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Question Number 66470 by mathmax by abdo last updated on 15/Aug/19

$$calculate{\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^n+8)}^3}withn>1$$

Commented by ~ À ® @ 237 ~ last updated on 16/Aug/19

$$letstate\mathrm{\forall }a>0f\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^n+a)}^2}$$$$f\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{dx}{a^2{[{\left(\frac{x}{a^{\frac{1}{n}}}\right)}^n+1]}^2}$$$$whenchangingx=a^{\frac{1}{n}}uweget$$$$f\left(a\right)=a^{\frac{1}{n}-2}{\int }_0^{\mathrm{\infty }}\frac{du}{{(u^n+1)}^2}$$$$nowchangev=\frac{1}{u^n+1}\Rightarrow u={(\frac{1}{v}-1)}^{\frac{1}{n}}\Rightarrow du=\frac{1}{n}.\left(\frac{-1}{v^2}\right){(\frac{1}{v}-1)}^{\frac{1}{n}-1}dv$$$$then$$$$f\left(a\right)=\frac{1}{n}.a^{\frac{1}{n}-2}{\int }_0^1{v}^2.\left(\frac{1}{v^2}\right){(\frac{1}{v}-1)}^{\frac{1}{n}-1}dv$$$$=\frac{a^{\frac{1}{n}-2}}{n}{\int }_0^1{v}^{1-\frac{1}{n}}{(1-v)}^{\frac{1}{n}-1}dv$$$$=\frac{a^{\frac{1}{n}-2}}{n}B(2-\frac{1}{n},\frac{1}{n})$$$$=\frac{a^{\frac{1}{n}-2}}{n}.\frac{\mathrm{\Gamma }(2-\frac{1}{n})\mathrm{\Gamma }\left(\frac{1}{n}\right)}{\mathrm{\Gamma }\left(2\right)}$$$$as\mathrm{\Gamma }(x+1)=x\mathrm{\Gamma }\left(x\right)wehave\mathrm{\Gamma }(2-\frac{1}{n})=(1-\frac{1}{n})\mathrm{\Gamma }(1-\frac{1}{n})and\mathrm{\Gamma }\left(2\right)=\mathrm{\Gamma }\left(1\right)=1$$$$Sof\left(a\right)=\frac{a^{\frac{1}{n}-2}}{n}.(1-\frac{1}{n})\mathrm{\Gamma }(1-\frac{1}{n})\mathrm{\Gamma }\left(\frac{1}{n}\right)$$$$as\mathrm{\Gamma }(1-z)\mathrm{\Gamma }\left(z\right)=\frac{\pi }{sin\left(\pi z\right)}wefinallyget$$$$$$$$f\left(a\right)=\frac{1}{n}(1-\frac{1}{n})a^{\frac{1}{n}-2}.\frac{\pi }{sin\left(\frac{\pi }{n}\right)}$$$$Nowwecandeducef\left(8\right),f\left(3\right),f\left(n\right)$$

Commented by ~ À ® @ 237 ~ last updated on 16/Aug/19

$$$$$$$$$$$$$$if\mathrm{\forall }p>2g_p\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^n+a)}^p}$$$$startasthepreviousandreachto$$$$g_p\left(a\right)=\frac{a^{\frac{1}{n}-p}}{n}{\int }_0^1{v}^p.\left(\frac{1}{v^2}\right){(\frac{1}{v}-1)}^{\frac{1}{n}-1}dv$$$$=\frac{a^{\frac{1}{n}-p}}{n}{\int }_0^1{v}^{p-2+1-\frac{1}{n}}{(1-v)}^{\frac{1}{n}-1}dv$$$$=\frac{a^{\frac{1}{n}-p}}{n}B(p-\frac{1}{n},\frac{1}{n})$$$$=\frac{a^{\frac{1}{n}-p}}{n}.\frac{\mathrm{\Gamma }(p-\frac{1}{n})\mathrm{\Gamma }\left(\frac{1}{n}\right)}{\mathrm{\Gamma }\left(p\right)}$$$$as\mathrm{\Gamma }(p+z)=z(z+1).......(z+p-1)\mathrm{\Gamma }\left(z\right)wehave\mathrm{\Gamma }(p-\frac{1}{n})=\mathrm{\Gamma }(p-1+(1-\frac{1}{n}))=\underset{k=0}{\prod ^{p-2}}[(1-\frac{1}{n})+k]\mathrm{\Gamma }(1-\frac{1}{n})$$$$Now$$$$g_p\left(a\right)=\frac{a^{\frac{1}{n}-p}}{n.(p-1)!}.\left(\underset{k=0}{\prod ^{p-2}}[(1-\frac{1}{n})+k]\right)\mathrm{\Gamma }(1-\frac{1}{n})\mathrm{\Gamma }\left(\frac{1}{n}\right)$$$$=\frac{a^{\frac{1}{n}-p}}{n.(p-1)!}.\frac{\pi }{sin\left(\frac{\pi }{n}\right)}.\underset{k=0}{\prod ^{p-2}}[(1-\frac{1}{n})+k]$$$$nowwecandeduce{g}_3\left(8\right),\cdot .......$$

Commented by mathmax by abdo last updated on 16/Aug/19

$$thankyousir.$$

Commented by mathmax by abdo last updated on 16/Aug/19

$$letg\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{dx}{{(a+x^n)}^2}witha>0wehaveprovedthat$$$$g\left(a\right)=\frac{\pi (1-\frac{1}{n})a^{\frac{1}{n}-2}}{nsin\left(\frac{\pi }{n}\right)}alsowehave$$$$g^{{}^{\prime }}\left(a\right)=-{\int }_0^{\mathrm{\infty }}\frac{2(a+x^n)}{{(a+x^n)}^4}dx=-2{\int }_0^{\mathrm{\infty }}\frac{dx}{{(a+x^n)}^3}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{{(a+x^n)}^3}=-\frac{1}{2}{g}^{{}^{\prime }}\left(a\right)wehave$$$$g^{{}^{\prime }}\left(a\right)=\frac{\pi }{nsin\left(\frac{\pi }{n}\right)}(1-\frac{1}{n})(\frac{1}{n}-2)a^{\frac{1}{n}-3}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{{(a+x^n)}^3}=\frac{2\pi }{nsin\left(\frac{\pi }{n}\right)}(1-\frac{1}{n})(2-\frac{1}{n})a^{\frac{1}{n}-3}$$$$a=8\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{dx}{{(8+x^n)}^3}=\frac{2\pi }{nsin\left(\frac{\pi }{n}\right)}(1-\frac{1}{n})(2-\frac{1}{n})8^{\frac{1}{n}-3}$$

Commented by mathmax by abdo last updated on 16/Aug/19

$$erroratfinalline{\int }_0^{\mathrm{\infty }}\frac{dx}{{(a+x^n)}^3}=\frac{\pi }{2nsin\left(\frac{\pi }{n}\right)}(1-\frac{1}{n})(2-\frac{1}{n})a^{\frac{1}{n}-3}$$$$a=8\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{dx}{{(8+x^n)}^3}=\frac{\pi }{2nsin\left(\frac{\pi }{n}\right)}(1-\frac{1}{n})(2-\frac{1}{n})8^{\frac{1}{n}-3}$$

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