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Question Number 45075 by arvinddayama01@gmail.com last updated on 08/Oct/18
∫_0 ^∞ (t^(a−1) /(1+t))dt=(π/(sin(πa))) please prove that
$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\mathrm{a}−\mathrm{1}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt}=\frac{\pi}{\mathrm{sin}\left(\pi\mathrm{a}\right)}\:\:\: \\ $$$$\mathrm{please}\:\mathrm{prove}\:\mathrm{that} \\ $$
Commented by abdo.msup.com last updated on 08/Oct/18
Commented by maxmathsup by imad last updated on 08/Oct/18
let ntegrate the complex funtion f(z) =(z^(a−1) /(1+z)) on this contour shown with 0<a<1 residus theorem give ∫_Γ f(z)dz + ∫_(AB) f(z)dz + ∫_γ f(z)dz + ∫_(A^′ B^′ ) f(z)dz =2iπ Res(f,−1) on Γ z =R e^(iθ) ⇒ ∫_Γ f(z)dz = ∫_0 ^(2π) f(R e^(iθ) )Ri e^(iθ) dθ = ∫_0 ^(2π) (((R e^(iθ) )^(a−1) )/(1+Re^(iθ) )) R i e^(iθ) dθ ⇒ ∫_Γ ∣f(z)∣dz_(R→+∞) →0 on AB z =t e^(i2π) ⇒ ∫_(AB) f(z)dz = ∫_R ^r (((t e^(i2π) )^(a−1) )/(1+t))dt_(r→0 andR→+∞) →−∫_0 ^(+∞) (t^(a−1) /(1+t)) e^(i2π(a−1)) dt on γ z =r e^(iθ) and ∫_γ f(z)dz =∫_0 ^(2π) f(r e^(iθ) )dθ →0 when r→0 on A^′ B^′ z =t and ∫_(A^′ B^′ ) f(z)dz =∫_r ^R (t^(a−1) /(1+t)) dt →∫_0 ^∞ (t^(a−1) /(1+t))dt (r→0 and R→+∞) so we get ∫_0 ^∞ (t^(a−1) /(1+t))dt −e^(i2π(a−1)) ∫_0 ^∞ (t^(a−1) /(1+t)) dt =2iπ Res(f,−1) ⇒ (1−e^(i2πa) ) ∫_0 ^∞ (t^(a−1) /(1+t))dt =2iπ (−1)^(a−1) ⇒∫_0 ^∞ (t^(a−1) /(1+t))dt =((2iπe^(iπ(a−1)) )/(1−e^(i2πa) )) = ((2iπ e^(−iπa) e^(iπ(a−1)) )/(e^(−iπa) (1−e^(i2πa) ) )) = ((2iπ (−1))/(e^(iπa) −e^(iπa) )) = ((2iπ)/(e^(iπa) −e^(−iπa) )) =((2iπ)/(2i sin(πa))) =(π/(sin(πa))) ⇒ we have proved that for 0<a<1 ∫_0 ^∞ (t^(a−1) /(1+t)) dt =(π/(sin(πa))) .
$${let}\:{ntegrate}\:{the}\:{complex}\:{funtion}\:{f}\left({z}\right)\:=\frac{{z}^{{a}−\mathrm{1}} }{\mathrm{1}+{z}}\:{on}\:{this}\:{contour}\:{shown} \\ $$$${with}\:\mathrm{0}<{a}<\mathrm{1}\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{\Gamma} {f}\left({z}\right){dz}\:+\:\int_{{AB}} \:{f}\left({z}\right){dz}\:+\:\int_{\gamma} {f}\left({z}\right){dz}\:+\:\int_{{A}^{'} {B}^{'} } \:\:{f}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({f},−\mathrm{1}\right)\: \\ $$$${on}\:\Gamma\:\:{z}\:={R}\:{e}^{{i}\theta} \:\:\:\:\Rightarrow\:\int_{\Gamma} {f}\left({z}\right){dz}\:\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} {f}\left({R}\:{e}^{{i}\theta} \right){Ri}\:{e}^{{i}\theta} {d}\theta \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\left({R}\:{e}^{{i}\theta} \right)^{{a}−\mathrm{1}} }{\mathrm{1}+{Re}^{{i}\theta} }\:{R}\:{i}\:{e}^{{i}\theta} {d}\theta\:\:\Rightarrow\:\int_{\Gamma} \mid{f}\left({z}\right)\mid{dz}_{{R}\rightarrow+\infty} \rightarrow\mathrm{0}\: \\ $$$${on}\:{AB}\:\:{z}\:={t}\:{e}^{{i}\mathrm{2}\pi} \Rightarrow\:\int_{{AB}} {f}\left({z}\right){dz}\:=\:\int_{{R}} ^{{r}} \:\:\frac{\left({t}\:{e}^{{i}\mathrm{2}\pi} \right)^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}_{{r}\rightarrow\mathrm{0}\:{andR}\rightarrow+\infty} \rightarrow−\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}\:{e}^{{i}\mathrm{2}\pi\left({a}−\mathrm{1}\right)} {dt} \\ $$$${on}\:\gamma\:\:{z}\:={r}\:{e}^{{i}\theta} \:\:\:\:{and}\:\int_{\gamma} \:\:{f}\left({z}\right){dz}\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:{f}\left({r}\:{e}^{{i}\theta} \right){d}\theta\:\rightarrow\mathrm{0}\:{when}\:{r}\rightarrow\mathrm{0} \\ $$$${on}\:{A}^{'} {B}^{'} \:\:\:{z}\:={t}\:\:\:{and}\:\int_{{A}^{'} {B}^{'} } \:\:\:{f}\left({z}\right){dz}\:=\int_{{r}} ^{{R}} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}\:{dt}\:\rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:\left({r}\rightarrow\mathrm{0}\:{and}\:{R}\rightarrow+\infty\right) \\ $$$${so}\:{we}\:{get}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:−{e}^{{i}\mathrm{2}\pi\left({a}−\mathrm{1}\right)} \int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}\:{dt}\:=\mathrm{2}{i}\pi\:{Res}\left({f},−\mathrm{1}\right)\:\Rightarrow \\ $$$$\left(\mathrm{1}−{e}^{{i}\mathrm{2}\pi{a}} \right)\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:\:=\mathrm{2}{i}\pi\:\left(−\mathrm{1}\right)^{{a}−\mathrm{1}} \:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\frac{\mathrm{2}{i}\pi{e}^{{i}\pi\left({a}−\mathrm{1}\right)} }{\mathrm{1}−{e}^{{i}\mathrm{2}\pi{a}} } \\ $$$$=\:\frac{\mathrm{2}{i}\pi\:{e}^{−{i}\pi{a}} \:{e}^{{i}\pi\left({a}−\mathrm{1}\right)} }{{e}^{−{i}\pi{a}} \left(\mathrm{1}−{e}^{{i}\mathrm{2}\pi{a}} \right)\:}\:=\:\frac{\mathrm{2}{i}\pi\:\left(−\mathrm{1}\right)}{{e}^{{i}\pi{a}} −{e}^{{i}\pi{a}} }\:=\:\frac{\mathrm{2}{i}\pi}{{e}^{{i}\pi{a}} −{e}^{−{i}\pi{a}} }\:=\frac{\mathrm{2}{i}\pi}{\mathrm{2}{i}\:{sin}\left(\pi{a}\right)}\:=\frac{\pi}{{sin}\left(\pi{a}\right)}\:\Rightarrow \\ $$$${we}\:{have}\:{proved}\:{that}\:{for}\:\mathrm{0}<{a}<\mathrm{1}\:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}\:{dt}\:=\frac{\pi}{{sin}\left(\pi{a}\right)}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 08/Oct/18
t=tan^2 θ dt=2tanθsec^2 θ dθ ∫_0 ^(π/2) (((tan^2 θ)^(a−1) )/(sec^2 θ))×2tanθsec^2 θdθ ∫_0 ^(π/2) ((sin^(2a−2) θ)/(cos^(2a−2) θ))×2((sinθ)/(cosθ))dθ ∫_0 ^(π/2) 2sin^(2a−1) ×cos^(−2a+1) θ 2a−1=2p−1 p=a 2q−1=−2a+1 2q=−2a+2 q=−a+1 ∫_0 ^(π/2) 2sin^(2a−1) cos^(2(1−a)−1) dθ formula ∫_0 ^(π/2) 2sin^(2p−1) θcos^(2q−1) θ dθ =((⌈(p)⌈(q))/(⌈(p+q))) ((⌈a)⌈1−a))/(⌈(a+1−a)))=⌈(a)⌈(1−a)=(π/(sin(πa))) proved formula ⌈(p)⌈(1−p)=(π/(sinpπ))
$${t}={tan}^{\mathrm{2}} \theta\:\:{dt}=\mathrm{2}{tan}\theta{sec}^{\mathrm{2}} \theta\:{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left({tan}^{\mathrm{2}} \theta\right)^{{a}−\mathrm{1}} }{{sec}^{\mathrm{2}} \theta}×\mathrm{2}{tan}\theta{sec}^{\mathrm{2}} \theta{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}^{\mathrm{2}{a}−\mathrm{2}} \theta}{{cos}^{\mathrm{2}{a}−\mathrm{2}} \theta}×\mathrm{2}\frac{{sin}\theta}{{cos}\theta}{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}{sin}^{\mathrm{2}{a}−\mathrm{1}} ×{cos}^{−\mathrm{2}{a}+\mathrm{1}} \theta \\ $$$$\mathrm{2}{a}−\mathrm{1}=\mathrm{2}{p}−\mathrm{1}\:\:\:\:{p}={a} \\ $$$$\mathrm{2}{q}−\mathrm{1}=−\mathrm{2}{a}+\mathrm{1} \\ $$$$\mathrm{2}{q}=−\mathrm{2}{a}+\mathrm{2}\:\:\:\:{q}=−{a}+\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}{sin}^{\mathrm{2}{a}−\mathrm{1}} {cos}^{\mathrm{2}\left(\mathrm{1}−{a}\right)−\mathrm{1}} {d}\theta \\ $$$$ \\ $$$${formula}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}{sin}^{\mathrm{2}{p}−\mathrm{1}} \theta{cos}^{\mathrm{2}{q}−\mathrm{1}} \theta\:{d}\theta \\ $$$$=\frac{\lceil\left({p}\right)\lceil\left({q}\right)}{\lceil\left({p}+{q}\right)} \\ $$$$ \\ $$$$ \\ $$$$\frac{\left.\lceil\left.{a}\right)\lceil\mathrm{1}−{a}\right)}{\lceil\left({a}+\mathrm{1}−{a}\right)}=\lceil\left({a}\right)\lceil\left(\mathrm{1}−{a}\right)=\frac{\pi}{{sin}\left(\pi{a}\right)}\:{proved} \\ $$$$ \\ $$$$ \\ $$$${formula}\:\:\lceil\left({p}\right)\lceil\left(\mathrm{1}−{p}\right)=\frac{\pi}{{sinp}\pi} \\ $$$$ \\ $$
Commented by arvinddayama01@gmail.com last updated on 09/Oct/18
thanks sir
$$\mathrm{thanks}\:\mathrm{sir} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Oct/18
most welcome...
$${most}\:{welcome}… \\ $$

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