let-0-lt-a-lt-1-find-the-valueof-0-t-a-1-1-t-2-dt-

Question Number 63661 by mathmax by abdo last updated on 06/Jul/19

let 0<a<1 find the valueof ∫_0 ^∞ (t^(a−1) /(1+t^2 ))dt

$${let}\:\mathrm{0}<{a}<\mathrm{1}\:{find}\:{the}\:{valueof}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$

Commented by mathmax by abdo last updated on 07/Jul/19

changement t^2 =x give ∫_0 ^∞ (t^(a−1) /(1+t^2 ))dt =∫_0 ^∞ (((x^(1/2) )^(a−1) )/(1+x))(1/2)x^(−(1/2)) dx =(1/2)∫_0 ^∞ (x^(((a−1)/2)−(1/2)) /(1+x))dx =(1/2)∫_0 ^∞ (x^((a/2)−1) /(1+x))dx =(1/2) (π/(sin(((πa)/2)))) ⇒ ∫_0 ^∞ (t^(a−1) /(1+t^2 ))dt =(π/(2sin(((πa)/2)))) .

$${changement}\:{t}^{\mathrm{2}} \:={x}\:{give}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\left({x}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)^{{a}−\mathrm{1}} }{\mathrm{1}+{x}}\frac{\mathrm{1}}{\mathrm{2}}{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\frac{{a}−\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}+{x}}{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\frac{{a}}{\mathrm{2}}−\mathrm{1}} }{\mathrm{1}+{x}}{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\pi}{{sin}\left(\frac{\pi{a}}{\mathrm{2}}\right)}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:=\frac{\pi}{\mathrm{2}{sin}\left(\frac{\pi{a}}{\mathrm{2}}\right)}\:. \\ $$

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