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Question Number 142338 by rs4089 last updated on 30/May/21
∫_0 ^∞ ((sinx)/x^μ )dx =?
$$\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}^{\mu} }{dx}\:\:=?\:\:\: \\ $$
Answered by Dwaipayan Shikari last updated on 30/May/21
(1/x^μ )=(1/(Γ(μ)))∫_0 ^∞ e^(−xt) t^(μ−1) dt ∫_0 ^∞ ((sin(x))/x^μ )dx=(1/(Γ(μ)))∫_0 ^∞ ∫_0 ^∞ e^(−xt) t^(μ−1) sin(x)dxdt =(1/(2iΓ(μ)))∫_0 ^∞ (t^(μ−1) /(t−i))−(t^(μ−1) /(t+i))dt =(1/(2Γ(μ)))∫_0 ^∞ (g^((μ/2)−1) /(1+g))dg=(1/(2Γ(μ)))B((μ/2),1−(μ/2)) =(π/(2Γ(μ)sin(((πμ)/2))))
$$\frac{\mathrm{1}}{{x}^{\mu} }=\frac{\mathrm{1}}{\Gamma\left(\mu\right)}\int_{\mathrm{0}} ^{\infty} {e}^{−{xt}} {t}^{\mu−\mathrm{1}} {dt} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right)}{{x}^{\mu} }{dx}=\frac{\mathrm{1}}{\Gamma\left(\mu\right)}\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {e}^{−{xt}} {t}^{\mu−\mathrm{1}} {sin}\left({x}\right){dxdt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}\Gamma\left(\mu\right)}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mu−\mathrm{1}} }{{t}−{i}}−\frac{{t}^{\mu−\mathrm{1}} }{{t}+{i}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\Gamma\left(\mu\right)}\int_{\mathrm{0}} ^{\infty} \frac{{g}^{\frac{\mu}{\mathrm{2}}−\mathrm{1}} }{\mathrm{1}+{g}}{dg}=\frac{\mathrm{1}}{\mathrm{2}\Gamma\left(\mu\right)}\boldsymbol{{B}}\left(\frac{\mu}{\mathrm{2}},\mathrm{1}−\frac{\mu}{\mathrm{2}}\right) \\ $$$$=\frac{\pi}{\mathrm{2}\Gamma\left(\mu\right){sin}\left(\frac{\pi\mu}{\mathrm{2}}\right)} \\ $$
Commented by rs4089 last updated on 30/May/21
thanks a lot sir
$${thanks}\:{a}\:{lot}\:{sir}\: \\ $$
Answered by mathmax by abdo last updated on 30/May/21
f(a)=∫_0 ^∞ ((sinx)/x^a )dx ⇒f(a)=−Im(∫_0 ^∞ x^(−a) e^(−ix) dx) we have ∫_0 ^∞ x^(−a) .e^(−ix) dx =_(ix=t) ∫_0 ^∞ (−it)^(−a) e^(−t) (dt/i) =(−i)^(−a+1 ) ∫_0 ^∞ t^(−a) e^(−t) dt =(e^(−((iπ)/2)) )^(1−a) .Γ(1−a) =Γ(1−a).e^((a−1)(π/2)) =Γ(1−a){cos(a−1)(π/2)+isin(a−1)(π/2)} =Γ(1−a){sin(((πa)/2))−icos(((πa)/2))} ⇒ f(a)=cos(((πa)/2)).Γ(1−a) we know Γ(a).Γ(1−a)=(π/(sin(πa))) ⇒ Γ(1−a)=(π/(sin(πa).Γ(a))) ⇒f(a)=((πcos(((πa)/2)))/(sin(πa).Γ(a))) =(π/(Γ(a))).((cos(((πa)/2)))/(sin(πa))) with 0<a<1
$$\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sinx}}{\mathrm{x}^{\mathrm{a}} }\mathrm{dx}\:\Rightarrow\mathrm{f}\left(\mathrm{a}\right)=−\mathrm{Im}\left(\int_{\mathrm{0}} ^{\infty} \mathrm{x}^{−\mathrm{a}} \mathrm{e}^{−\mathrm{ix}} \mathrm{dx}\right)\:\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{−\mathrm{a}} .\mathrm{e}^{−\mathrm{ix}} \:\mathrm{dx}\:=_{\mathrm{ix}=\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\left(−\mathrm{it}\right)^{−\mathrm{a}} \:\mathrm{e}^{−\mathrm{t}} \:\:\frac{\mathrm{dt}}{\mathrm{i}} \\ $$$$=\left(−\mathrm{i}\right)^{−\mathrm{a}+\mathrm{1}\:} \int_{\mathrm{0}} ^{\infty} \:\:\mathrm{t}^{−\mathrm{a}} \:\:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt}\:\:=\left(\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{2}}} \right)^{\mathrm{1}−\mathrm{a}} \:.\Gamma\left(\mathrm{1}−\mathrm{a}\right) \\ $$$$=\Gamma\left(\mathrm{1}−\mathrm{a}\right).\mathrm{e}^{\left(\mathrm{a}−\mathrm{1}\right)\frac{\pi}{\mathrm{2}}} =\Gamma\left(\mathrm{1}−\mathrm{a}\right)\left\{\mathrm{cos}\left(\mathrm{a}−\mathrm{1}\right)\frac{\pi}{\mathrm{2}}+\mathrm{isin}\left(\mathrm{a}−\mathrm{1}\right)\frac{\pi}{\mathrm{2}}\right\} \\ $$$$=\Gamma\left(\mathrm{1}−\mathrm{a}\right)\left\{\mathrm{sin}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right)−\mathrm{icos}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right)\right\}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\mathrm{cos}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right).\Gamma\left(\mathrm{1}−\mathrm{a}\right)\:\:\mathrm{we}\:\mathrm{know}\:\Gamma\left(\mathrm{a}\right).\Gamma\left(\mathrm{1}−\mathrm{a}\right)=\frac{\pi}{\mathrm{sin}\left(\pi\mathrm{a}\right)}\:\Rightarrow \\ $$$$\Gamma\left(\mathrm{1}−\mathrm{a}\right)=\frac{\pi}{\mathrm{sin}\left(\pi\mathrm{a}\right).\Gamma\left(\mathrm{a}\right)}\:\Rightarrow\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi\mathrm{cos}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right)}{\mathrm{sin}\left(\pi\mathrm{a}\right).\Gamma\left(\mathrm{a}\right)} \\ $$$$=\frac{\pi}{\Gamma\left(\mathrm{a}\right)}.\frac{\mathrm{cos}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right)}{\mathrm{sin}\left(\pi\mathrm{a}\right)}\:\:\:\mathrm{with}\:\mathrm{0}<\mathrm{a}<\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 30/May/21
but sin(πa)=2sin(((πa)/2))cos(((πa)/2)) ⇒ ∫_0 ^∞ ((sinx)/x^a )dx =(π/(2 Γ(a)sin(((πa)/2))))
$$\mathrm{but}\:\mathrm{sin}\left(\pi\mathrm{a}\right)=\mathrm{2sin}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sinx}}{\mathrm{x}^{\mathrm{a}} }\mathrm{dx}\:=\frac{\pi}{\mathrm{2}\:\Gamma\left(\mathrm{a}\right)\mathrm{sin}\left(\frac{\pi\mathrm{a}}{\mathrm{2}}\right)} \\ $$$$ \\ $$

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