advanced-calculus-0-x-b-1-1-x-a-s-dx-b-a-s-b-a-a-s-

FacebookTweetPin

Question Number 131104 by mnjuly1970 last updated on 01/Feb/21

$$\dots advancedcalculus\dots$$$${\int }_0^{\mathrm{\infty }}\frac{x^{b-1}}{{(1+x^a)}^s}dx\stackrel{???}{=}\frac{\mathrm{\Gamma }\left(\frac{b}{a}\right)\mathrm{\Gamma }(s-\frac{b}{a})}{a\mathrm{\Gamma }\left(s\right)}$$

Answered by Ar Brandon last updated on 01/Feb/21

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{b}-1}}{{(1+{\mathrm{x}}^{\mathrm{a}})}^{\mathrm{s}}}\mathrm{dx},{\mathrm{x}}^{\mathrm{a}}=\mathrm{t}\Rightarrow {\mathrm{ax}}^{\mathrm{a}-1}\mathrm{dx}=\mathrm{dt}$$$$=\frac{1}{\mathrm{a}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{b}-\mathrm{a}}}{{(1+\mathrm{t})}^{\mathrm{s}}}\mathrm{dt}=\frac{1}{\mathrm{a}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{\mathrm{b}-\mathrm{a}}{\mathrm{a}}}}{{(1+\mathrm{t})}^{\mathrm{s}}}\mathrm{dt}$$$$=\frac{1}{\mathrm{a}}\beta (\frac{\mathrm{b}}{\mathrm{a}},\frac{\mathrm{as}-\mathrm{b}}{\mathrm{a}})=\frac{\mathrm{\Gamma }\left(\frac{\mathrm{b}}{\mathrm{a}}\right)\mathrm{\Gamma }\left(\frac{\mathrm{as}-\mathrm{b}}{\mathrm{a}}\right)}{\mathrm{a}\mathrm{\Gamma }\left(\mathrm{s}\right)}$$$$=\frac{\mathrm{\Gamma }\left(\frac{\mathrm{b}}{\mathrm{a}}\right)\mathrm{\Gamma }(\mathrm{s}-\frac{\mathrm{b}}{\mathrm{a}})}{\mathrm{a}\mathrm{\Gamma }\left(\mathrm{s}\right)}$$

Commented by mnjuly1970 last updated on 01/Feb/21

$$thankyoumrbrandon..$$

Commented by Ar Brandon last updated on 01/Feb/21

With pleasure Sir��

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

FacebookTweetPin