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Question Number 45075 by arvinddayama01@gmail.com last updated on 08/Oct/18

$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}-1}}{1+\mathrm{t}}\mathrm{dt}=\frac{\pi }{\sin \left(\pi \mathrm{a}\right)}$$$$\mathrm{please}\mathrm{prove}\mathrm{that}$$

Commented by abdo.msup.com last updated on 08/Oct/18

Commented by maxmathsup by imad last updated on 08/Oct/18

$$letntegratethecomplexfuntionf\left(z\right)=\frac{z^{a-1}}{1+z}onthiscontourshown$$$$with0<a<1residustheoremgive$$$${\int }_{\mathrm{\Gamma }}f\left(z\right)dz+{\int }_{AB}f\left(z\right)dz+{\int }_{\gamma }f\left(z\right)dz+{\int }_{A^{{}^{\prime }}{B}^{{}^{\prime }}}f\left(z\right)dz=2i\pi Res(f,-1)$$$$on\mathrm{\Gamma }z=R{e}^{i\theta }\Rightarrow {\int }_{\mathrm{\Gamma }}f\left(z\right)dz={\int }_0^{2\pi }f\left(R{e}^{i\theta }\right)Ri{e}^{i\theta }d\theta$$$$={\int }_0^{2\pi }\frac{{\left(R{e}^{i\theta }\right)}^{a-1}}{1+{Re}^{i\theta }}Ri{e}^{i\theta }d\theta \Rightarrow {\int }_{\mathrm{\Gamma }}\mid f\left(z\right)\mid {dz}_{R\to +\mathrm{\infty }}\to 0$$$$onABz=t{e}^{i2\pi }\Rightarrow {\int }_{AB}f\left(z\right)dz={\int }_R^r\frac{{\left(t{e}^{i2\pi }\right)}^{a-1}}{1+t}{dt}_{r\to 0andR\to +\mathrm{\infty }}\to -{\int }_0^{+\mathrm{\infty }}\frac{t^{a-1}}{1+t}{e}^{i2\pi (a-1)}dt$$$$on\gamma z=r{e}^{i\theta }and{\int }_{\gamma }f\left(z\right)dz={\int }_0^{2\pi }f\left(r{e}^{i\theta }\right)d\theta \to 0whenr\to 0$$$$on{A}^{{}^{\prime }}{B}^{{}^{\prime }}z=tand{\int }_{A^{{}^{\prime }}{B}^{{}^{\prime }}}f\left(z\right)dz={\int }_r^R\frac{t^{a-1}}{1+t}dt\to {\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt(r\to 0andR\to +\mathrm{\infty })$$$$soweget{\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt-e^{i2\pi (a-1)}{\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt=2i\pi Res(f,-1)\Rightarrow$$$$(1-e^{i2\pi a}){\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt=2i\pi {(-1)}^{a-1}\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt=\frac{2i\pi e^{i\pi (a-1)}}{1-e^{i2\pi a}}$$$$=\frac{2i\pi e^{-i\pi a}{e}^{i\pi (a-1)}}{e^{-i\pi a}(1-e^{i2\pi a})}=\frac{2i\pi (-1)}{e^{i\pi a}-e^{i\pi a}}=\frac{2i\pi }{e^{i\pi a}-e^{-i\pi a}}=\frac{2i\pi }{2isin\left(\pi a\right)}=\frac{\pi }{sin\left(\pi a\right)}\Rightarrow$$$$wehaveprovedthatfor0<a<1{\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt=\frac{\pi }{sin\left(\pi a\right)}.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 08/Oct/18

$$t={tan}^2\theta dt=2tan\theta {sec}^2\theta d\theta$$$${\int }_0^{\frac{\pi }{2}}\frac{{\left({tan}^2\theta \right)}^{a-1}}{{sec}^2\theta }\times 2tan\theta {sec}^2\theta d\theta$$$${\int }_0^{\frac{\pi }{2}}\frac{{sin}^{2a-2}\theta }{{cos}^{2a-2}\theta }\times 2\frac{sin\theta }{cos\theta }d\theta$$$${\int }_0^{\frac{\pi }{2}}2{sin}^{2a-1}\times {cos}^{-2a+1}\theta$$$$2a-1=2p-1p=a$$$$2q-1=-2a+1$$$$2q=-2a+2q=-a+1$$$${\int }_0^{\frac{\pi }{2}}2{sin}^{2a-1}{cos}^{2(1-a)-1}d\theta$$$$$$$$formula{\int }_0^{\frac{\pi }{2}}2{sin}^{2p-1}\theta {cos}^{2q-1}\theta d\theta$$$$=\frac{\lceil \left(p\right)\lceil \left(q\right)}{\lceil (p+q)}$$$$$$$$$$$$\frac{\lceil a)\lceil 1-a)}{\lceil (a+1-a)}=\lceil \left(a\right)\lceil (1-a)=\frac{\pi }{sin\left(\pi a\right)}proved$$$$$$$$$$$$formula\lceil \left(p\right)\lceil (1-p)=\frac{\pi }{sinp\pi }$$$$$$

Commented by arvinddayama01@gmail.com last updated on 09/Oct/18

$$\mathrm{thanks}\mathrm{sir}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Oct/18

$$mostwelcome...$$

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