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Question Number 100590 by Rohit@Thakur last updated on 27/Jun/20

$${\int }_0^{\mathrm{\infty }}\frac{dx}{{(1+x^{18})}^2}$$

Answered by mathmax by abdo last updated on 27/Jun/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{t}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{\mathrm{t}+{\mathrm{x}}^{18}}\mathrm{with}\mathrm{t}>0\mathrm{we}\mathrm{have}\frac{\mathrm{df}}{\mathrm{dt}}\left(\mathrm{t}\right)=-{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{(\mathrm{t}+{\mathrm{x}}^{18})}^2}$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{(\mathrm{t}+{\mathrm{x}}^{18})}^2}=-{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{t}\right)\mathrm{and}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{(1+{\mathrm{x}}^{18})}^2}=-{\mathrm{f}}^{{}^{\prime }}\left(1\right)$$$$\mathrm{f}\left(\mathrm{t}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{\mathrm{t}(1+\frac{{\mathrm{x}}^{18}}{\mathrm{t}})}=\frac{1}{\mathrm{t}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\left(\frac{\mathrm{x}}{\alpha }\right)}^{18}}\mathrm{with}\alpha ={\mathrm{t}}^{\frac{1}{18}}$$$${=}_{\frac{\mathrm{x}}{\alpha }=\mathrm{u}}\frac{1}{\mathrm{t}}{\int }_0^{\mathrm{\infty }}\frac{\alpha \mathrm{du}}{1+{\mathrm{u}}^{18}}=\frac{\alpha }{\mathrm{t}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{du}}{1+{\mathrm{u}}^{18}}{=}_{\mathrm{u}={\mathrm{z}}^{\frac{1}{18}}}{\mathrm{t}}^{\frac{1}{18}-1}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{z}}^{\frac{1}{18}-1}}{18(1+\mathrm{z})}\mathrm{dz}$$$$=\frac{{\mathrm{t}}^{\frac{1}{18}-1}}{18}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{z}}^{\frac{1}{18}-1}}{1+\mathrm{z}}\mathrm{dz}=\frac{{\mathrm{t}}^{\frac{1}{18}-1}}{18}\times \frac{\pi }{\sin \left(\frac{\pi }{18}\right)}=\frac{\pi }{18\sin \left(\frac{\pi }{18}\right)}\times {\mathrm{t}}^{-\frac{17}{18}}$$$$\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{t}\right)=-\frac{17}{18}{\mathrm{t}}^{-\frac{17}{18}-1}\times \frac{\pi }{18\sin \left(\frac{\pi }{18}\right)}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(1\right)=-\frac{17}{18}\times \frac{\pi }{18\sin \left(\frac{\pi }{18}\right)}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{(1+{\mathrm{x}}^{18})}^2}=-{\mathrm{f}}^{{}^{\prime }}\left(1\right)=\frac{17\pi }{{18}^2\sin \left(\frac{\pi }{18}\right)}$$

Commented by Ar Brandon last updated on 27/Jun/20

Understood, impressionnant ! ��

Commented by 1549442205 last updated on 27/Jun/20

$$\mathrm{Great}\mathrm{method}!\mathrm{But}\mathrm{how}\mathrm{do}\mathrm{you}\mathrm{know}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{z}}^{\frac{1}{18}-1}}{\mathrm{z}+1}\mathrm{dz}=\frac{\pi }{18\sin \frac{\pi }{18}}$$$$\mathbf{Please},\mathbf{can}\mathbf{you}\mathbf{show}\mathbf{me}?$$

Commented by Coronavirus last updated on 27/Jun/20

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Commented by abdomathmax last updated on 27/Jun/20

$$\mathrm{i}\mathrm{have}\mathrm{proved}\mathrm{this}\mathrm{relation}\mathrm{take}\mathrm{alook}\mathrm{in}\mathrm{the}$$$$\mathrm{platform}...\mathrm{i}\mathrm{have}\mathrm{used}\mathrm{residus}\mathrm{for}\mathrm{that}...$$

