Downwash of a circle For an aerodynamic application similar to that of a windmill, we are currently looking what the downwash is at every point of a loop with circulation. 


From aerodynamics, we know that the equation for downwash is:
\begin{equation}
\vec{V} = \int_{-\infty}^{\infty}\frac{\Gamma}{4\pi} \frac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}
\end{equation}
Now we have looked for references, but have not been very successful, and thus our attempt at deriving it is put at question, especially because we were surprised by the result given. Thus we hoped that this community here could provide us with some further feedback. Additionally, the integral expression seems like it can only be solved numerically. If this could be avoided, that would be great.
Below the derivation we have done:

First we can rearrange the down wash equation, since the circulation is constant:
\begin{equation}
\vec{V} = \frac{\Gamma}{4\pi} \int_{-\infty}^{\infty} \frac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}
\end{equation}
Where the magnitude of the cross product can be rewritten to:
\begin{equation}
|\vec{r}\times\vec{dl}| = |\vec{r}| |\vec{dl}| sin(\frac{\pi}{2}-\alpha-\theta)
\end{equation}
Thus our down wash equation ends up as:
\begin{equation}
V = \frac{\Gamma}{4\pi} \int_{0}^{2\pi R} \frac{sin(\frac{\pi}{2}-\alpha-\theta)}{r^2}dl
\end{equation}
We can further improve the expression by adding the transformations to $r$ and $\theta$ and using polar coordinates:
\begin{equation}
a = T\cos(\gamma) ; b=T\sin(\gamma)
\end{equation}
\begin{equation}
r = \sqrt{(a + R\cos(\alpha))^2 + (b + R\sin(\alpha))^2} = \sqrt{R^2 + 2RT\cos(\alpha - \gamma) + T^2}
\end{equation}
\begin{equation}
\theta = \arctan \Big(\frac{T\sin(\gamma) + R\sin(\alpha)}{T\cos(\gamma) + R\cos(\alpha)}\Big)
\end{equation}
\begin{equation}
R d\alpha = dl
\end{equation}
Since R is invariant across $\alpha$, we can make transformations in one ideal direction to reduce the equations even further by setting $\gamma = 0$.
\begin{equation}
r = \sqrt{R^2 + 2RT\cos(\alpha) + T^2}
\end{equation}
\begin{equation}
\theta = \arctan \Big(\frac{R\sin(\alpha)}{T + R\cos(\alpha)}\Big)
\end{equation}
\begin{equation}
R d\alpha = dl
\end{equation}
After rearranging, one arrives at the following final downwash expression:
\begin{equation}
V = \frac{\Gamma}{4\pi} \int_{0}^{2\pi} \frac{\cos(\alpha+\arctan \Big(\frac{R\sin(\alpha)}{T + R\cos(\alpha)}\Big)}{R^2 + 2RT\cos(\alpha) + T^2} R d\alpha
\end{equation}
Now the strange part of the equation derived here is that when taking $T=0$, so we are taking the downwash at a point in the centre, it equals to zero. However from the two first figures, it seems like circulation is equal in direction there and of equal magnitude, and thus not equal to zero...?
Anyways, any help or suggestion with this is much appreciated, as our project would benefit a lot from an accurate simulation, which must include the downwash here described. 
 A: As you wrote, assuming that $\Gamma$ is constant, Biot-Savart law states that:
$$\mathbf{V} = \frac{\Gamma}{4\pi} \int \frac{d\,\mathbf{l}\times\mathbf{r}'}{|\mathbf{r}'|^3}$$
For a point $P$ inside the circle, this results in:
$$\mathbf{V}_P=\frac{\Gamma}{4\pi}\oint\frac{d\,\mathbf{l}\times\mathbf{r}'}{|\mathbf{r}'|^3}$$

According to the picture:
$$\begin{align}
d\,\mathbf{l}&=r d\theta\,\hat{\boldsymbol{\theta}}\\\hat{\boldsymbol{\theta}}\times\mathbf{r}'&=\mathbf{n}\,r'\sin(\alpha+\frac{\pi}2)=
\mathbf{n}\,r'\cos\alpha=\mathbf{n}(r-x\cos\theta)
\end{align}$$
where $\mathbf{n}$ is the normal vector of the plane and $\hat{\boldsymbol{\theta}}$ is the tangential unit vector. Now let $\xi=\frac x r$ and use the cosine theorem to get:
$$\begin{align}
\frac{4\pi}\Gamma\mathbf{V}=\oint\frac{d\,\mathbf{l}\times\mathbf{r}'}{|\mathbf{r}'|^3}
&=\mathbf{n}\int_0^{2\pi}\frac{r(r-x\cos\theta)d\theta}{{r'}^3}\\
&=\mathbf{n}\int_0^{2\pi}\frac{r^2(1-\xi\cos\theta)}{(r^2+x^2-2r x\cos\theta)^{3/2}}d\theta\\
&=\frac{\mathbf{n}}r\int_0^{2\pi}\frac{1-\xi\cos\theta}{(1+\xi^2-2\xi\cos\theta)^{3/2}}d\theta
\end{align}$$
The above integral is generally hard to evaluate. Note that $0\le\xi<1$, and the integral is $2\pi$ for $\xi=0$ and $\infty$ for $\xi=1$. I personally prefer numerical methods over analytic ones:

But if you insist on finding an analytic solution, you may get the idea by checking out Olivier's great answer.
