# Harmonic conjugate with tough integral

I am trying to find the harmonic conjugate of

$u(x,y) = \dfrac{y}{(x^2 + y^2)}$

I have got $Ux = Vy = \dfrac{-2xy}{(x^2 + y^2)^2}$

And now I need to integrate Vy with respect to y to find V.

However this is a tough integral. In our tutorial our tutor gave some method for doing this by 'switching x with -y'...something along those lines. I didnt get time to take down what he was saying properly.

How am I supposed to get from Vy to V?

-

If you recognize that the analytic function ${\displaystyle {1 \over z}}$ is the same as ${\displaystyle {x - iy \over x^2 + y^2}}$, then since ${\displaystyle {-iy \over x^2 + y^2}}$ is the harmonic conjugate of ${\displaystyle {x \over x^2 + y^2}}$, the function ${\displaystyle {ix \over x^2 + y^2}}$ is the harmonic conjugate of ${\displaystyle i ({-iy \over x^2 + y^2}) = {y \over x^2 + y^2}}$.

-

You get from $v_y$ to $v$ by integrating with respect to $y$:

$$v(x,y)-v(x,y_0) = \int_{y_0}^y v_y(x,y) \, dy$$

Hint 1: Assuming $x\ne 0$, we have

$$\int\frac{xy}{(x^2 + y^2)^2} \; dy = \int\frac{y/x}{x(1+(y/x)^2)^2} \, d\left(\frac{y}{x}\right)$$

Hint 2: This is essentially of the form $$\int f'(g(u)) g'(u) \, du = \int (f\circ g)'(u) \, du$$

-

Try integrating in polar coordinates. Use $x=r\cos(a)$ , $y=r\sin(a)$ Then the denominator is just $r^2$ Don't for the Jacobian - $r$. Then it's really simple and you get the desired v without having to guess the original function $f(Z)$.

-