# let-f-x-arctan-ax-1-with-a-real-1-calculate-f-n-x-and-f-n-0-2-developp-f-at-integr-serie-3-calculate-f-x-x-2-4-dx-

Question Number 68243 by mathmax by abdo last updated on 07/Sep/19
$${let}\:{f}\left({x}\right)\:={arctan}\left({ax}\:+\mathrm{1}\right)\:\:{with}\:{a}\:{real} \\$$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\$$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie} \\$$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{−\infty} ^{+\infty} \:\frac{{f}\left({x}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx} \\$$
Commented by mathmax by abdo last updated on 08/Sep/19
$$\left.\mathrm{1}\right)\:{f}\left({x}\right)={arctan}\left({ax}+\mathrm{1}\right)\:\Rightarrow{f}^{'} \left({x}\right)\:=\frac{{a}}{\mathrm{1}+\left({ax}+\mathrm{1}\right)^{\mathrm{2}} } \\$$$$=\frac{{a}}{\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \:+\mathrm{2}{ax}\:+\mathrm{1}}\:=\frac{{a}}{{a}^{\mathrm{2}} {x}^{\mathrm{2}} \:+\mathrm{2}{ax}\:+\mathrm{2}}\:\Rightarrow \\$$$${f}^{\left({n}\right)} \left({x}\right)\:={a}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} {x}^{\mathrm{2}\:} +\mathrm{2}{ax}\:+\mathrm{2}}\right)^{\left({n}−\mathrm{1}\right)} \\$$$${a}^{\mathrm{2}} {x}^{\mathrm{2}\:} +\mathrm{2}{ax}\:+\mathrm{2}=\mathrm{0}\rightarrow\Delta^{'} ={a}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} =−{a}^{\mathrm{2}} \:=\left({ia}\right)^{\mathrm{2}} \:\Rightarrow \\$$$${x}_{\mathrm{1}} =\frac{−{a}+{ia}}{{a}^{\mathrm{2}} }\:\:=\frac{−\mathrm{1}+{i}}{{a}}\:\:\:\left({we}\:{suppose}\:{a}\neq\mathrm{0}\right) \\$$$${x}_{\mathrm{2}} =\frac{−\mathrm{1}−{i}}{{a}}\:\Rightarrow\frac{\mathrm{1}}{{a}^{\mathrm{2}} {x}^{\mathrm{2}} \:+\mathrm{2}{ax}\:+\mathrm{2}}\:=\frac{\mathrm{1}}{{a}^{\mathrm{2}} \left({x}−\frac{−\mathrm{1}+{i}}{{a}}\right)\left({x}+\frac{\mathrm{1}+{i}}{{a}}\right)} \\$$$$=\frac{\mathrm{1}}{{a}^{\mathrm{2}} \left({x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)}=\frac{\mathrm{1}}{{a}^{\mathrm{2}} \frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{2}{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)}\left\{\frac{\mathrm{1}}{{x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }−\frac{\mathrm{1}}{{x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{+\frac{{i}\pi}{\mathrm{4}}} }\right\} \\$$$$=\frac{\mathrm{1}}{{ia}^{\mathrm{2}} }\left\{\frac{\mathrm{1}}{{x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }−\frac{\mathrm{1}}{{x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }\right\}\:\Rightarrow \\$$$${f}^{\left({n}\right)} \left({x}\right)=\frac{\mathrm{1}}{{ia}}\left\{\:\:\:\left(\frac{\mathrm{1}}{{x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }\right)^{\left({n}−\mathrm{1}\right)} −\left(\frac{\mathrm{1}}{{x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }\right)^{\left.\right)\left.{n}−\mathrm{1}\right)} \right\} \\$$$$=\frac{\mathrm{1}}{{ia}}\left\{\:\frac{\left.\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \right)\left({n}−\mathrm{1}\right)!}{\left({x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }−\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }\right\}\:\Rightarrow \\$$$${f}^{\left({n}\right)} \left({x}\right)=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{{ia}}\left\{\frac{\left({x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} −\left({x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{2}{x}}{{a}}+\:\frac{\mathrm{2}}{{a}^{\mathrm{2}} }\right)^{{n}} }\right\} \\$$$${x}=\mathrm{0}\:\Rightarrow{f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{{ia}}\left\{\frac{\mathrm{2}{iIm}\left(\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }{\left(\frac{\mathrm{2}}{{a}^{\mathrm{2}} }\right)^{{n}} }\right\} \\$$$${we}\:{have}\:\:\left(\frac{\sqrt{\mathrm{2}}}{{a}}\right)^{{n}} \:{e}^{\frac{{in}\pi}{\mathrm{4}}} \:=\frac{\mathrm{2}^{\frac{{n}}{\mathrm{2}}} }{{a}^{{n}} }\left({cos}\left(\frac{{n}\pi}{\mathrm{4}}\right)+{isin}\left(\frac{{n}\pi}{\mathrm{4}}\right)\right)\:\Rightarrow \\$$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{{a}}×\mathrm{2}×\frac{\mathrm{2}^{\frac{{n}}{\mathrm{2}}} }{{a}^{{n}} }{sin}\left(\frac{{n}\pi}{\mathrm{4}}\right) \\$$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{{a}^{{n}+\mathrm{1}} }×\mathrm{2}^{\frac{{n}}{\mathrm{2}}+\mathrm{1}} \:{sin}\left(\frac{{n}\pi}{\mathrm{4}}\right) \\$$
Commented by mathmax by abdo last updated on 08/Sep/19
$$\left.\mathrm{2}\right)\:{f}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \:={f}\left(\mathrm{0}\right)\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \\$$$$=\frac{\pi}{\mathrm{4}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \left\{\:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{na}^{{n}+\mathrm{1}} }×\mathrm{2}^{\frac{{n}}{\mathrm{2}}+\mathrm{1}} \:{sin}\left(\frac{{n}\pi}{\mathrm{4}}\right)\right\}{x}^{{n}} \\$$$$\\$$
Commented by mathmax by abdo last updated on 08/Sep/19
$$\left.\mathrm{3}\right)\:{let}\:\:{I}\:=\int_{−\infty} ^{+\infty\:} \:\frac{{f}\left({x}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx}\:\Rightarrow\:{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left({ax}+\mathrm{1}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx}\:=\varphi\left({a}\right) \\$$$${we}\:{have}\:\varphi^{'} \left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{a}}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)\left(\mathrm{1}+\left({ax}+\mathrm{1}\right)^{\mathrm{2}} \right)}{dx} \\$$$$={a}\:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)\left({a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{ax}\:+\mathrm{2}\right)}\:{let}\:{W}\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} +\mathrm{4}\right)\left({a}^{\mathrm{2}} {z}^{\mathrm{2}} \:+\mathrm{2}{az}\:+\mathrm{2}\right)} \\$$$${poles}\:{of}\:{W}\:?\:{a}^{\mathrm{2}} {x}^{\mathrm{2}\:} +\mathrm{2}{ax}\:+\mathrm{2}\:=\mathrm{0}\rightarrow\Delta^{'} =−{a}^{\mathrm{2}} =\left({ia}\right)^{\mathrm{2}} \:\Rightarrow{z}_{\mathrm{1}} =\frac{−{a}+{ia}}{{a}^{\mathrm{2}} } \\$$$$=\frac{−\mathrm{1}+{i}}{{a}}\:=−\frac{\mathrm{1}−{i}}{{a}}\:=−\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:\:{andz}_{\mathrm{2}} =\frac{−{a}−{ia}}{{a}^{\mathrm{2}} }\:=\frac{−\mathrm{1}−{i}}{{a}} \\$$$$=−\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:\:\:\left({we}\:{suppose}\:{a}>\mathrm{0}\right)\:\Rightarrow \\$$$${W}\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−\mathrm{2}{i}\right)\left({z}+\mathrm{2}{i}\right){a}^{\mathrm{2}} \left({z}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)}=\frac{\mathrm{1}}{{a}^{\mathrm{2}} \left({z}^{\mathrm{2}} +\mathrm{4}\right)\left({z}^{\mathrm{2}} \:+\frac{\mathrm{2}{z}}{{a}}+\frac{\mathrm{2}}{{a}^{\mathrm{2}} }\right)} \\$$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left({W},\mathrm{2}{i}\right)\:+{Res}\left({W},−\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\$$$${Res}\left({W},\mathrm{2}{i}\right)\:={lim}_{{z}\rightarrow\mathrm{2}{i}} \left({z}−\mathrm{2}{i}\right){W}\left({z}\right)=\frac{\mathrm{1}}{\left(\mathrm{4}{i}\right){a}^{\mathrm{2}} \left(−\mathrm{4}\:+\frac{\mathrm{4}{i}}{{a}}\:+\frac{\mathrm{2}}{{a}^{\mathrm{2}} }\right)} \\$$$$=\frac{\mathrm{1}}{\mathrm{4}{i}\left(−\mathrm{4}{a}^{\mathrm{2}\:} +\mathrm{4}{ia}\:+\mathrm{2}\right)}\:….{be}\:{continued}…. \\$$$$\\$$