Integral of polynomial times rational function of trig function over multiple periods

Problem: Compute

$$\int_0^{12\pi} dx \frac{x}{6+\cos 8x}$$

It looks like the OP is trying to compute the antiderivative and use the fundamental theorem of calculus. With multiple periods, that approach is paved with all sorts of difficulty that is really an artifice related to the functional form of the antiderivative. In truth, there should be no such difficulty. A better approach involves the residue theorem, which I will outline below. Note that the presence of the linear term in the numerator provides a slight complication, but one that has been treated before in this site several times.

$$\int_{0}^{12 \pi} dx \frac{x}{6+\cos{8 x}} = \frac1{64} \int_0^{96 \pi} du \frac{u}{6+\cos{u}}$$

$$\int_{n 2 \pi}^{(n+1) 2 \pi} du \frac{u}{6+\cos{u}} = \int_0^{2 \pi} dv \frac{v+2 \pi n}{6+\cos{v}} = \int_0^{2 \pi} dv \frac{v}{6+\cos{v}} + 2 \pi n \int_0^{2 \pi} \frac{dv}{6+\cos{v}}$$

Thus, summing over $n$:

$$\int_{0}^{12 \pi} dx \frac{x}{6+\cos{8 x}} = \frac{48}{64} \int_0^{2 \pi} dv \frac{v}{6+\cos{v}} + \frac{48 (47) \pi}{64} \int_0^{2 \pi} \frac{dv}{6+\cos{v}}$$

We may now compute each of these integrals. The second integral is straightforward by the residue theorem, i.e., let $z=e^{i v}$, then

$$\int_0^{2 \pi} \frac{dv}{6+\cos{v}} = -i 2 \oint_{|z|=1} \frac{dz}{z^2+12 z+1}$$

The only pole of the integrand inside the unit circle is at $z=-6+\sqrt{35}$. The integral is then $i 2 \pi$ times the residue of the integrand at this pole, or $2 \pi/\sqrt{35}$.

To compute the first integral, we consider the complex integral

$$ \oint_C dz \frac{\log{z}}{z^2+12 z+1} $$

where $C$ is the unit circle with a detour up and back about the positive real axis. (A circular piece about the origin vanishes.) This integral is equal to

$$-\frac12 \int_0^{2 \pi} dv \frac{v}{6+\cos{v}} + \int_1^0 dx \frac{\log{x}+i 2 \pi}{x^2+12 x+1} + \int_0^1 dx \frac{\log{x}}{x^2+12 x+1}$$

or, simplifying,

$$-\frac12 \int_0^{2 \pi} dv \frac{v}{6+\cos{v}} – i 2 \pi \int_0^1 \frac{dx}{x^2+12 x+1} $$

The contour integral is also equal to $i 2 \pi$ times the residue at the pole $z=-6+\sqrt{35}$. Note that the negative sign is taken to be $e^{i \pi}$. Thus we have

$$-\frac12 \int_0^{2 \pi} dv \frac{v}{6+\cos{v}} = i 2 \pi \int_0^1 \frac{dx}{x^2+12 x+1} + i 2 \pi \frac{-\log{\left ( 6+\sqrt{35} \right )}+i \pi}{2 \sqrt{35}} $$


$$\frac1{x^2+12 x+1} = \frac1{2 \sqrt{35}} \left (\frac1{x+6-\sqrt{35}} – \frac1{x+6+\sqrt{35}} \right ) $$

so that

$$\int_0^1 \frac{dx}{x^2+12 x+1} = \frac1{2 \sqrt{35}} \log{\left (\frac{7+\sqrt{35}}{7-\sqrt{35}} \right )} = \frac1{2 \sqrt{35}} \log{\left (6+\sqrt{35}\right )}$$

Note the cancellation with the real part of the residue. Thus,

$$\int_0^{2 \pi} dv \frac{v}{6+\cos{v}} = \frac{2 \pi^2}{\sqrt{35}} $$

Putting these results altogether, we have

>$$\int_0^{12 \pi} dx \frac{x}{6+\cos{8 x}} = \frac{72 \pi^2}{\sqrt{35}} $$

One Comment

  • Hello Ron,

    I think the second bit is somewhat more convoluted than necessary since one could just reflect the integral across the bounds to cancel out the linear term on the numerator.

    Apart from that, this simplification is great! 🙂

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