Evaluation of $\int \limits _0 ^{2 \pi} \frac {(r \cos \phi +x) \cos n\phi} {r^2+2xr \cos \phi +x^2} d\phi$

Question

How to compute

$$\int \limits _0 ^{2 \pi} \frac {(r \cos \phi +x) \cos n\phi} {r^2+2xr \cos \phi +x^2} d\phi ?$$

The answer I am provided with is $\dfrac {(-1)^n\pi r^n} {x^{n+1}}$ for $\ x>r$, but I have no idea whether this is actually correct and how to get this.

Answer 1

If I may add my two cents. A 'real' method to consider.

There are many Poisson-like series identities. A 'family' if you will.

One of those sums that can be derived from the geometric series:

$$\sum_{n=0}^{\infty}\frac{(-1)^{n}r^{n}}{x^{n+1}}\cos(n\theta)=\Re\left(\frac{1}{x}\cdot \frac{1}{1+\frac{re^{i\theta}}{x}}\right)=\frac{x+r\cos(\theta)}{r^{2}+2rx\cos(\theta)+x^{2}}, \;\ r<x$$

EDIT:

Something I just thought of that may be another approach.

If we instead find, by whatever means (residues is a good approach, of course), $$\int_{0}^{2\pi}\frac{\cos(n\theta)}{r^{2}+2xr\cos(\theta)+x^{2}}d\theta=\frac{2(-1)^{n}\pi r^{n}}{x^{n}(x^{2}-r^{2})}**$$

then use it with the identity:

$$\frac{2x\cos(n\theta)}{x^{2}-r^{2}}\left(\frac{x+r\cos(\theta)}{r^{2}+2rx\cos(\theta)+x^{2}}-\frac{1}{2x}\right)=\frac{\cos(n\theta)}{r^{2}+2rx\cos(\theta)+x^{2}}$$

Expand and integrate both sides:

$$\frac{2x}{x^{2}-r^{2}}\int_{0}^{2\pi}\frac{(x+r\cos(\theta))\cos(n\theta)}{r^{2}+2rx\cos(\theta)+x^{2}}d\theta-\frac{1}{x^{2}-r^{2}}\int_{0}^{2\pi}\cos(n\theta)d\theta=\frac{2(-1)^{n}\pi r^{n}}{x^{n}(x^{2}-r^{2})}$$

The rightmost integral with just the cos term evaluates to 0. Multiply by $\frac{x^{2}-r^{2}}{2x}$ and obtain:

$$\int_{0}^{2\pi}\frac{(x+r\cos(\theta))\cos(n\theta)}{r^{2}+2rx\cos(\theta)+x^{2}}d\theta=\frac{2(-1)^{n}\pi r^{n}}{x^{n}(r^{2}-x^{2})}\cdot \frac{r^{2}-x^{2}}{2x}=\frac{(-1)^{n}\pi r^{n}}{x^{n+1}}$$

Please excuse any typos and let me know of them. Really easy in all that :)

** one way is to consider $$f(z)=\frac{z^{n}}{(1-rz)(1-r/z)}$$ and note the residue at $z=r$. Then, do some manipulating by letting $r\to -r/x$.

Ergo, the residue at $z=r$ is $\frac{r^{n}}{1-r^{2}}$.

Thus, $$\int_{0}^{2\pi}\frac{\cos(n\theta)}{1-2r\cos*\theta)+r^{2}}d\theta=\frac{2\pi r^{n}}{1-r^{2}}$$

Let $r\to -r/x$ and obtain:

$$\int_{0}^{2\pi}\frac{x^{2}\cos(n\theta)}{x^{2}+2rx\cos(\theta)+r^{2}}d\theta=\frac{2\pi\cdot x^{2}(-1)^{n}r^{n}}{x^{n}(x^{2}-r^{2})}$$.

Multiply by $1/x^{2}$:

$$\int_{0}^{2\pi}\frac{\cos(n\theta)}{x^{2}+2rx\cos(\theta)+r^{2}}d\theta=\frac{2\pi (-1)^{n}r^{n}}{x^{n}(x^{2}-r^{2})}$$

From above, multiplying by the $\frac{x^{2}-r^{2}}{2x}$ gives the result needed.

