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Question-65729




Question Number 65729 by aliesam last updated on 02/Aug/19
Commented by mathmax by abdo last updated on 03/Aug/19
a)  let f(α) =∫_0 ^∞  e^(ix)  x^(−α)  dx  ⇒f(α) =_(ix =t)    ∫_0 ^∞   e^t  ((t/i))^(−α)  (dt/i)  =(1/i^(1−α) ) ∫_0 ^∞   e^t  t^(1−α−1)  dt =(1/i^(1−α) )Γ(1−α) the convergence of f(α)  is assured  b)∫_0 ^∞  cosx x^(−α)  dx =Re(∫_0 ^∞  e^(ix)  x^(−α) dx) =Re((1/i^(α−1) )Γ(1−α)) but  (1/i^(α−1) )Γ(1−α) =(1/((e^(i(π/2)) )^(α−1) ))Γ(1−α) =e^(−((iπ)/2)(α−1)) Γ(1−α)  =e^((iπ)/2)  e^(−((iπ)/2)α)  Γ(1−α) =i(cos(((πα)/2))−isin(((πα)/2)))Γ(1−α)  =sin(((πα)/2))Γ(1−α)+i cos(((πα)/2))Γ(1−α) ⇒  ∫_0 ^∞ cosx x^(−α) dx =sin(((πα)/2))Γ(1−α)
$$\left.{a}\right)\:\:{let}\:{f}\left(\alpha\right)\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{{ix}} \:{x}^{−\alpha} \:{dx}\:\:\Rightarrow{f}\left(\alpha\right)\:=_{{ix}\:={t}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{{t}} \:\left(\frac{{t}}{{i}}\right)^{−\alpha} \:\frac{{dt}}{{i}} \\ $$$$=\frac{\mathrm{1}}{{i}^{\mathrm{1}−\alpha} }\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{{t}} \:{t}^{\mathrm{1}−\alpha−\mathrm{1}} \:{dt}\:=\frac{\mathrm{1}}{{i}^{\mathrm{1}−\alpha} }\Gamma\left(\mathrm{1}−\alpha\right)\:{the}\:{convergence}\:{of}\:{f}\left(\alpha\right) \\ $$$${is}\:{assured} \\ $$$$\left.{b}\right)\int_{\mathrm{0}} ^{\infty} \:{cosx}\:{x}^{−\alpha} \:{dx}\:={Re}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{{ix}} \:{x}^{−\alpha} {dx}\right)\:={Re}\left(\frac{\mathrm{1}}{{i}^{\alpha−\mathrm{1}} }\Gamma\left(\mathrm{1}−\alpha\right)\right)\:{but} \\ $$$$\frac{\mathrm{1}}{{i}^{\alpha−\mathrm{1}} }\Gamma\left(\mathrm{1}−\alpha\right)\:=\frac{\mathrm{1}}{\left({e}^{{i}\frac{\pi}{\mathrm{2}}} \right)^{\alpha−\mathrm{1}} }\Gamma\left(\mathrm{1}−\alpha\right)\:={e}^{−\frac{{i}\pi}{\mathrm{2}}\left(\alpha−\mathrm{1}\right)} \Gamma\left(\mathrm{1}−\alpha\right) \\ $$$$={e}^{\frac{{i}\pi}{\mathrm{2}}} \:{e}^{−\frac{{i}\pi}{\mathrm{2}}\alpha} \:\Gamma\left(\mathrm{1}−\alpha\right)\:={i}\left({cos}\left(\frac{\pi\alpha}{\mathrm{2}}\right)−{isin}\left(\frac{\pi\alpha}{\mathrm{2}}\right)\right)\Gamma\left(\mathrm{1}−\alpha\right) \\ $$$$={sin}\left(\frac{\pi\alpha}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−\alpha\right)+{i}\:{cos}\left(\frac{\pi\alpha}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−\alpha\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} {cosx}\:{x}^{−\alpha} {dx}\:={sin}\left(\frac{\pi\alpha}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−\alpha\right) \\ $$$$ \\ $$
Commented by aliesam last updated on 03/Aug/19
thank you sir.but the first question   ineed to prove the integral is ineed  convergent before computing its value  concerning .the second question,i meant by complex integration  is by using the residue theorem
$${thank}\:{you}\:{sir}.{but}\:{the}\:{first}\:{question} \\ $$$$\:{ineed}\:{to}\:{prove}\:{the}\:{integral}\:{is}\:{ineed} \\ $$$${convergent}\:{before}\:{computing}\:{its}\:{value} \\ $$$${concerning}\:.{the}\:{second}\:{question},{i}\:{meant}\:{by}\:{complex}\:{integration} \\ $$$${is}\:{by}\:{using}\:{the}\:{residue}\:{theorem} \\ $$$$ \\ $$
Commented by aliesam last updated on 04/Aug/19