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Question Number 68243 by mathmax by abdo last updated on 07/Sep/19

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

$${let}\:{f}\left({x}\right)\:={arctan}\left({ax}\:+\mathrm{1}\right)\:\:{with}\:{a}\:{real} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{−\infty} ^{+\infty} \:\frac{{f}\left({x}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx} \\ $$

Commented by mathmax by abdo last updated on 08/Sep/19

1) f(x)=arctan(ax+1) ⇒f^′ (x) =(a/(1+(ax+1)^2 )) =(a/(1+a^2 x^2 +2ax +1)) =(a/(a^2 x^2 +2ax +2)) ⇒ f^((n)) (x) =a((1/(a^2 x^(2 ) +2ax +2)))^((n−1)) a^2 x^(2 ) +2ax +2=0→Δ^′ =a^2 −2a^2 =−a^2 =(ia)^2 ⇒ x_1 =((−a+ia)/a^2 ) =((−1+i)/a) (we suppose a≠0) x_2 =((−1−i)/a) ⇒(1/(a^2 x^2 +2ax +2)) =(1/(a^2 (x−((−1+i)/a))(x+((1+i)/a)))) =(1/(a^2 (x+((√2)/a)e^((iπ)/4) )(x+((√2)/a)e^(−((iπ)/4)) )))=(1/(a^2 ((√2)/2)(2i((√2)/2)))){(1/(x+((√2)/a)e^(−((iπ)/4)) ))−(1/(x+((√2)/a)e^(+((iπ)/4)) ))} =(1/(ia^2 )){(1/(x+((√2)/a)e^(−((iπ)/4)) ))−(1/(x+((√2)/a)e^((iπ)/4) ))} ⇒ f^((n)) (x)=(1/(ia)){ ((1/(x+((√2)/a)e^(−((iπ)/4)) )))^((n−1)) −((1/(x+((√2)/a)e^((iπ)/4) )))^()n−1)) } =(1/(ia)){ (((−1)^(n−1) )(n−1)!)/((x+((√2)/a)e^(−((iπ)/4)) )^n ))−(((−1)^(n−1) (n−1)!)/((x+((√2)/a)e^((iπ)/4) )^n ))} ⇒ f^((n)) (x)=(((−1)^(n−1) (n−1)!)/(ia)){(((x+((√2)/a)e^((iπ)/4) )^n −(x+((√2)/a)e^(−((iπ)/4)) )^n )/((x^2 +((2x)/a)+ (2/a^2 ))^n ))} x=0 ⇒f^((n)) (0) =(((−1)^(n−1) (n−1)!)/(ia)){((2iIm(((√2)/a)e^((iπ)/4) )^n )/(((2/a^2 ))^n ))} we have (((√2)/a))^n e^((inπ)/4) =(2^(n/2) /a^n )(cos(((nπ)/4))+isin(((nπ)/4))) ⇒ f^((n)) (0) =(((−1)^(n−1) (n−1)!)/a)×2×(2^(n/2) /a^n )sin(((nπ)/4)) =(((−1)^(n−1) (n−1)!)/a^(n+1) )×2^((n/2)+1) sin(((nπ)/4))

