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Question Number 134474 by rs4089 last updated on 04/Mar/21
Evaluate ∫_0 ^∞ ((sinx)/x^a )dx
$${Evaluate}\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}^{{a}} }{dx} \\ $$
Answered by Dwaipayan Shikari last updated on 04/Mar/21
∫_0 ^∞ ((sinx)/x^a )dx =(1/(Γ(a)))∫_0 ^∞ ∫_0 ^∞ t^(a−1) e^(−xt) sinxdtdx =(1/(2iΓ(a)))∫_0 ^∞ ∫_0 ^∞ t^(a−1) e^(−x(t−i)) −t^(a−1) e^(−x(t+i)) dx dt =(1/(2iΓ(a)))∫_0 ^∞ (t^(a−1) /(t−i))−(t^(a−1) /(t+i))dt =(1/(Γ(a)))∫_0 ^∞ (t^(a−1) /(t^2 +1))dt =(1/(2Γ(a)))∫_0 ^∞ (u^((a/2)−1) /(u+1))du t^2 =u =(1/(2Γ(a))).(π/(sin((π/2)a)))
$$\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}^{{a}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\Gamma\left({a}\right)}\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {t}^{{a}−\mathrm{1}} {e}^{−{xt}} {sinxdtdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}\Gamma\left({a}\right)}\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {t}^{{a}−\mathrm{1}} {e}^{−{x}\left({t}−{i}\right)} −{t}^{{a}−\mathrm{1}} {e}^{−{x}\left({t}+{i}\right)} {dx}\:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}\Gamma\left({a}\right)}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{a}−\mathrm{1}} }{{t}−{i}}−\frac{{t}^{{a}−\mathrm{1}} }{{t}+{i}}{dt} \\ $$$$=\frac{\mathrm{1}}{\Gamma\left({a}\right)}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{a}−\mathrm{1}} }{{t}^{\mathrm{2}} +\mathrm{1}}{dt}\:\:=\frac{\mathrm{1}}{\mathrm{2}\Gamma\left({a}\right)}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{{a}}{\mathrm{2}}−\mathrm{1}} }{{u}+\mathrm{1}}{du}\:\:\:\:\:\:{t}^{\mathrm{2}} ={u} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\Gamma\left({a}\right)}.\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{2}}{a}\right)} \\ $$
Answered by mnjuly1970 last updated on 04/Mar/21
𝛗=−im∫_0 ^( ∞) (e^(−ix) /x^a )dx=−im∫_(0 ) ^( ∞) e^(−ix) x^(−a) dx =−im(L [x^(−a) ])∣_(s=i) =−im{((Γ(1−a))/s^(1−a) )∣_(s=i) } =−im{((Γ(a)Γ(1−a))/(Γ(a)e^(((iπ)/2)(1−a)) ))} =−im{(π/(Γ(a)sin(πa)(sin((π/2)a)+icos((π/2)a)))) =−im{((π(sin((π/2)a)−icos((π/2)a)))/(Γ(a)sin(πa)))} =((πcos((π/2)a))/(2Γ(a)sin((π/2)a)cos((π/2)a)))=(π/(2Γ(a)sin((π/2)a))) ...