Commented by maths mind last updated on 27/Jun/20

$${\int }_0^1{t}^{x-1}{(1-t)}^{y-1}dt=\beta (x,y)..E$$$$\frac{t}{1-t}=u\Rightarrow t=\frac{u}{1+u}\Rightarrow dt=\frac{du}{{(1+u)}^2}$$$$E\iff {\int }_0^{+\mathrm{\infty }}\frac{u^{x-1}}{{(1+u)}^{x-1}}.\frac{1}{{(1+u)}^{y-1}}.\frac{du}{{(1+u)}^2}$$$$={\int }_0^{+\mathrm{\infty }}\frac{u^{x-1}}{{(1+u)}^{x+y}}du$$$$x=\frac{1}{18},y=\frac{17}{18}$$$$\Rightarrow \beta (\frac{1}{18},1-\frac{1}{18})=\frac{\pi }{sin\left(\frac{\pi }{18}\right)}$$$$\beta (a,b)=\frac{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }(a+b)},ifa+b=1\Rightarrow \mathrm{\Gamma }(a+b)=\mathrm{\Gamma }\left(1\right)=1,$$$$\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }(1-a)=\frac{\pi }{sin\left(\pi a\right)}....$$

Commented by 1549442205 last updated on 27/Jun/20

$$\mathrm{Thankyou},\mathrm{sir}\mathrm{a}\mathrm{lot}$$

Commented by mathmax by abdo last updated on 27/Jun/20

$$\mathrm{let}\mathrm{prove}\mathrm{that}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}-1}}{1+\mathrm{t}}\mathrm{dt}=\frac{\pi }{\sin \left(\pi \mathrm{a}\right)}\mathrm{with}0<\mathrm{a}<1\mathrm{changement}\mathrm{t}={\tan }^2\theta \mathrm{give}$$$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}-1}}{1+\mathrm{t}}\mathrm{dt}={\int }_0^{\frac{\pi }{2}}\frac{{\tan }^{2\mathrm{a}-2}}{1+{\tan }^2\theta }\times 2\tan \theta (1+{\tan }^2\theta )\mathrm{d}\theta$$$$=2{\int }_0^{\frac{\pi }{2}}{\tan }^{2\mathrm{a}-1}\theta \mathrm{d}\theta =2{\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2\mathrm{a}-1}\theta }{{\cos }^{2\mathrm{a}-1}\theta }\mathrm{d}\theta =2{\int }_0^{\frac{\pi }{2}}{\sin }^{2\mathrm{a}-1}\theta {\cos }^{1-2\mathrm{a}}\theta \mathrm{d}\theta$$$$\mathrm{we}\mathrm{know}2{\int }_0^{\frac{\pi }{2}}{\cos }^{2\mathrm{p}-1}\theta {\sin }^{2\mathrm{q}-1}\theta =\mathrm{B}(\mathrm{p},\mathrm{q})=\frac{\mathrm{\Gamma }\left(\mathrm{p}\right).\mathrm{\Gamma }\left(\mathrm{q}\right)}{\mathrm{\Gamma }(\mathrm{p}+\mathrm{q})}$$$$2\mathrm{q}-1=1-2\mathrm{a}\Rightarrow 2\mathrm{q}=2-2\mathrm{a}\Rightarrow \mathrm{q}=1-\mathrm{a}\Rightarrow$$$$2{\int }_0^{\frac{\pi }{2}}{\sin }^{2\mathrm{a}-1}\theta {\cos }^{1-2\mathrm{a}}\theta \mathrm{d}\theta =2{\int }_0^{\frac{\pi }{2}}{\sin }^{2\mathrm{a}-1}\theta {\cos }^{2(1-\mathrm{a})-1}\theta \mathrm{d}\theta$$$$=\mathrm{B}(\mathrm{a},1-\mathrm{a})=\frac{\mathrm{\Gamma }\left(\mathrm{a}\right).\mathrm{\Gamma }(1-\mathrm{a})}{\mathrm{\Gamma }(\mathrm{a}+1-\mathrm{a})}=\mathrm{\Gamma }\left(\mathrm{a}\right).\mathrm{\Gamma }(1-\mathrm{a})=\frac{\pi }{\sin \left(\pi \mathrm{a}\right)}\left(\mathrm{compoliment}\mathrm{formulae}\right)$$$$\mathrm{the}\mathrm{value}\mathrm{is}\mathrm{proved}.$$

