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Question Number 166956 by mnjuly1970 last updated on 03/Mar/22

$$$$$$calculate$$$${\int }_0^{\mathrm{\infty }}\frac{\sqrt{x}arctan\left(x\right)}{1+x^2}dx=?$$

Answered by ArielVyny last updated on 03/Mar/22

$$dt=\frac{arctan\left(x\right)}{1+x^2}dx\to t=\frac{1}{2}{\left(arctanx\right)}^2$$$$\sqrt{2t}=arctanx\to x=arctan\left(\sqrt{2}t\right)$$$${\int }_0^{\frac{{\pi }^2}{8}}arctan\left(\sqrt{2}t\right)dt=$$$$\frac{1}{\sqrt{2}}{\int }_0^{\frac{{\pi }^2}{8}}arctan\left(u\right)du$$$$wecantstophere$$$$=\sum _{n⩾0}\frac{{(-1)}^n{\left(\sqrt{2}\right)}^{2n+1}}{2n+1}{\int }_0^{\frac{{\pi }^2}{8}}{t}^{2n+1}dt$$$$=\sum _{n⩾0}\frac{{(-1)}^n{\left(\sqrt{2}\right)}^{2n+1}}{2n+1}\left[\frac{1}{2n+2}\frac{{\pi }^2}{8}\right]$$$$=\frac{{\pi }^2}{8}\sum _{n⩾0}\frac{{(-1)}^n{\left(\sqrt{2}\right)}^{2n+1}}{(2n+1)(2n+2)}$$$$=\frac{{\pi }^2}{8}\left[\mathrm{\Sigma }\frac{{(-1)}^n{\left(\sqrt{2}\right)}^{2n+1}}{2n+1}\right]-\frac{{\pi }^2}{8}\sum _{n⩾0}\frac{{(-1)}^n{\left(\sqrt{2}\right)}^{2n+1}}{2n+2}$$$$=\frac{{\pi }^2}{8}arctan\left(\sqrt{2}\right)\frac{{\pi }^2}{16}\sum _{n⩾1}\frac{{(-1)}^{n+1}{\left(2\right)}^n}{n}$$$$=\frac{{\pi }^2}{8}arctan\left(\sqrt{2}\right)+\frac{{\pi }^2}{16}ln\left(3\right)$$$$$$

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