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Question Number 143808 by mnjuly1970 last updated on 18/Jun/21

Answered by Dwaipayan Shikari last updated on 18/Jun/21

$$\xi \left(a\right)={\int }_0^{\mathrm{\infty }}\frac{1}{{(1+x^a)}^a}dx$$$$=\frac{1}{a}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{1}{a}-1}}{{(1+u)}^a}du$$$$=\frac{1}{a}.\frac{\mathrm{\Gamma }(a-\frac{1}{a})\mathrm{\Gamma }\left(\frac{1}{a}\right)}{\mathrm{\Gamma }\left(a\right)}=\frac{\mathrm{\Gamma }(a-\frac{1}{a})\mathrm{\Gamma }\left(\frac{1}{a}\right)}{\mathrm{\Gamma }(a+1)}$$$$\xi (1+\sqrt{2})=\frac{\mathrm{\Gamma }(\sqrt{2}+1-\sqrt{2}+1)\mathrm{\Gamma }(\sqrt{2}-1)}{(\sqrt{2}+1)\mathrm{\Gamma }(\sqrt{2}+1)}=\frac{\mathrm{\Gamma }\left(\sqrt{2}\right)}{\mathrm{\Gamma }(\sqrt{2}+1)}=\frac{1}{\sqrt{2}}$$$$$$

Commented by mnjuly1970 last updated on 18/Jun/21

$$thankyousomuch..$$

Answered by mathmax by abdo last updated on 18/Jun/21

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{(1+{\mathrm{x}}^{1+\sqrt{2}})}^{1+\sqrt{2}}}\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{(1+{\mathrm{x}}^{\mathrm{a}})}^{\mathrm{a}}}\Rightarrow \mathrm{I}=\mathrm{f}(1+\sqrt{2})$$$$\mathrm{f}\left(\mathrm{a}\right){=}_{{\mathrm{x}}^{\mathrm{a}}=\mathrm{t}}\frac{1}{\mathrm{a}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{\mathrm{a}}-1}}{{(1+\mathrm{t})}^{\mathrm{a}}}\mathrm{dt}\mathrm{we}\mathrm{know}\mathrm{B}(\mathrm{x},\mathrm{y})={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{x}-1}}{{(1+\mathrm{t})}^{\mathrm{x}+\mathrm{y}}}\mathrm{dt}$$$$\Rightarrow \mathrm{x}=\frac{1}{\mathrm{a}}\mathrm{and}\mathrm{x}+\mathrm{y}=\mathrm{a}\Rightarrow \mathrm{y}=\mathrm{a}-\frac{1}{\mathrm{a}}\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\mathrm{B}(\frac{1}{\mathrm{a}},\mathrm{a}-\frac{1}{\mathrm{a}})=\frac{\mathrm{\Gamma }\left(\frac{1}{\mathrm{a}}\right).\mathrm{\Gamma }(\mathrm{a}-\frac{1}{\mathrm{a}})}{\mathrm{\Gamma }\left(\mathrm{a}\right)}\Rightarrow$$$$\mathrm{I}=\mathrm{f}(1+\sqrt{2})=\frac{\mathrm{\Gamma }\left(\frac{1}{1+\sqrt{2}}\right).\mathrm{\Gamma }(1+\sqrt{2}-\frac{1}{1+\sqrt{2}})}{\mathrm{\Gamma }(1+\sqrt{2})}$$$$=\frac{\mathrm{\Gamma }(\sqrt{2}-1).\mathrm{\Gamma }(1+\sqrt{2}-(\sqrt{2}-1))}{\mathrm{\Gamma }(1+\sqrt{2})}=\frac{\mathrm{\Gamma }(\sqrt{2}-1).\mathrm{\Gamma }\left(2\right)}{\mathrm{\Gamma }(1+\sqrt{2})}$$$$=\frac{\mathrm{\Gamma }(\sqrt{2}-1)}{\mathrm{\Gamma }(\sqrt{2}+1)}$$

Commented by mathmax by abdo last updated on 18/Jun/21

$$\mathrm{sorry}\mathrm{f}\left(\mathrm{a}\right)=\frac{1}{\mathrm{a}}\mathrm{B}(\frac{1}{\mathrm{a}},\mathrm{a}-\frac{1}{\mathrm{a}})=\frac{\mathrm{\Gamma }\left(\frac{1}{\mathrm{a}}\right).\mathrm{\Gamma }(\mathrm{a}-\frac{1}{\mathrm{a}})}{\mathrm{\Gamma }(\mathrm{a}+1)}\Rightarrow$$$$\mathrm{I}=\mathrm{f}(1+\sqrt{2})=\frac{\mathrm{\Gamma }\left(\frac{1}{1+\sqrt{2}}\right).\mathrm{\Gamma }(1+\sqrt{2}-\frac{1}{1+\sqrt{2}})}{\mathrm{\Gamma }(1+\sqrt{2}+1)}$$$$=\frac{\mathrm{\Gamma }(\sqrt{2}-1).\mathrm{\Gamma }\left(2\right)}{\mathrm{\Gamma }(2+\sqrt{2})}=\frac{\mathrm{\Gamma }(\sqrt{2}-1)}{\mathrm{\Gamma }(2+\sqrt{2})}$$

Commented by akolade last updated on 19/Jun/21

$$greatworksie$$

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