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I-n-0-1-1-x-2-n-dx-prove-that-n-1-I-n-n-pi-

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Question Number 208194 by universe last updated on 07/Jun/24
I_n = ∫_(0 ) ^∞ (1/((1+x^2 )^n ))dx prove that Σ_(n=1) ^∞ (I_n /n) = π
$$\:\:\:\:\:\:\:\:{I}_{{n}} \:=\:\:\int_{\mathrm{0}\:} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }{dx} \\ $$$$\:\:{prove}\:{that}\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{I}_{{n}} }{{n}}\:\:=\:\:\pi \\ $$
Answered by Berbere last updated on 07/Jun/24
I_n =∫_0 ^∞ ((y^(−(1/2)) dy)/(2(1+y)^n ))=(1/2)∫_0 ^∞ (y^((1/2)−1) /((1+y)^((1/2)+1−(1/2)) ))=(1/2)β((1/2),n−(1/2)) Σ_(n≥1) (I/n)=Σ_(n≥1) (1/(2n)).((Γ((1/2))Γ(n−(1/2)))/(Γ(n)))=Σ(((1/2)Γ((1/2))Γ(n−(1/2)))/(Γ(n+1)))=Σ((Γ((3/2))Γ(n−(1/2)))/(Γ(n−(1/2)+(3/2)))) =Σ_(n≥1) β((3/2),n−(1/2))=^(n=1+n) Σ_(n≥0) ∫_0 ^1 t^(n+(1/2)) (1−t)^(3/2) dt =∫_0 ^1 (√t)(1−t)^(3/2) .(1/(1−t))dt=∫_0 ^1 (√(t(1−t)))dt=β((1/2),(1/2))=(π/(sin((π/2))))=π Σ_(n≥1) (I_n /n)=π
$${I}_{{n}} =\int_{\mathrm{0}} ^{\infty} \frac{{y}^{−\frac{\mathrm{1}}{\mathrm{2}}} {dy}}{\mathrm{2}\left(\mathrm{1}+{y}\right)^{{n}} }=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{y}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\left(\mathrm{1}+{y}\right)^{\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}} }=\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\frac{\mathrm{1}}{\mathrm{2}},{n}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{I}}{{n}}=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{2}{n}}.\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({n}\right)}=\Sigma\frac{\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({n}+\mathrm{1}\right)}=\Sigma\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({n}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}\right)} \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\beta\left(\frac{\mathrm{3}}{\mathrm{2}},{n}−\frac{\mathrm{1}}{\mathrm{2}}\right)\overset{{n}=\mathrm{1}+{n}} {=}\underset{{n}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{n}+\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} {dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{t}}\left(\mathrm{1}−{t}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} .\frac{\mathrm{1}}{\mathrm{1}−{t}}{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{t}\left(\mathrm{1}−{t}\right)}{dt}=\beta\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{2}}\right)}=\pi \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{I}_{{n}} }{{n}}=\pi \\ $$
Answered by namphamduc last updated on 07/Jun/24
No need to use Beta function I_n =∫_0 ^∞ (1/((1+x^2 )^n ))dx ⇒S=Σ_(n=1) ^∞ (I_n /n)=∫_0 ^∞ (Σ_(n=1) ^∞ (1/(n(1+x^2 )^n )))dx=−∫_0 ^∞ ln(1−(1/(1+x^2 )))dx =−∫_0 ^∞ ln((x^2 /(1+x^2 )))dx { ((u=ln((x^2 /(1+x^2 ))))),((dv=dx)) :}⇒ { ((du=(2/(x(1+x^2 )))dx)),((v=x)) :} ⇒S=(−xln((x^2 /(1+x^2 )))+2tan^(−1) (x))∣_0 ^∞ =2.(π/2)=π
$$\mathrm{No}\:\mathrm{need}\:\mathrm{to}\:\mathrm{use}\:\mathrm{Beta}\:\mathrm{function} \\ $$$${I}_{{n}} =\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }{dx} \\ $$$$\Rightarrow{S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{I}_{{n}} }{{n}}=\int_{\mathrm{0}} ^{\infty} \left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }\right){dx}=−\int_{\mathrm{0}} ^{\infty} \mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \mathrm{ln}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx} \\ $$$$\begin{cases}{{u}=\mathrm{ln}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\right)}\\{{dv}={dx}}\end{cases}\Rightarrow\begin{cases}{{du}=\frac{\mathrm{2}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}}\\{{v}={x}}\end{cases} \\ $$$$\Rightarrow{S}=\left(−{x}\mathrm{ln}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\right)+\mathrm{2tan}^{−\mathrm{1}} \left({x}\right)\right)\mid_{\mathrm{0}} ^{\infty} =\mathrm{2}.\frac{\pi}{\mathrm{2}}=\pi \\ $$

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