Answer 2

Suppose we seek to evaluate $$\int_0^{2\pi} \frac{(r\cos\phi+x)\cos(n\phi)}{r^2+2xr\cos\phi+x^2} \; d\phi.$$

Introduce $z=\exp(i\phi)$ so that $dz=iz \; d\phi$ to get $$\int_{|z|=1} \frac{(r(z+1/z)/2+x)(z^n+1/z^n)/2}{r^2+xr(z+1/z)+x^2} \frac{dz}{iz} \\ = \frac{1}{4i} \int_{|z|=1} \frac{(r(z+1/z)+2x)(z^n+1/z^n)}{(r^2+x^2)z + xrz^2 + xr} \; dz.$$

The denominator may be factored manually and we get $$(xz+r)(rz+x)$$ so the poles are at $z=0$ and $$\rho_0 = -r/x \quad\text{and}\quad \rho_1 = -x/r$$ and with $r \lt x$ only $\rho_0$ is inside the contour.

We get for the residue at $\rho_0$ $$\frac{1}{4i} \left. \frac{(r(z+1/z)+2x)(z^n+1/z^n)}{(r^2+x^2) + 2xrz} \right|_{z=-r/x} \\ = \frac{(-1)^n}{4i} \frac{(-r^2/x - x + 2x)((r/x)^n+(x/r)^n)} {r^2+x^2-2r^2} \\ = \frac{(-1)^n}{4ix} ((r/x)^n+(x/r)^n).$$

Supposing that $n$ is a non-negative integer the only remaining contribution is from the pole at zero of

$$\frac{1}{4i} \int_{|z|=\epsilon} \frac{1}{z^n} \frac{r(z+1/z)+2x}{(r^2+x^2)z + xrz^2 + xr} \; dz.$$

Observe that $$\frac{1}{(xz+r)(rz+x)} = \frac{1}{x^2-r^2}\frac{1}{z+r/x} - \frac{1}{x^2-r^2}\frac{1}{z+x/r}.$$

and that $$[z^n] \frac{1}{z-\beta} = \frac{1}{\beta} [z^n] \frac{1}{z/\beta-1} = - \frac{1}{\beta^{n+1}}.$$

so that on performing coefficient extraction we obtain $$\frac{1}{4i} \frac{1}{x^2-r^2} \left(-r(-x/r)^{n-1}+r(-r/x)^{n-1} - r(-x/r)^{n+1}+r(-r/x)^{n+1} \\ - 2x(-x/r)^{n}+2x(-r/x)^{n}\right).$$

This is $$\frac{1}{4i} \frac{1}{x^2-r^2} (-x/r)^{n-1} (-r-r(-x/r)^2-2x(-x/r)) \\+ \frac{1}{4i} \frac{1}{x^2-r^2} (-r/x)^{n-1} (r+r(-r/x)^2+2x(-r/x)) \\ = \frac{1}{4i} \frac{1}{x^2-r^2} (-x/r)^{n-1} (-r+x^2/r) \\+ \frac{1}{4i} \frac{1}{x^2-r^2} (-r/x)^{n-1} (-r+r^3/x^2) \\ = \frac{1}{4i} \frac{1}{r} (-x/r)^{n-1} - \frac{1}{4i} \frac{r}{x^2} (-r/x)^{n-1} \\ = - \frac{1}{4i} \frac{1}{x} (-x/r)^{n} + \frac{1}{4i} \frac{1}{x} (-r/x)^{n}.$$

Finally collect the contributions from both poles to get

$$2\pi i \times \frac{(-1)^n}{4ix} \left(-(x/r)^n + (r/x)^n + (r/x)^n + (x/r)^n\right) \\ = 2\pi i \times \frac{(-1)^n}{4ix} \times 2 (r/x)^n \\ = \frac{\pi (-1)^n}{x} \left(\frac{r}{x}\right)^n.$$

Answer 3

Utilize \begin{align} I(a) =& \int_0^{2\pi}\ln(1+2a\cos\phi+a^2) \cos n\phi\ d\phi\\ = & \int_0^{2\pi} \bigg(2\sum_{k=1}^\infty \frac{(-1)^{k+1}}{ka^k}\ \cos k \phi\bigg) \cos n\phi \ d\phi=\frac{2\pi(-1)^{n+1}}{n a^n} \end{align} with $|a|>1$ to evaluate \begin{align} &\int_0 ^{2 \pi} \frac {(r \cos \phi +x) \cos n\phi} {r^2+2xr \cos \phi +x^2} \ d\phi = \frac1{2r}\frac{dI(a)}{da}\bigg|_{a=\frac xt }= \frac {(-1)^n\pi r^n} {x^{n+1}} \end{align}