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Question Number 63892 by mathmax by abdo last updated on 10/Jul/19

$$calculateA={\int }_0^{\mathrm{\infty }}\frac{x^{2017}}{1+x^{2019}}dxandB={\int }_0^{\mathrm{\infty }}\frac{x^{2019}}{1+x^{2021}}dx$$$$calculatethefraction\frac{A}{B}$$

Commented by Prithwish sen last updated on 11/Jul/19

$${\mathrm{A}}^{\prime }\left(\mathrm{t}\right)={\int }_0^{\mathrm{\infty }}\frac{\partial }{\partial \mathrm{t}}\left[\frac{{\mathrm{x}}^{\mathrm{t}-2}}{1+{\mathrm{x}}^{\mathrm{t}}}\right]\mathrm{dx}={\int }_0^{\mathrm{\infty }}\frac{1}{{\mathrm{x}}^2}\frac{\partial }{\partial \mathrm{t}}[1-\frac{1}{1+{\mathrm{x}}^{\mathrm{t}}}]\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{1}{{\mathrm{x}}^2}.\frac{1}{{(1+{\mathrm{x}}^{\mathrm{t}})}^2}\mathrm{dx}$$

Commented by mathmax by abdo last updated on 12/Jul/19

$$letf\left(t\right)={\int }_0^{\mathrm{\infty }}\frac{x^{t-2}}{1+x^t}dtwitht>2changement{x}^t=ugive$$$$x=u^{\frac{1}{t}}\Rightarrow f\left(t\right)={\int }_0^{\mathrm{\infty }}\frac{{\left(u^{\frac{1}{t}}\right)}^{t-2}}{1+u}\frac{1}{t}{u}^{\frac{1}{t}-1}du$$$$=\frac{1}{t}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{t-2}{t}+\frac{1}{t}-1}}{1+u}du=\frac{1}{t}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{t-1}{t}-1}}{1+u}duwehavet>2\Rightarrow$$$$\frac{1}{t}<\frac{1}{2}\Rightarrow -\frac{1}{t}>-\frac{1}{2}\Rightarrow 1-\frac{1}{t}>\frac{1}{2}\Rightarrow \frac{t-1}{t}>\frac{1}{2}and\frac{t-1}{t}=1-\frac{1}{t}<1\Rightarrow$$$$f\left(t\right)=\frac{1}{t}\frac{\pi }{sin\left(\pi \left(\frac{t-1}{t}\right)\right)}=\frac{\pi }{tsin(\pi -\frac{\pi }{t})}=\frac{\pi }{tsin\left(\frac{\pi }{t}\right)}$$$${\int }_0^{\mathrm{\infty }}\frac{x^{2017}}{1+x^{2019}}={\int }_0^{\mathrm{\infty }}\frac{x^{2019-2}}{1+x^{2919}}=f\left(2019\right)=\frac{\pi }{2019sin\left(\frac{\pi }{2019}\right)}=A$$$$B=f\left(2021\right)=\frac{\pi }{2021sin\left(\frac{\pi }{2021}\right)}$$

Commented by mathmax by abdo last updated on 12/Jul/19

$$\frac{A}{B}=\frac{\pi }{2019sin\left(\frac{\pi }{2019}\right)}\times \frac{2021sin\left(\frac{\pi }{2021}\right)}{\pi }=\frac{2021}{2019}\times \frac{sin\left(\frac{\pi }{2021}\right)}{sin\left(\frac{\pi }{2019}\right)}$$

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