Integration Questions

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

$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\mathrm{a}−\mathrm{1}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt}=\frac{\pi}{\mathrm{sin}\left(\pi\mathrm{a}\right)}\:\:\: \\$$$$\mathrm{please}\:\mathrm{prove}\:\mathrm{that} \\$$

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

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

$${let}\:{ntegrate}\:{the}\:{complex}\:{funtion}\:{f}\left({z}\right)\:=\frac{{z}^{{a}−\mathrm{1}} }{\mathrm{1}+{z}}\:{on}\:{this}\:{contour}\:{shown} \\$$$${with}\:\mathrm{0}<{a}<\mathrm{1}\:\:{residus}\:{theorem}\:{give} \\$$$$\int_{\Gamma} {f}\left({z}\right){dz}\:+\:\int_{{AB}} \:{f}\left({z}\right){dz}\:+\:\int_{\gamma} {f}\left({z}\right){dz}\:+\:\int_{{A}^{'} {B}^{'} } \:\:{f}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({f},−\mathrm{1}\right)\: \\$$$${on}\:\Gamma\:\:{z}\:={R}\:{e}^{{i}\theta} \:\:\:\:\Rightarrow\:\int_{\Gamma} {f}\left({z}\right){dz}\:\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} {f}\left({R}\:{e}^{{i}\theta} \right){Ri}\:{e}^{{i}\theta} {d}\theta \\$$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\left({R}\:{e}^{{i}\theta} \right)^{{a}−\mathrm{1}} }{\mathrm{1}+{Re}^{{i}\theta} }\:{R}\:{i}\:{e}^{{i}\theta} {d}\theta\:\:\Rightarrow\:\int_{\Gamma} \mid{f}\left({z}\right)\mid{dz}_{{R}\rightarrow+\infty} \rightarrow\mathrm{0}\: \\$$$${on}\:{AB}\:\:{z}\:={t}\:{e}^{{i}\mathrm{2}\pi} \Rightarrow\:\int_{{AB}} {f}\left({z}\right){dz}\:=\:\int_{{R}} ^{{r}} \:\:\frac{\left({t}\:{e}^{{i}\mathrm{2}\pi} \right)^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}_{{r}\rightarrow\mathrm{0}\:{andR}\rightarrow+\infty} \rightarrow−\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}\:{e}^{{i}\mathrm{2}\pi\left({a}−\mathrm{1}\right)} {dt} \\$$$${on}\:\gamma\:\:{z}\:={r}\:{e}^{{i}\theta} \:\:\:\:{and}\:\int_{\gamma} \:\:{f}\left({z}\right){dz}\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:{f}\left({r}\:{e}^{{i}\theta} \right){d}\theta\:\rightarrow\mathrm{0}\:{when}\:{r}\rightarrow\mathrm{0} \\$$$${on}\:{A}^{'} {B}^{'} \:\:\:{z}\:={t}\:\:\:{and}\:\int_{{A}^{'} {B}^{'} } \:\:\:{f}\left({z}\right){dz}\:=\int_{{r}} ^{{R}} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}\:{dt}\:\rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:\left({r}\rightarrow\mathrm{0}\:{and}\:{R}\rightarrow+\infty\right) \\$$$${so}\:{we}\:{get}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:−{e}^{{i}\mathrm{2}\pi\left({a}−\mathrm{1}\right)} \int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}\:{dt}\:=\mathrm{2}{i}\pi\:{Res}\left({f},−\mathrm{1}\right)\:\Rightarrow \\$$$$\left(\mathrm{1}−{e}^{{i}\mathrm{2}\pi{a}} \right)\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:\:=\mathrm{2}{i}\pi\:\left(−\mathrm{1}\right)^{{a}−\mathrm{1}} \:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\frac{\mathrm{2}{i}\pi{e}^{{i}\pi\left({a}−\mathrm{1}\right)} }{\mathrm{1}−{e}^{{i}\mathrm{2}\pi{a}} } \\$$$$=\:\frac{\mathrm{2}{i}\pi\:{e}^{−{i}\pi{a}} \:{e}^{{i}\pi\left({a}−\mathrm{1}\right)} }{{e}^{−{i}\pi{a}} \left(\mathrm{1}−{e}^{{i}\mathrm{2}\pi{a}} \right)\:}\:=\:\frac{\mathrm{2}{i}\pi\:\left(−\mathrm{1}\right)}{{e}^{{i}\pi{a}} −{e}^{{i}\pi{a}} }\:=\:\frac{\mathrm{2}{i}\pi}{{e}^{{i}\pi{a}} −{e}^{−{i}\pi{a}} }\:=\frac{\mathrm{2}{i}\pi}{\mathrm{2}{i}\:{sin}\left(\pi{a}\right)}\:=\frac{\pi}{{sin}\left(\pi{a}\right)}\:\Rightarrow \\$$$${we}\:{have}\:{proved}\:{that}\:{for}\:\mathrm{0}<{a}<\mathrm{1}\:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}\:{dt}\:=\frac{\pi}{{sin}\left(\pi{a}\right)}\:. \\$$

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

$${t}={tan}^{\mathrm{2}} \theta\:\:{dt}=\mathrm{2}{tan}\theta{sec}^{\mathrm{2}} \theta\:{d}\theta \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left({tan}^{\mathrm{2}} \theta\right)^{{a}−\mathrm{1}} }{{sec}^{\mathrm{2}} \theta}×\mathrm{2}{tan}\theta{sec}^{\mathrm{2}} \theta{d}\theta \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}^{\mathrm{2}{a}−\mathrm{2}} \theta}{{cos}^{\mathrm{2}{a}−\mathrm{2}} \theta}×\mathrm{2}\frac{{sin}\theta}{{cos}\theta}{d}\theta \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}{sin}^{\mathrm{2}{a}−\mathrm{1}} ×{cos}^{−\mathrm{2}{a}+\mathrm{1}} \theta \\$$$$\mathrm{2}{a}−\mathrm{1}=\mathrm{2}{p}−\mathrm{1}\:\:\:\:{p}={a} \\$$$$\mathrm{2}{q}−\mathrm{1}=−\mathrm{2}{a}+\mathrm{1} \\$$$$\mathrm{2}{q}=−\mathrm{2}{a}+\mathrm{2}\:\:\:\:{q}=−{a}+\mathrm{1} \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}{sin}^{\mathrm{2}{a}−\mathrm{1}} {cos}^{\mathrm{2}\left(\mathrm{1}−{a}\right)−\mathrm{1}} {d}\theta \\$$$$\\$$$${formula}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}{sin}^{\mathrm{2}{p}−\mathrm{1}} \theta{cos}^{\mathrm{2}{q}−\mathrm{1}} \theta\:{d}\theta \\$$$$=\frac{\lceil\left({p}\right)\lceil\left({q}\right)}{\lceil\left({p}+{q}\right)} \\$$$$\\$$$$\\$$$$\frac{\left.\lceil\left.{a}\right)\lceil\mathrm{1}−{a}\right)}{\lceil\left({a}+\mathrm{1}−{a}\right)}=\lceil\left({a}\right)\lceil\left(\mathrm{1}−{a}\right)=\frac{\pi}{{sin}\left(\pi{a}\right)}\:{proved} \\$$$$\\$$$$\\$$$${formula}\:\:\lceil\left({p}\right)\lceil\left(\mathrm{1}−{p}\right)=\frac{\pi}{{sinp}\pi} \\$$$$\\$$

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

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

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

$${most}\:{welcome}... \\$$