Question-142689

Question Number 142689 by rs4089 last updated on 04/Jun/21

Answered by Dwaipayan Shikari last updated on 04/Jun/21

∫_0 ^∞ ((sin(x))/x)f(x)dx=∫_0 ^(π/2) f(x)dx f(x±π)=f(x) let f(x)=sin^(2n−2) x ∫_0 ^(π/2) sin^(2n−2) x dx=((Γ(n−(1/2))Γ((1/2)))/(2Γ(n)))=((√π)/2).((Γ(n−(1/2)))/((n−1)!))

$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right)}{{x}}{f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({x}\right){dx}\:\:\:\:{f}\left({x}\pm\pi\right)={f}\left({x}\right) \\ $$$${let}\:{f}\left({x}\right)={sin}^{\mathrm{2}{n}−\mathrm{2}} {x}\:\:\:\:\:\: \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{n}−\mathrm{2}} {x}\:{dx}=\frac{\Gamma\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left({n}\right)}=\frac{\sqrt{\pi}}{\mathrm{2}}.\frac{\Gamma\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\left({n}−\mathrm{1}\right)!} \\ $$

Commented by mindispower last updated on 04/Jun/21

∫_0 ^∞ ((sin(x))/x)f(x)dx=∫_0 ^(π/2) f(x)dx,let f(−x)=f(x),f(x+π)=f(x) ∫_0 ^∞ ((sin(x))/x)f(x)=(1/2)Σ_(k=−∞) ^∞ ∫_(kπ) ^((k+1)π) ((sin(x))/x)f(x)dx x→kπ+t =(1/2)Σ_(−∞) ^∞ ∫_0 ^π (((−1)^k sin(t))/(kπ+t))f(t)dt=A Σ_(−∞) ^∞ (((−1)^k )/(kπ+t))=(1/(sin(t))) (1/2)∫_0 ^π f(t)=(1/2)(.∫_(−(π/2)) ^0 f(t)dt+∫_0 ^(π/2) f(t)dt)=A =(1/2).2∫_0 ^(π/2) f(t)dt=∫_0 ^(π/2) f(t).dt

$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right)}{{x}}{f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({x}\right){dx},{let}\:{f}\left(−{x}\right)={f}\left({x}\right),{f}\left({x}+\pi\right)={f}\left({x}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right)}{{x}}{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=−\infty} {\overset{\infty} {\sum}}\int_{{k}\pi} ^{\left({k}+\mathrm{1}\right)\pi} \frac{{sin}\left({x}\right)}{{x}}{f}\left({x}\right){dx} \\ $$$${x}\rightarrow{k}\pi+{t} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{−\infty} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\pi} \frac{\left(−\mathrm{1}\right)^{{k}} {sin}\left({t}\right)}{{k}\pi+{t}}{f}\left({t}\right){dt}={A} \\ $$$$\underset{−\infty} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}\pi+{t}}=\frac{\mathrm{1}}{{sin}\left({t}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} {f}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(.\int_{−\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} {f}\left({t}\right){dt}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({t}\right){dt}\right)={A} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({t}\right){dt}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({t}\right).{dt} \\ $$

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

we have ∫_0 ^(π/2) cos^(2p−1) x sin^(2q−1) x dx =(1/2)((Γ(p).Γ(q))/(Γ(p+q))) ⇒ ∫_0 ^(π/2) cos^0 x sin^(2n−2) x dx =∫_0 ^(π/2) cos^(2.(1/2)−1) x .sin^(2(n−(1/2))−1) x dx =(1/2)((Γ((1/2)).Γ(n−(1/2)))/(Γ((1/2)+n−(1/2)))) =((√π)/(2(n−1)!))Γ(n−(1/2))=((√π)/2)×(((n−(3/2))!)/((n−1)!))

$$\mathrm{we}\:\mathrm{have}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{2p}−\mathrm{1}} \mathrm{x}\:\mathrm{sin}^{\mathrm{2q}−\mathrm{1}} \mathrm{x}\:\mathrm{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\frac{\Gamma\left(\mathrm{p}\right).\Gamma\left(\mathrm{q}\right)}{\Gamma\left(\mathrm{p}+\mathrm{q}\right)}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{0}} \mathrm{x}\:\mathrm{sin}^{\mathrm{2n}−\mathrm{2}} \mathrm{x}\:\mathrm{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \mathrm{x}\:.\mathrm{sin}^{\mathrm{2}\left(\mathrm{n}−\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{1}} \mathrm{x}\:\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right).\Gamma\left(\mathrm{n}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{n}−\frac{\mathrm{1}}{\mathrm{2}}\right)}\:=\frac{\sqrt{\pi}}{\mathrm{2}\left(\mathrm{n}−\mathrm{1}\right)!}\Gamma\left(\mathrm{n}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}×\frac{\left(\mathrm{n}−\frac{\mathrm{3}}{\mathrm{2}}\right)!}{\left(\mathrm{n}−\mathrm{1}\right)!} \\ $$

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