Solving $\int dx/(x^{2}+1)$ with polynomial interpolation on the complex plane and with trigonometric substitution

Question

Using the Lagrange interpolation theory for $x_{0}=-i$ and $x_{1}=i$ we have

$\displaystyle \int \frac{dx}{(x^{2}+1)}= \frac{-1}{2i}\int \frac{dx}{x+i} + \frac{1}{2i}\int\frac{dx}{x-i}=\frac{1}{2i}(\ln|x-i|+\ln|x+i|)+ C$

So I was wondering in how to prove the equivalence with the trigonometric solution given when we do $x=tg( \theta)$ that we obtain

$arc tg(x)+ C_{1}$

Answer 1

You can do this algebraically. Set

$$y=\arctan x$$

$$\tan y =x$$

$$\frac{e^{iy} - e^{-iy}}{i(e^{iy} + e^{-iy})} = x$$

$$e^{iy} - e^{-iy} = i(e^{iy}x + e^{-iy}x)$$

$$e^{2iy}-1=e^{2iy}ix+ix$$

$$e^{2iy}(1-ix) = 1+ix$$

$$e^{2iy} = \frac{1+ix}{1-ix}$$

$$2iy = \ln\left(\frac{i(-i+x)}{i(-1-x)}\right)$$

$$y = \frac{1}{2i}\ln\left(-\frac{x-i}{x+i}\right)$$

$$y =\frac{1}{2i}\left( \ln(x-i) -\ln(x+i) +\ln(-1)\right)$$

Answer 2

Actually the sign is minus between the two $\ln$'s.

Hence you can arrange them as:$$\ln \frac{x+i}{x-i}$$ Now these are complex conjugates, hence when going over to polar coordinates, $r$ is cancelled and there remains: $$\ln (\exp 2it) = 2it$$ Recall that $t$ is $\arctan x$, and you are done.

Answer 3

Caution, in the complex,

$$\int\frac{dx}{x+i}=\log(x+i),$$ and not $$\log|x+i|.$$

More precisely, there comes an imaginary part,

$$\log(x+i)=\log|x+i|+i\left(\arctan\frac1x+2k\pi\right).$$

This is how the arc tangent term appears. Also notice that

$$\log|x+i|-\log|x-i|=0$$ so that the integral remains real.