# How to prove a fomula is real or not?

The formula is:

$$f(u):=\sqrt{1+i}\arctan\left(\dfrac{\sqrt{2}}{\sqrt{(-1+i)u}}\right)+\sqrt{1-i}\arctan\left(\frac{(-1)^{3/4}\sqrt{(-1-i)u}}{u}\right)\tag{1}$$

I obtained it from the following definite integral: $$f(u):=\int_u^{\infty } \frac{\sqrt{x}}{x^2-2 x+2} \, dx\quad {\text{ where }} u>0\tag{2}$$

Numercial estimation indicates that, when $u>0$, $f(u)\in \mathbb{R}$ because the imaginary part of the results are close to numerical epsilon.

How to prove or disprove it? If the conclusion is true, is it possible to eliminate the imaginary unit $i$ from the fomula of $f(u)$ in $(1)$ and obtain a relatively simpler form of it which does not contain or implies any imaginary number?

update

$$f^*(u):=\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{\sqrt{2}}} \left(\left(\sqrt{2}-1\right) \left(\ln \left(u-\sqrt{2 \left(1+\sqrt{2}\right)} \sqrt{u}+\sqrt{2}\right)-\ln \left(u+\sqrt{2 \left(1+\sqrt{2}\right)} \sqrt{u}+\sqrt{2}\right)\right)-2 \arctan\left(-\sqrt{2 \left(1+\sqrt{2}\right)} \sqrt{u}+\sqrt{2}+1\right)+2 \arctan\left(\sqrt{2 \left(1+\sqrt{2}\right)} \sqrt{u}+\sqrt{2}+1\right)-2 \pi \right)$$

• As an integral can be expressed as an area the result must be a real number. But as for how to express it without the $i$ I don't think you can write it in a closed form without $i$. Commented Aug 13, 2016 at 9:36
• The integrand is quite nice (just plot it) for $x\geq 0$; so the result must be real (think about the area under the curve). By the way, I got something different but as awful as your formula but I did not find any simplification. Sorry ! Commented Aug 13, 2016 at 9:37
• I suspect that for $u>0$, the second term is just the complex conjugate of the first. Since $z+\bar z = 2\Re(x)$, this would imply that $f(u)$ is real. However there are some complexities [no pun intended] with the definition of the square root and $(-1)^{3/4}$. Commented Aug 13, 2016 at 10:04
• @user6043040 I converted your primitive $\frac{\sqrt{x}}{x^2-2 x+2}$ using only real functions (see my answer). I followed the procedure described here: math.stackexchange.com/questions/1848283/… Commented Aug 13, 2016 at 11:56

Inspired by How to simplify $\Re\left[\sqrt2 \tan^{-1} {x\over \sqrt i}\right]$? I manage to write a primitive of $\frac{\sqrt{x}}{x^2-2 x+2}$ using real functions. Let $$F(x):=\frac{1}{\sqrt{2\sqrt{2}-2}}\cdot \arctan\left(\frac{\sqrt{2\sqrt{2}-2}\cdot \sqrt{x}}{x-\sqrt{2}}\right)$$ $$G(x):=\frac{1}{\sqrt{2\sqrt{2}+2}}\cdot \mbox{arctanh}\left(\frac{\sqrt{2\sqrt{2}+2}\cdot\sqrt{x}}{x+\sqrt{2}}\right),$$ then you can verify that $$\frac{d}{dx}\left(-F(x)-G(x)\right)= \frac{\sqrt{x}}{x^2-2 x+2}.$$