$$\left.\mathrm{1}\right)\:{f}\left({x}\right)={arctan}\left({ax}+\mathrm{1}\right)\:\Rightarrow{f}^{'} \left({x}\right)\:=\frac{{a}}{\mathrm{1}+\left({ax}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{{a}}{\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \:+\mathrm{2}{ax}\:+\mathrm{1}}\:=\frac{{a}}{{a}^{\mathrm{2}} {x}^{\mathrm{2}} \:+\mathrm{2}{ax}\:+\mathrm{2}}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:={a}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} {x}^{\mathrm{2}\:} +\mathrm{2}{ax}\:+\mathrm{2}}\right)^{\left({n}−\mathrm{1}\right)} \\ $$$${a}^{\mathrm{2}} {x}^{\mathrm{2}\:} +\mathrm{2}{ax}\:+\mathrm{2}=\mathrm{0}\rightarrow\Delta^{'} ={a}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} =−{a}^{\mathrm{2}} \:=\left({ia}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$${x}_{\mathrm{1}} =\frac{−{a}+{ia}}{{a}^{\mathrm{2}} }\:\:=\frac{−\mathrm{1}+{i}}{{a}}\:\:\:\left({we}\:{suppose}\:{a}\neq\mathrm{0}\right) \\ $$$${x}_{\mathrm{2}} =\frac{−\mathrm{1}−{i}}{{a}}\:\Rightarrow\frac{\mathrm{1}}{{a}^{\mathrm{2}} {x}^{\mathrm{2}} \:+\mathrm{2}{ax}\:+\mathrm{2}}\:=\frac{\mathrm{1}}{{a}^{\mathrm{2}} \left({x}−\frac{−\mathrm{1}+{i}}{{a}}\right)\left({x}+\frac{\mathrm{1}+{i}}{{a}}\right)} \\ $$$$=\frac{\mathrm{1}}{{a}^{\mathrm{2}} \left({x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)}=\frac{\mathrm{1}}{{a}^{\mathrm{2}} \frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{2}{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)}\left\{\frac{\mathrm{1}}{{x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }−\frac{\mathrm{1}}{{x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{+\frac{{i}\pi}{\mathrm{4}}} }\right\} \\ $$$$=\frac{\mathrm{1}}{{ia}^{\mathrm{2}} }\left\{\frac{\mathrm{1}}{{x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }−\frac{\mathrm{1}}{{x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }\right\}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\frac{\mathrm{1}}{{ia}}\left\{\:\:\:\left(\frac{\mathrm{1}}{{x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }\right)^{\left({n}−\mathrm{1}\right)} −\left(\frac{\mathrm{1}}{{x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }\right)^{\left.\right)\left.{n}−\mathrm{1}\right)} \right\} \\ $$$$=\frac{\mathrm{1}}{{ia}}\left\{\:\frac{\left.\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \right)\left({n}−\mathrm{1}\right)!}{\left({x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }−\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }\right\}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{{ia}}\left\{\frac{\left({x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} −\left({x}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{2}{x}}{{a}}+\:\frac{\mathrm{2}}{{a}^{\mathrm{2}} }\right)^{{n}} }\right\} \\ $$$${x}=\mathrm{0}\:\Rightarrow{f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{{ia}}\left\{\frac{\mathrm{2}{iIm}\left(\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }{\left(\frac{\mathrm{2}}{{a}^{\mathrm{2}} }\right)^{{n}} }\right\} \\ $$$${we}\:{have}\:\:\left(\frac{\sqrt{\mathrm{2}}}{{a}}\right)^{{n}} \:{e}^{\frac{{in}\pi}{\mathrm{4}}} \:=\frac{\mathrm{2}^{\frac{{n}}{\mathrm{2}}} }{{a}^{{n}} }\left({cos}\left(\frac{{n}\pi}{\mathrm{4}}\right)+{isin}\left(\frac{{n}\pi}{\mathrm{4}}\right)\right)\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{{a}}×\mathrm{2}×\frac{\mathrm{2}^{\frac{{n}}{\mathrm{2}}} }{{a}^{{n}} }{sin}\left(\frac{{n}\pi}{\mathrm{4}}\right) \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{{a}^{{n}+\mathrm{1}} }×\mathrm{2}^{\frac{{n}}{\mathrm{2}}+\mathrm{1}} \:{sin}\left(\frac{{n}\pi}{\mathrm{4}}\right) \\ $$

Commented by mathmax by abdo last updated on 08/Sep/19

2) f(x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n =f(0) +Σ_(n=1) ^∞ ((f^((n)) (0))/(n!)) x^n =(π/4) +Σ_(n=1) ^∞ { (((−1)^(n−1) )/(na^(n+1) ))×2^((n/2)+1) sin(((nπ)/4))}x^n

$$\left.\mathrm{2}\right)\:{f}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \:={f}\left(\mathrm{0}\right)\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \\ $$$$=\frac{\pi}{\mathrm{4}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \left\{\:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{na}^{{n}+\mathrm{1}} }×\mathrm{2}^{\frac{{n}}{\mathrm{2}}+\mathrm{1}} \:{sin}\left(\frac{{n}\pi}{\mathrm{4}}\right)\right\}{x}^{{n}} \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 08/Sep/19