$$\:\:\:\boldsymbol{\phi}=−{im}\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{−{ix}} }{{x}^{{a}} }{dx}=−{im}\int_{\mathrm{0}\:} ^{\:\infty} {e}^{−{ix}} {x}^{−{a}} {dx} \\ $$$$\:\:\:\:=−{im}\left(\mathscr{L}\:\left[{x}^{−{a}} \right]\right)\mid_{{s}={i}} \\ $$$$\:\:\:\:\:=−{im}\left\{\frac{\Gamma\left(\mathrm{1}−{a}\right)}{{s}^{\mathrm{1}−{a}} }\mid_{{s}={i}} \right\} \\ $$$$\:\:\:\:=−{im}\left\{\frac{\Gamma\left({a}\right)\Gamma\left(\mathrm{1}−{a}\right)}{\Gamma\left({a}\right){e}^{\frac{{i}\pi}{\mathrm{2}}\left(\mathrm{1}−{a}\right)} }\right\}\:\:\: \\ $$$$\:\:\:\:\:\:\:=−{im}\left\{\frac{\pi}{\Gamma\left({a}\right){sin}\left(\pi{a}\right)\left({sin}\left(\frac{\pi}{\mathrm{2}}{a}\right)+{icos}\left(\frac{\pi}{\mathrm{2}}{a}\right)\right)}\right. \\ $$$$=−{im}\left\{\frac{\pi\left({sin}\left(\frac{\pi}{\mathrm{2}}{a}\right)−{icos}\left(\frac{\pi}{\mathrm{2}}{a}\right)\right)}{\Gamma\left({a}\right){sin}\left(\pi{a}\right)}\right\} \\ $$$$=\frac{\pi{cos}\left(\frac{\pi}{\mathrm{2}}{a}\right)}{\mathrm{2}\Gamma\left({a}\right){sin}\left(\frac{\pi}{\mathrm{2}}{a}\right){cos}\left(\frac{\pi}{\mathrm{2}}{a}\right)}=\frac{\pi}{\mathrm{2}\Gamma\left({a}\right){sin}\left(\frac{\pi}{\mathrm{2}}{a}\right)}\:… \\ $$$$\:\:\:\:\:\: \\ $$
Answered by mathmax by abdo last updated on 27/Mar/21
Φ=∫_0 ^∞ ((sinx)/x^a )dx =−Im(∫_0 ^∞ x^(−a) e^(−ix) dx) we have ∫_0 ^∞ x^(−a) e^(−ix) dx =_(ix=t→x=−it) ∫_0 ^∞ (−it)^(−a) e^(−t) (−i)dt =(−i)^(−a+1) ∫_0 ^∞ t^(1−a−1) e^(−t) dt =(e^(−((iπ)/2)) )^((1−a)) Γ(1−a) =e^(i(π/2)(a−1)) .Γ(1−a) =Γ(1−a) {cos(((π(a−1))/2))+isin(((π(a−1))/2))} ⇒ Φ=−Γ(1−a)sin(((π(a−1))/2)) =sin((π/2)(1−a))×Γ(1−a)
$$\Phi=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sinx}}{\mathrm{x}^{\mathrm{a}} }\mathrm{dx}\:=−\mathrm{Im}\left(\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{−\mathrm{a}} \:\mathrm{e}^{−\mathrm{ix}} \:\mathrm{dx}\right)\:\mathrm{we}\:\mathrm{have} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{−\mathrm{a}} \:\mathrm{e}^{−\mathrm{ix}} \:\mathrm{dx}\:\:=_{\mathrm{ix}=\mathrm{t}\rightarrow\mathrm{x}=−\mathrm{it}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\left(−\mathrm{it}\right)^{−\mathrm{a}} \:\mathrm{e}^{−\mathrm{t}} \:\left(−\mathrm{i}\right)\mathrm{dt} \\ $$$$=\left(−\mathrm{i}\right)^{−\mathrm{a}+\mathrm{1}} \:\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{1}−\mathrm{a}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt}\:=\left(\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{2}}} \right)^{\left(\mathrm{1}−\mathrm{a}\right)} \:\Gamma\left(\mathrm{1}−\mathrm{a}\right) \\ $$$$=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}\left(\mathrm{a}−\mathrm{1}\right)} \:.\Gamma\left(\mathrm{1}−\mathrm{a}\right)\:=\Gamma\left(\mathrm{1}−\mathrm{a}\right)\:\left\{\mathrm{cos}\left(\frac{\pi\left(\mathrm{a}−\mathrm{1}\right)}{\mathrm{2}}\right)+\mathrm{isin}\left(\frac{\pi\left(\mathrm{a}−\mathrm{1}\right)}{\mathrm{2}}\right)\right\}\:\Rightarrow \\ $$$$\Phi=−\Gamma\left(\mathrm{1}−\mathrm{a}\right)\mathrm{sin}\left(\frac{\pi\left(\mathrm{a}−\mathrm{1}\right)}{\mathrm{2}}\right)\:=\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\mathrm{a}\right)\right)×\Gamma\left(\mathrm{1}−\mathrm{a}\right) \\ $$

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