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Question Number 116672 by mnjuly1970 last updated on 05/Oct/20

$$...nicecalculus...$$$$$$$$provethat:$$$$$$$$\mathrm{I}={\int }_0^1(\frac{\pi }{4}-\mathcal{A}rctan\left(x\right))\frac{dx}{1-x^2}=\frac{\mathrm{G}}{2}✓$$$$$$$$\mathrm{G}iscatalanconstant...$$$$\mathcal{M}.\mathcal{N}.1970$$$$$$$$$$$$$$

Answered by mnjuly1970 last updated on 05/Oct/20

$$solution:$$$$\mathrm{I}={\int }_0^1(\mathcal{A}rctan\left(1\right)-\mathcal{A}rctan\left(x\right))dx$$$$\mathrm{I}={\int }_0^1\mathcal{A}rctan\left(\frac{1-x}{1+x}\right)\frac{dx}{1-x^2}$$$$\mathrm{I}\stackrel{(\frac{1-x}{1+x}=t)}{=}-2{\int }_1^0\mathcal{A}rctan\left(t\right)\frac{dt}{{(1+t)}^2[1-{\left(\frac{1-t}{1+t}\right)}^2]}$$$$\mathrm{I}=2{\int }_0^1\frac{\mathcal{A}rctan\left(t\right)}{4t}dt=\frac{1}{2}{\int }_0^1\frac{\mathcal{A}rctan\left(t\right)}{t}dt$$$$\mathrm{I}=\frac{1}{2}{\int }_0^1\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{t}^{2n}}{(2n+1)}dt=\frac{1}{2}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(2n+1)}^2}$$$$\mathrm{I}:=\frac{\mathrm{G}}{2}✓✓$$$$...m.n.july.1970...$$$$$$

Answered by Bird last updated on 06/Oct/20

$$I={\int }_0^1(\frac{\pi }{4}-arctanx)\frac{dx}{1-x^2}$$$$wehavetan(\frac{\pi }{4}-arctsnx)=$$$$\frac{1-x}{1+x}\Rightarrow \frac{\pi }{4}-arctanx=arctan\left(\frac{1-x}{1+x}\right)\Rightarrow$$$$I={\int }_0^1\frac{arctan\left(\frac{1-x}{1+x}\right)}{1-x^2}dx$$$$ch\frac{1-x}{1+x}=tgive1-x=t+tx\Rightarrow$$$$1-t=(1+t)x\Rightarrow x=\frac{1-t}{1+t}\Rightarrow$$$$\frac{dx}{dt}=\frac{-(1+t)-(1-t)}{{(1+t)}^2}=\frac{-2}{{(1+t)}^2}\Rightarrow$$$$I=-{\int }_0^1\frac{arctan\left(t\right)}{1-\frac{{(1-t)}^2}{{(1+t)}^2}}\times \frac{-2}{{(1+t)}^2}dt$$$$=2{\int }_0^1\frac{arctant}{{(1+t)}^2-{(1-t)}^2}dt$$$$=2{\int }_0^1\frac{arctan\left(t\right)}{1+2t+t^2-1+2t-t^2}dt$$$$=\frac{1}{2}{\int }_0^1\frac{arctant}{t}dtwehave$$$${\left(arctant\right)}^{{}^{\prime }}=\frac{1}{1+t^2}=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{t}^{2n}$$$$\Rightarrow arctan\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{t}^{2n+1}}{2n+1}$$$$\Rightarrow \frac{arctant}{t}=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{2n+1}{t}^{2n}\Rightarrow$$$${\int }_0^1\frac{arctant}{t}dt=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(2n+1)}^2}=K$$$$\Rightarrow I=\frac{K}{2}\left(katalanconstante\right)$$$$$$

Commented by mnjuly1970 last updated on 06/Oct/20

$$thankyouverymuch...$$

Commented by Bird last updated on 07/Oct/20

$$youarewelcome$$

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