3) let I =∫_(−∞) ^(+∞ ) ((f(x))/(x^2 +4))dx ⇒ I =∫_(−∞) ^(+∞) ((arctan(ax+1))/(x^2 +4))dx =ϕ(a) we have ϕ^′ (a) =∫_(−∞) ^(+∞) (a/((x^2 +4)(1+(ax+1)^2 )))dx =a ∫_(−∞) ^(+∞) (dx/((x^2 +4)(a^2 x^2 +2ax +2))) let W(z)=(1/((z^2 +4)(a^2 z^2 +2az +2))) poles of W ? a^2 x^(2 ) +2ax +2 =0→Δ^′ =−a^2 =(ia)^2 ⇒z_1 =((−a+ia)/a^2 ) =((−1+i)/a) =−((1−i)/a) =−((√2)/a)e^(−((iπ)/4)) andz_2 =((−a−ia)/a^2 ) =((−1−i)/a) =−((√2)/a)e^((iπ)/4) (we suppose a>0) ⇒ W(z) =(1/((z−2i)(z+2i)a^2 (z+((√2)/a)e^(−((iπ)/4)) )(z+((√2)/a)e^((iπ)/4) )))=(1/(a^2 (z^2 +4)(z^2 +((2z)/a)+(2/a^2 )))) ∫_(−∞) ^(+∞) W(z)dz =2iπ { Res(W,2i) +Res(W,−((√2)/a)e^(−((iπ)/4)) )} Res(W,2i) =lim_(z→2i) (z−2i)W(z)=(1/((4i)a^2 (−4 +((4i)/a) +(2/a^2 )))) =(1/(4i(−4a^(2 ) +4ia +2))) ....be continued....

$$\left.\mathrm{3}\right)\:{let}\:\:{I}\:=\int_{−\infty} ^{+\infty\:} \:\frac{{f}\left({x}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx}\:\Rightarrow\:{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left({ax}+\mathrm{1}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx}\:=\varphi\left({a}\right) \\ $$$${we}\:{have}\:\varphi^{'} \left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{a}}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)\left(\mathrm{1}+\left({ax}+\mathrm{1}\right)^{\mathrm{2}} \right)}{dx} \\ $$$$={a}\:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)\left({a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{ax}\:+\mathrm{2}\right)}\:{let}\:{W}\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} +\mathrm{4}\right)\left({a}^{\mathrm{2}} {z}^{\mathrm{2}} \:+\mathrm{2}{az}\:+\mathrm{2}\right)} \\ $$$${poles}\:{of}\:{W}\:?\:{a}^{\mathrm{2}} {x}^{\mathrm{2}\:} +\mathrm{2}{ax}\:+\mathrm{2}\:=\mathrm{0}\rightarrow\Delta^{'} =−{a}^{\mathrm{2}} =\left({ia}\right)^{\mathrm{2}} \:\Rightarrow{z}_{\mathrm{1}} =\frac{−{a}+{ia}}{{a}^{\mathrm{2}} } \\ $$$$=\frac{−\mathrm{1}+{i}}{{a}}\:=−\frac{\mathrm{1}−{i}}{{a}}\:=−\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:\:{andz}_{\mathrm{2}} =\frac{−{a}−{ia}}{{a}^{\mathrm{2}} }\:=\frac{−\mathrm{1}−{i}}{{a}} \\ $$$$=−\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:\:\:\left({we}\:{suppose}\:{a}>\mathrm{0}\right)\:\Rightarrow \\ $$$${W}\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−\mathrm{2}{i}\right)\left({z}+\mathrm{2}{i}\right){a}^{\mathrm{2}} \left({z}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)}=\frac{\mathrm{1}}{{a}^{\mathrm{2}} \left({z}^{\mathrm{2}} +\mathrm{4}\right)\left({z}^{\mathrm{2}} \:+\frac{\mathrm{2}{z}}{{a}}+\frac{\mathrm{2}}{{a}^{\mathrm{2}} }\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left({W},\mathrm{2}{i}\right)\:+{Res}\left({W},−\frac{\sqrt{\mathrm{2}}}{{a}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$${Res}\left({W},\mathrm{2}{i}\right)\:={lim}_{{z}\rightarrow\mathrm{2}{i}} \left({z}−\mathrm{2}{i}\right){W}\left({z}\right)=\frac{\mathrm{1}}{\left(\mathrm{4}{i}\right){a}^{\mathrm{2}} \left(−\mathrm{4}\:+\frac{\mathrm{4}{i}}{{a}}\:+\frac{\mathrm{2}}{{a}^{\mathrm{2}} }\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{i}\left(−\mathrm{4}{a}^{\mathrm{2}\:} +\mathrm{4}{ia}\:+\mathrm{2}\right)}\:....{be}\:{continued}.... \\ $$$$ \\ $$

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