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Question Number 200619 by sonukgindia last updated on 21/Nov/23

Answered by witcher3 last updated on 21/Nov/23

$${\mathrm{I}}_9=2{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{a}}}{\mathrm{c}+{\mathrm{kx}}^{\mathrm{b}}}\mathrm{dx}$$$$\mathrm{in}\mathrm{\infty }\mathrm{c}+{\mathrm{kx}}^{\mathrm{b}}\sim {\mathrm{x}}^{\mathrm{b}};\frac{{\mathrm{x}}^{\mathrm{a}}}{\mathrm{c}+{\mathrm{kx}}^{\mathrm{b}}}\sim {\mathrm{x}}^{\mathrm{a}-\mathrm{b}}\mathrm{integrabl}\Rightarrow \mathrm{a}-\mathrm{b}<-1$$$$\Rightarrow \mathrm{b}>1+\mathrm{a}\Rightarrow \mathrm{b}⩾\mathrm{a}+2$$$$\frac{\mathrm{k}}{\mathrm{c}}{\mathrm{x}}^{\mathrm{b}}=\mathrm{t}\Rightarrow \mathrm{x}={\left(\frac{\mathrm{ct}}{\mathrm{k}}\right)}^{\frac{1}{\mathrm{b}}}\Rightarrow \mathrm{dx}={\left(\frac{\mathrm{c}}{\mathrm{k}}\right)}^{\frac{1}{\mathrm{b}}}\frac{{\mathrm{t}}^{\frac{1}{\mathrm{b}}-1}}{\mathrm{b}}$$$${\mathrm{I}}_9=2{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{\mathrm{b}}-1}.{\left(\frac{\mathrm{ct}}{\mathrm{k}}\right)}^{\frac{\mathrm{a}}{\mathrm{b}}}.{\left(\frac{\mathrm{c}}{\mathrm{k}}\right)}^{\frac{1}{\mathrm{b}}}}{\mathrm{c}(1+\mathrm{t})\mathrm{b}}$$$$=\frac{2}{\mathrm{cb}}.{\left(\frac{\mathrm{c}}{\mathrm{k}}\right)}^{\frac{\mathrm{a}+1}{\mathrm{b}}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1+\mathrm{a}}{\mathrm{b}}-1}}{1+\mathrm{t}}\mathrm{dt}$$$$=\frac{2}{\mathrm{cb}}{\left(\frac{\mathrm{c}}{\mathrm{k}}\right)}^{\frac{\mathrm{a}+1}{\mathrm{b}}}.\beta (\frac{1+\mathrm{a}}{\mathrm{b}},1-\frac{1+\mathrm{a}}{\mathrm{b}})$$$$=\frac{2}{\mathrm{cb}}{\left(\frac{\mathrm{c}}{\mathrm{k}}\right)}^{\frac{\mathrm{a}+1}{\mathrm{b}}}.\frac{\pi }{\sin \left(\frac{1+\mathrm{a}}{\mathrm{b}}\pi \right)}$$$$$$$$$$$$$$

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