Commented by mathmax by abdo last updated on 27/Jun/20

$$\mathrm{error}\mathrm{of}\mathrm{typo}\left(\mathrm{compliment}\mathrm{formulae}\right)$$

Commented by Rohit@Thakur last updated on 28/Jun/20

$$thankyousir$$

Answered by smridha last updated on 27/Jun/20

$$\mathit{l}\mathit{e}\mathit{t}{\mathit{x}}^9=\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }\mathit{s}\mathit{o}\mathit{w}\mathit{e}\mathit{g}\mathit{e}\mathit{t}$$$$\frac{1}{9}{\int }_0^{\frac{\mathit{\pi }}{2}}{\left(\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\right)}^{-\frac{8}{9}}.{\left(\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\right)}^{\frac{26}{9}}\mathit{d}\mathit{\theta }$$$$=\frac{1}{18}.\frac{\mathbf{\Gamma }\left(\frac{1}{18}\right).\mathrm{\Gamma }\left(\frac{35}{18}\right)}{\mathbf{\Gamma }\left(2\right)}=\frac{17}{324}.\mathbf{\Gamma }\left(\frac{1}{18}\right).\mathrm{\Gamma }(1-\frac{1}{18})$$$$=\frac{17\mathit{\pi }}{324}.\frac{1}{\mathit{s}\mathit{i}\mathit{n}\left(\frac{\mathit{\pi }}{18}\right)}.$$

