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How to Find the value of $$I=\int_{0}^{1}\log \Big(2\arctan x+\frac{2x}{1+x^2}\Big)\Big(1+x\arctan x\Big)dx$$

I tried to subsititution:x=tant, but the process became very complicated,I don't know how to deal with it, and any help will be appreciated.

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  • $\begingroup$ Mathematica gives: $\frac{1}{4} \left(\pi \left(\log \left(\frac{2+\pi }{2}\right)-1\right)+\log \left(\frac{1}{4} (2+\pi )^2\right)\right)$ $\endgroup$ Commented Jan 10, 2018 at 20:20

1 Answer 1

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Well, we have:

$$\mathcal{I}:=\int\limits_0^1\ln\left(2\cdot\arctan\left(x\right)+\frac{2x}{1+x^2}\right)\cdot\left(1+x\cdot\arctan\left(x\right)\right)\space\text{d}x\tag1$$

Using integration by parts:

$$\mathcal{I}=\frac{2+\pi}{4}\cdot\ln\left(\frac{2+\pi}{2}\right)-\int\limits_0^1\frac{1}{1+x^2}\space\text{d}x=\frac{2+\pi}{4}\cdot\ln\left(\frac{2+\pi}{2}\right)-\frac{\pi}{4}\tag2$$


Integration by parts is used as follows:

$$\int\text{f}\left(x\right)\cdot\text{g}\space'\left(x\right)\space\text{d}x=\text{f}\left(x\right)\cdot\text{g}\left(x\right)-\int\text{f}\space'\left(x\right)\cdot\text{g}\left(x\right)\space\text{d}x\tag3$$

Where:

  • $$\text{f}\left(x\right):=\ln\left(2\cdot\arctan\left(x\right)+\frac{2x}{1+x^2}\right)\tag4$$
  • $$\text{g}\space'\left(x\right):=1+x\cdot\arctan\left(x\right)\tag5$$

And then you'll find:

  • \begin{equation} \begin{split} \left[\text{f}\left(x\right)\cdot\text{g}\left(x\right)\right]_0^1&=\lim_{x\space\to\space1}\left(\text{f}\left(x\right)\cdot\text{g}\left(x\right)-\text{f}\left(x-1\right)\cdot\text{g}\left(x-1\right)\right)\\ \\ &=\text{f}\left(1\right)\cdot\text{g}\left(1\right)-\lim_{x\space\to\space0}\text{f}\left(x\right)\cdot\text{g}\left(x\right)\\ \\ &=\frac{2+\pi}{4}\cdot\ln\left(\frac{2+\pi}{2}\right) \end{split}\tag6 \end{equation}

  • $$\text{f}\space'\left(x\right)\cdot\text{g}\left(x\right)=\frac{1}{1+x^2}\tag7$$

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  • $\begingroup$ I still don't understand why:f '(x)g(x)=1/(1+x^2)? $\endgroup$
    – FofX
    Commented Jan 11, 2018 at 5:26
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    $\begingroup$ I see now. thank you very much for your answer. $\endgroup$
    – FofX
    Commented Jan 11, 2018 at 5:41
  • $\begingroup$ @FofX You're welcome, I'm glad that I could help. $\endgroup$ Commented Jan 11, 2018 at 7:35

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