\overset{−} {+}\mathrm{2}{i} \\ $$$${z}^{\mathrm{4}} −\mathrm{2}{tz}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \:+\mathrm{1}\:=\mathrm{0}\:\Rightarrow{u}^{\mathrm{2}} −\mathrm{2}{tu}\:+{t}^{\mathrm{2}\:} +\mathrm{1}\:=\mathrm{0}\:{with}\:{u}={z}^{\mathrm{2}} \\ $$$$\Delta^{'} ={t}^{\mathrm{2}} −\left({t}^{\mathrm{2}} +\mathrm{1}\right)\:=−\mathrm{1}\:\Rightarrow{u}_{\mathrm{1}} ={t}+{i}\:\:{and}\:{u}_{\mathrm{2}} ={t}−{i}\:\Rightarrow \\ $$$${u}_{\mathrm{1}} =\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:{e}^{{iarctan}\left(\frac{\mathrm{1}}{{t}}\right)} \:\:{and}\:{u}_{\mathrm{2}} =\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:{e}^{−{iarctan}\left(\frac{\mathrm{1}}{{t}}\right)} \\ $$$${z}^{\mathrm{2}} ={u}_{\mathrm{1}} \Rightarrow{z}\:=\overset{−} {+}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{t}}\right)} \\ $$$$\mathrm{z}^{\mathrm{2}} =\mathrm{u}_{\mathrm{2}} \:\Rightarrow{z}\:=\overset{−} {+}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{t}\:}\right)} \:\Rightarrow \\ $$$${W}\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−\mathrm{2}{i}\right)\left({z}+\mathrm{2}{i}\right)\left({z}−\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} {e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{t}}\right)} \left({z}+\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} {e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{t}}\right)} \right)\left({z}−\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} {e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{t}}\right)} \right)\left({z}+\left(\mathrm{1}+{t}^{\mathrm{2}} \right){e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{t}}\right)} \right)\right.} \\ $$$${residustheorem}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left({W},\mathrm{2}{i}\right)\:+{Res}\left({W},\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{t}}\right)} \right.\right. \\ $$$$+\mathrm{R}{es}\left({W},−\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}}{\mathrm{2}}{afctan}\left(\frac{\mathrm{1}}{{t}}\right)} \right\}...{be}\:{continued}... \\ $$ | ||