# Question-142689

Question Number 142689 by rs4089 last updated on 04/Jun/21
Answered by Dwaipayan Shikari last updated on 04/Jun/21
$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right)}{{x}}{f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({x}\right){dx}\:\:\:\:{f}\left({x}\pm\pi\right)={f}\left({x}\right) \\$$$${let}\:{f}\left({x}\right)={sin}^{\mathrm{2}{n}−\mathrm{2}} {x}\:\:\:\:\:\: \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{n}−\mathrm{2}} {x}\:{dx}=\frac{\Gamma\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left({n}\right)}=\frac{\sqrt{\pi}}{\mathrm{2}}.\frac{\Gamma\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\left({n}−\mathrm{1}\right)!} \\$$
Commented by mindispower last updated on 04/Jun/21
$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right)}{{x}}{f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({x}\right){dx},{let}\:{f}\left(−{x}\right)={f}\left({x}\right),{f}\left({x}+\pi\right)={f}\left({x}\right) \\$$$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right)}{{x}}{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=−\infty} {\overset{\infty} {\sum}}\int_{{k}\pi} ^{\left({k}+\mathrm{1}\right)\pi} \frac{{sin}\left({x}\right)}{{x}}{f}\left({x}\right){dx} \\$$$${x}\rightarrow{k}\pi+{t} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{−\infty} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\pi} \frac{\left(−\mathrm{1}\right)^{{k}} {sin}\left({t}\right)}{{k}\pi+{t}}{f}\left({t}\right){dt}={A} \\$$$$\underset{−\infty} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}\pi+{t}}=\frac{\mathrm{1}}{{sin}\left({t}\right)} \\$$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} {f}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(.\int_{−\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} {f}\left({t}\right){dt}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({t}\right){dt}\right)={A} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({t}\right){dt}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({t}\right).{dt} \\$$
Commented by Dwaipayan Shikari last updated on 04/Jun/21
$${Thanks}\:{for}\:{proving} \\$$
Commented by mindispower last updated on 04/Jun/21
$${pleasur}\:{nice}\:{answer}\:{too} \\$$
Commented by mathmax by abdo last updated on 06/Jun/21
$$\mathrm{we}\:\mathrm{have}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{2p}−\mathrm{1}} \mathrm{x}\:\mathrm{sin}^{\mathrm{2q}−\mathrm{1}} \mathrm{x}\:\mathrm{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\frac{\Gamma\left(\mathrm{p}\right).\Gamma\left(\mathrm{q}\right)}{\Gamma\left(\mathrm{p}+\mathrm{q}\right)}\:\Rightarrow \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{0}} \mathrm{x}\:\mathrm{sin}^{\mathrm{2n}−\mathrm{2}} \mathrm{x}\:\mathrm{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \mathrm{x}\:.\mathrm{sin}^{\mathrm{2}\left(\mathrm{n}−\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{1}} \mathrm{x}\:\mathrm{dx} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right).\Gamma\left(\mathrm{n}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{n}−\frac{\mathrm{1}}{\mathrm{2}}\right)}\:=\frac{\sqrt{\pi}}{\mathrm{2}\left(\mathrm{n}−\mathrm{1}\right)!}\Gamma\left(\mathrm{n}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}×\frac{\left(\mathrm{n}−\frac{\mathrm{3}}{\mathrm{2}}\right)!}{\left(\mathrm{n}−\mathrm{1}\right)!} \\$$