Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

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

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

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