Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 169797 by mathocean1 last updated on 09/May/22

Calculate for n∈ N^∗ :∫_0 ^(+∞) (dt/((t^2 +1)^n )) (Notice: 1=(1+t^2 )−t^2 )

$${Calculate}\:{for}\:{n}\in\:\mathbb{N}^{\ast} :\int_{\mathrm{0}} ^{+\infty} \frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} } \\ $$$$\left({Notice}:\:\mathrm{1}=\left(\mathrm{1}+{t}^{\mathrm{2}} \right)−{t}^{\mathrm{2}} \right) \\ $$

Answered by floor(10²Eta[1]) last updated on 09/May/22

t=tgθ⇒dt=sec^2 θdθ ∫_0 ^(π/2) ((sec^2 dθ)/(sec^(2n) θ))=∫_0 ^(π/2) cos^(2n−2) θdθ, use reduction formula.

$$\mathrm{t}=\mathrm{tg}\theta\Rightarrow\mathrm{dt}=\mathrm{sec}^{\mathrm{2}} \theta\mathrm{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{\mathrm{sec}^{\mathrm{2}} \mathrm{d}\theta}{\mathrm{sec}^{\mathrm{2n}} \theta}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{cos}^{\mathrm{2n}−\mathrm{2}} \theta\mathrm{d}\theta,\:\mathrm{use}\:\mathrm{reduction}\:\mathrm{formula}. \\ $$

Answered by Mathspace last updated on 09/May/22

u_n =∫_0 ^∞ (dt/((t^2 +1)^n )) ⇒ we have B(x,y)=∫_0 ^∞ (t^(x−1) /((1+t)^(x+y) ))dt so∫_0 ^∞ (dt/((1+t^2 )^n ))=_(t=(√x)) (1/2)∫_0 ^∞ (x^(−(1/2)) /((1+x)^n ))dx =(1/2)∫_0 ^∞ (t^((1/2)−1) /((1+t)^(n+(1/2)−(1/2)) ))dt =(1/2)∫_0 ^∞ (t^((1/2)−1) /((1+t)^((1/2)+n−(1/2)) ))dt =(1/2)B((1/2),n−(1/2)) =(1/2)×((Γ((1/2)).Γ(n−(1/2)))/(Γ(n))) =((√π)/(2(n−1)!))Γ(n−(1/2)) =((√π)/2)×(((n−(3/2))!)/((n−1)!)) (n>0)

$${u}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} }\:\Rightarrow \\ $$$${we}\:{have}\:{B}\left({x},{y}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{x}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{{x}+{y}} }{dt} \\ $$$${so}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}} }=_{{t}=\sqrt{{x}}} \:\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} }{\left(\mathrm{1}+{x}\right)^{{n}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{{n}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} }{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}+{n}−\frac{\mathrm{1}}{\mathrm{2}}} }{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{B}\left(\frac{\mathrm{1}}{\mathrm{2}},{n}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right).\Gamma\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({n}\right)} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}\left({n}−\mathrm{1}\right)!}\Gamma\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}}×\frac{\left({n}−\frac{\mathrm{3}}{\mathrm{2}}\right)!}{\left({n}−\mathrm{1}\right)!}\:\:\left({n}>\mathrm{0}\right) \\ $$

Answered by Mathspace last updated on 09/May/22

residus method ∫_0 ^∞ (dt/((t^2 +1)^n ))=(1/2)∫_(−∞) ^(+∞) (dt/((t^2 +1)^n )) ϕ(z)=(1/((z^2 +1)^n ))⇒ϕ(z)=(1/((z−i)^n (z+i)^n )) ∫_R ϕdz=2iπ Res(ϕ,i) Res(ϕ,i)=lim_(z→i) (1/((n−1)!)){(z−i)^n ϕ(z)}^((n−1)) =lim_(z→i) (1/((n−1)!)){(z+i)^(−n) }^((n−1)) we have (z+i)^p }^((1)) =p(z+i)^(p−1) ...(z+i)^p }^((k)) =p(p−1)...(p−k+1)(z+i)^(p−k) (z+i)^(−n) }^((n−1)) =(−n)(−n−1)...(−n−n+1+1)(z+i)^(−n−n+1) =(−1)^(n−1) n(n+1)....(2n+2)(z+i)^(−2n+1) .....

$${residus}\:{method} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} }=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} } \\ $$$$\varphi\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} }\Rightarrow\varphi\left({z}\right)=\frac{\mathrm{1}}{\left({z}−{i}\right)^{{n}} \left({z}+{i}\right)^{{n}} } \\ $$$$\int_{{R}} \varphi{dz}=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)={lim}_{{z}\rightarrow{i}} \frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{{n}} \varphi\left({z}\right)\right\}^{\left({n}−\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left({n}−\mathrm{1}\right)} \\ $$$$\left.{we}\:{have}\:\left({z}+{i}\right)^{{p}} \right\}^{\left(\mathrm{1}\right)} ={p}\left({z}+{i}\right)^{{p}−\mathrm{1}} \\ $$$$\left....\left({z}+{i}\right)^{{p}} \right\}^{\left({k}\right)} ={p}\left({p}−\mathrm{1}\right)...\left({p}−{k}+\mathrm{1}\right)\left({z}+{i}\right)^{{p}−{k}} \\ $$$$\left.\left({z}+{i}\right)^{−{n}} \right\}^{\left({n}−\mathrm{1}\right)} =\left(−{n}\right)\left(−{n}−\mathrm{1}\right)...\left(−{n}−{n}+\mathrm{1}+\mathrm{1}\right)\left({z}+{i}\right)^{−{n}−{n}+\mathrm{1}} \\ $$$$=\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}\left({n}+\mathrm{1}\right)....\left(\mathrm{2}{n}+\mathrm{2}\right)\left({z}+{i}\right)^{−\mathrm{2}{n}+\mathrm{1}} \\ $$$$..... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com