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Question Number 200691 by sonukgindia last updated on 22/Nov/23

Commented by aleks041103 last updated on 22/Nov/23

$$a,b\in \mathbb{Z}?$$$$orjusta,b\in \mathbb{R}$$

Answered by witcher3 last updated on 22/Nov/23

$$\mathrm{if}\mathrm{b}⩾0={\mathrm{x}}^{2\mathrm{b}}-{\mathrm{x}}^{\mathrm{b}}+1\sim 1$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^{\mathrm{a}}}{{\mathrm{x}}^{2\mathrm{b}}-{\mathrm{x}}^{\mathrm{b}}+1}\sim {\mathrm{x}}^{\mathrm{a}}\mathrm{integrable}\mathrm{if}\mathrm{a}+1>0$$$$\mathrm{a}>-1=\mathrm{in}\mathrm{\infty }{\mathrm{x}}^{2\mathrm{b}}-{\mathrm{x}}^{\mathrm{b}}+1\sim {\mathrm{x}}^{2\mathrm{b}}$$$$\mathrm{f}\left(\mathrm{x}\right)\sim {\mathrm{x}}^{\mathrm{a}-2\mathrm{b}}\mathrm{integrabl}\mathrm{if}2\mathrm{b}-\mathrm{a}>1$$$$2\mathrm{b}>\mathrm{a}+1\mathrm{since}\mathrm{b}>0\mid 2\mathrm{b}\mid >\mid \mathrm{a}+1\mid$$$$\mathrm{b}<0$$$${\mathrm{x}}^{2\mathrm{b}}-{\mathrm{x}}^{\mathrm{b}}+1\sim {\mathrm{x}}^{2\mathrm{b}}\mathrm{in}\mathrm{zero}$$$$\mathrm{f}\left(\mathrm{x}\right)\sim {\mathrm{x}}^{\mathrm{a}-2\mathrm{b}}\mathrm{integrabl}\iff \mathrm{a}-2\mathrm{b}>1$$$$\mathrm{in}\mathrm{\infty }\mathrm{f}\left(\mathrm{x}\right)=\sim {\mathrm{x}}^{\mathrm{a}}[\mathrm{integrabl}\mathrm{if}\mathrm{a}<-1$$$$2\mathrm{b}<\mathrm{a}+1<0$$$$\Rightarrow \mid 2\mathrm{b}\mid >\mid \mathrm{a}+1\mid \iff \mathrm{\forall }(\mathrm{a},\mathrm{b})\in \mathbb{R}\mid 2\mathrm{b}\mid >\mid \mathrm{a}+1\mid$$$$\mathrm{f}\left(\mathrm{x}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{a}}}{{\mathrm{x}}^{2\mathrm{b}}-{\mathrm{x}}^{\mathrm{b}}+1}\mathrm{dx}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{a}}({\mathrm{x}}^{\mathrm{b}}+1)}{{\mathrm{x}}^{3\mathrm{b}}+1}\mathrm{dx}$$$${\mathrm{x}}^{3\mathrm{b}}=\mathrm{t}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{\mathrm{a}+\mathrm{b}}{3\mathrm{b}}}+{\mathrm{t}}^{\frac{\mathrm{a}}{3\mathrm{b}}}}{1+\mathrm{t}}.\frac{{\mathrm{t}}^{\frac{1}{3\mathrm{b}}-1}}{1+\mathrm{t}}\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{\mathrm{a}+\mathrm{b}+1}{3\mathrm{b}}-1}+{\mathrm{t}}^{\frac{\mathrm{a}+1}{3\mathrm{b}}-1}}{1+\mathrm{t}}=\beta (\frac{\mathrm{a}+\mathrm{b}+1}{3\mathrm{b}},1-\frac{\mathrm{a}+\mathrm{b}+1}{3\mathrm{b}})$$$$+\beta (\frac{\mathrm{a}+1}{3\mathrm{b}},1-\frac{\mathrm{a}+1}{3\mathrm{b}})=\pi (\frac{1}{\sin \left(\frac{\pi (\mathrm{a}+1)}{3\mathrm{b}}\right)}+\frac{1}{\sin (\frac{\pi }{3\mathrm{b}}(\mathrm{a}+\mathrm{b}+1)})$$$$$$$$$$$$$$

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