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Question Number 63662 by mathmax by abdo last updated on 06/Jul/19

let A_n =∫_0 ^∞ (x^(a−1) /(1+x^n ))dx with n integr and n≥2 and 0<a<1 1) calculate A_n 2) find the values of ∫_0 ^∞ (x^(a−1) /(1+x^2 ))dx and ∫_0 ^∞ (x^(a−1) /(1+x^3 ))dx 3)calculate ∫_0 ^∞ (dx/((√x)(1+x^4 ))) and ∫_0 ^∞ (dx/((^3 (√x^2 ))(1+x^4 )))

$$\:{let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{{a}−\mathrm{1}} }{\mathrm{1}+{x}^{{n}} }{dx}\:\:{with}\:{n}\:{integr}\:{and}\:{n}\geqslant\mathrm{2}\:\:{and}\:\mathrm{0}<{a}<\mathrm{1} \\ $$ $$\left.\mathrm{1}\right)\:{calculate}\:{A}_{{n}} \\ $$ $$\left.\mathrm{2}\right)\:{find}\:{the}\:{values}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{a}−\mathrm{1}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:{and}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{a}−\mathrm{1}} }{\mathrm{1}+{x}^{\mathrm{3}} }{dx} \\ $$ $$\left.\mathrm{3}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\sqrt{{x}}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)}\:\:{and}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left(^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} }\right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)} \\ $$

Commented bymathmax by abdo last updated on 10/Jul/19

1) we have A_n =∫_0 ^∞ (x^(a−1) /(1+x^n ))dx changement x^n =t give x=t^(1/n) ⇒ A_n =∫_0 ^∞ (((t^(1/n) )^(a−1) )/(1+t)) (1/n)t^((1/n)−1) dt =(1/n)∫_0 ^∞ (t^(((a−1)/n)+(1/n)−1) /(1+t))dt =(1/n)∫_0 ^∞ (t^((a/n)−1) /(1+t))dt =(1/n) (π/(sin(((πa)/n)))) by use of result ∫_0 ^∞ (t^(α−1) /(1+t))dt =(π/(sin(πα))) 2)∫_0 ^∞ (x^(a−1) /(1+x^2 ))dx =A_2 =(π/(2sin(((πa)/2)))) ∫_0 ^∞ (x^(a−1) /(1+x^3 )) dx =A_3 =(π/(3sin(((πa)/3)))) 3)∫_0 ^∞ (dx/((√x)(1+x^4 ))) =∫_0 ^∞ (x^(−(1/2)) /(1+x^4 )) dx =∫_0 ^∞ (x^((1/2)−1) /(1+x^4 )) (a=(1/2) and n=4) = (π/(4sin((π/4)))) =(π/(4 ((√2)/2))) =(π/(2(√2))) . ∫_0 ^∞ (dx/((^3 (√x^2 ))(1+x^4 ))) =∫_0 ^∞ (x^(−(2/3)) /(1+x^4 ))dx =∫_0 ^∞ (x^((1/3)−1) /(1+x^4 ))(→n=4 and a=(1/3)) =(π/(4sin((π/6)))) =(π/(4.(1/2))) =(π/2)

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{{a}−\mathrm{1}} }{\mathrm{1}+{x}^{{n}} }{dx}\:\:{changement}\:{x}^{{n}} ={t}\:{give}\:{x}={t}^{\frac{\mathrm{1}}{{n}}} \:\Rightarrow \\ $$ $${A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{\left({t}^{\frac{\mathrm{1}}{{n}}} \right)^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}\:\frac{\mathrm{1}}{{n}}{t}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} {dt}\:=\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{{a}−\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt} \\ $$ $$=\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{{a}}{{n}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\frac{\mathrm{1}}{{n}}\:\frac{\pi}{{sin}\left(\frac{\pi{a}}{{n}}\right)}\:{by}\:{use}\:{of}\:{result}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{\alpha−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\frac{\pi}{{sin}\left(\pi\alpha\right)} \\ $$ $$\left.\mathrm{2}\right)\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{a}−\mathrm{1}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:={A}_{\mathrm{2}} =\frac{\pi}{\mathrm{2}{sin}\left(\frac{\pi{a}}{\mathrm{2}}\right)} \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{{a}−\mathrm{1}} }{\mathrm{1}+{x}^{\mathrm{3}} }\:{dx}\:={A}_{\mathrm{3}} =\frac{\pi}{\mathrm{3}{sin}\left(\frac{\pi{a}}{\mathrm{3}}\right)} \\ $$ $$\left.\mathrm{3}\right)\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\sqrt{{x}}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}+{x}^{\mathrm{4}} }\:{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\mathrm{1}+{x}^{\mathrm{4}} }\:\:\left({a}=\frac{\mathrm{1}}{\mathrm{2}}\:\:{and}\:{n}=\mathrm{4}\right) \\ $$ $$=\:\frac{\pi}{\mathrm{4}{sin}\left(\frac{\pi}{\mathrm{4}}\right)}\:=\frac{\pi}{\mathrm{4}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:. \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\left(^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} }\right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{−\frac{\mathrm{2}}{\mathrm{3}}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} }{\mathrm{1}+{x}^{\mathrm{4}} }\left(\rightarrow{n}=\mathrm{4}\:{and}\:{a}=\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$ $$=\frac{\pi}{\mathrm{4}{sin}\left(\frac{\pi}{\mathrm{6}}\right)}\:=\frac{\pi}{\mathrm{4}.\frac{\mathrm{1}}{\mathrm{2}}}\:=\frac{\pi}{\mathrm{2}} \\ $$

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