Commented by Rohit@Thakur last updated on 28/Jun/20

$$plzprovidethedetailedsolution$$

Commented by smridha last updated on 28/Jun/20

$${\int }_0^{\mathrm{\infty }}\frac{\mathit{d}\mathit{x}}{{(1+{\mathit{x}}^{18})}^2}$$$$\mathit{l}\mathit{e}\mathit{t}{\mathit{x}}^9=\mathit{t}\mathit{a}\mathit{n}\left(\mathit{\theta }\right)\mathit{s}\mathit{o}\mathit{w}\mathit{e}\mathit{g}\mathit{e}\mathit{t}$$$$\frac{1}{9}{\int }_0^{\frac{\mathit{\pi }}{2}}{\mathit{s}\mathit{e}\mathit{c}}^{-4}\left(\mathit{\theta }\right).{\mathit{t}\mathit{a}\mathit{n}}^{\frac{1}{9}-1}\left(\mathit{\theta }\right){\mathit{s}\mathit{e}\mathit{c}}^2\left(\mathit{\theta }\right)\mathit{d}\mathit{\theta }$$$$=\frac{1}{9}{\int }_0^{\frac{\mathit{\pi }}{2}}\frac{{\left(\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\right)}^{-\frac{8}{9}}}{{\left(\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\right)}^{-\frac{8}{9}}}.{\mathit{c}\mathit{o}\mathit{s}}^2\left(\mathit{\theta }\right)\mathit{d}\mathit{\theta }$$$$=\frac{1}{9}{\int }_0^{\frac{\mathit{\pi }}{2}}{\left(\mathit{s}\mathit{i}\mathit{n}\left(\mathit{\theta }\right)\right)}^{-\frac{8}{9}}{\left(\mathit{c}\mathit{o}\mathit{s}\left(\mathit{\theta }\right)\right)}^{\frac{26}{9}}$$$$=\frac{1}{9}.\frac{1}{2}.\frac{\mathbf{\Gamma }\left(\frac{-\frac{8}{9}+1}{2}\right).\mathbf{\Gamma }\left(\frac{\frac{26}{9}+1}{2}\right)}{\mathbf{\Gamma }\left(\frac{-\frac{8}{9}+\frac{26}{9}+2}{2}\right)}=\frac{1}{18}\frac{\mathbf{\Gamma }\left(\frac{1}{18}\right).\mathbf{\Gamma }\left(\frac{35}{18}\right)}{\mathbf{\Gamma }\left(2\right)}$$$$=\frac{1}{18}\frac{\mathbf{\Gamma }\left(\frac{1}{18}\right).\mathbf{\Gamma }(\frac{17}{18}+1)}{1!}=\frac{17}{{18}^2}.\mathbf{\Gamma }\left(\frac{1}{18}\right).\mathrm{\Gamma }\left(\frac{17}{18}\right)$$$$[\mathbf{\Gamma }\left(\mathit{n}\right)=(\mathit{n}-1)!\mathit{a}\mathit{n}\mathit{d}\mathbf{\Gamma }(\mathit{n}+1)=\mathit{n}\mathbf{\Gamma }\left(\mathit{n}\right)]$$$$=\frac{17}{324}\mathbf{\Gamma }\left(\frac{1}{18}\right).\mathbf{\Gamma }(1-\frac{1}{18})=\frac{17\mathit{\pi }}{324}.\frac{1}{\mathit{s}\mathit{i}\mathit{n}\left(\frac{\mathit{\pi }}{18}\right)}$$$$[\mathbf{\Gamma }\left(\mathit{m}\right).\mathbf{\Gamma }(1-\mathit{m})=\frac{\mathit{\pi }}{\mathit{s}\mathit{i}\mathit{n}\left(\mathit{m}\mathit{\pi }\right)}\mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathit{m}<1$$$$\mathit{t}\mathit{h}\mathit{i}\mathit{s}\mathit{i}\mathit{s}\mathit{c}\mathit{a}\mathit{l}\mathit{l}\mathit{e}\mathit{d}\mathit{r}\mathit{e}\mathit{f}\mathit{l}\mathit{e}\mathit{c}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{a}$$$$\mathit{f}\mathit{o}\mathit{r}\mathit{G}\mathit{a}\mathit{m}\mathit{m}\mathit{a}{\mathit{f}}^{\mathit{n}}]$$$$$$

Commented by smridha last updated on 28/Jun/20

$$\mathit{t}\mathit{h}\mathit{u}\mathit{s}\mathit{y}\mathit{o}\mathit{u}\mathit{p}\mathit{o}\mathit{s}\mathit{t}\mathit{e}\mathit{d}\mathit{s}\mathit{o}\mathit{m}\mathit{e}t\mathit{h}\mathit{i}\mathit{n}\mathit{g}$$$$\mathit{l}\mathit{i}\mathit{k}\mathit{e}\mathit{t}\mathit{h}\mathit{i}\mathit{s}!!\mathit{i}\mathit{t}\mathit{s}\mathit{h}\mathit{o}\mathit{u}\mathit{l}\mathit{d}\mathit{b}\mathit{e}\mathit{n}\mathit{e}\mathit{e}\mathit{d}\mathit{e}\mathit{d}$$$$\mathit{t}\mathit{o}\mathit{k}\mathit{n}\mathit{o}\mathit{w}\mathit{t}\mathit{h}\mathit{e}\mathit{b}\mathit{a}\mathit{s}\mathit{i}\mathit{c}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{i}\mathit{s}\mathit{m}$$$$\mathit{a}\mathit{t}\mathit{f}\mathit{i}\mathit{r}\mathit{s}\mathit{t}.$$

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