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I am trying to evaluate the following integrals:

$$\int\limits_{-\infty}^\infty \frac{x^2}{1+x^2+x^4}dx $$

$$\int\limits_{0}^\pi \frac{d\theta}{a\cos\theta+ b} \text{ where }0<a<b$$

My very limited text has the following substitution:

$$\int\limits_0^\infty \frac{\sin x}{x}dx = \frac{1}{2i}\int\limits_{\delta}^R \frac{e^{ix}-e^{-ix}}{x}dx \cdots $$

Is the same of substitution available for the polynomial? Thanks for any help. I apologize in advance for slow responses, I have a disability that limits me to an on-screen keyboard.

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up vote 1 down vote accepted

$f(z) := \frac{1}{a\cos z+b}$

Let $I$ be the integral in question. Let's double the integral, to get

$$2I =\int_0^{2\pi} f(z)\, dz$$

Let $C$ be the contour of the unit circle $|z|=1$ in the positive direction. Then using this, we have

$$2I = \oint_C \frac{1}{a\left(\frac{z+z^{-1}}{2}\right)+b}\frac{dz}{iz} = -2 i \oint_C \frac{dz}{az^2+2bz+a}$$

The last integral can be evaluated by residues. The poles of the function are at

$$b_\pm = \frac{-b\pm \sqrt{b^2-a^2}}{a}$$

Based on $b>a>0$, we find the only pole in the contour $C$ is $b_+$. The residue there is:

$$z_+ = \operatorname*{Res}_{z=b_+}\frac{1}{az^2+2bz+a} = \frac{1}{2\sqrt{b^2-a^2}}$$


$$2I = -2 i \left(\frac{2 \pi i}{2\sqrt{b^2-a^2}}\right) = \frac{2\pi}{\sqrt{b^2-a^2}}$$

Divide by $2$ and done.

$$f(z) := \frac{z^2}{1+z^2+z^4} = \frac{z^2}{(z^2-z+1)(z^2+z+1)}$$

Poles of $f$ occur at, using the quadratic formula:

$$b_{1} = \frac{1+\sqrt 3i}{2}$$ $$b_{2} = \frac{-1+\sqrt 3i}{2}$$ $$b_{3} = \frac{1-\sqrt 3i}{2}$$ $$b_{4} = \frac{-1-\sqrt 3i}{2}$$

Using the canonical semicircle contour ($Re^{i\theta}$ for $\theta \in [0,\pi]$) over the upper half plane, it is easily seen that the integral of $f(z)$ over the arc disappears as $R \to \infty$. Then we only need to find the residues of $b_1$ and $b_2$ (using the first or second formula here):

$$z_1 = \operatorname*{Res}_{z=b_1}f(z)= -\frac{1}{4}-\frac{i}{4\sqrt 3}$$ $$z_2 = \operatorname*{Res}_{z=b_2}f(z)= \frac{1}{4}-\frac{i}{4\sqrt 3}$$

$$\oint_C f(z)\, dz = \int_{-\infty}^\infty f(x)\, dx = 2 \pi i (z_1+z_2) =-2 \pi i \frac{i}{2\sqrt 3} = \frac{\pi}{\sqrt 3}$$

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Check your work - I think there's a typo in your original function definition, in the denominator. – Ron Gordon May 6 '13 at 23:37

For the first one, write $\dfrac{x^2}{1+x^2+x^4}$ as $\dfrac{x}{2(1-x+x^2)} - \dfrac{x}{2(1+x+x^2)}$. Now $$\dfrac{x}{(1-x+x^2)} = \dfrac{x-1/2}{\left(x-\dfrac12\right)^2 + \left(\dfrac{\sqrt3}2 \right)^2} + \dfrac{1/2}{\left(x-\dfrac12\right)^2 + \left(\dfrac{\sqrt3}2 \right)^2}$$ and $$\dfrac{x}{(1-x+x^2)} = \dfrac{x+1/2}{\left(x+\dfrac12\right)^2 + \left(\dfrac{\sqrt3}2 \right)^2} - \dfrac{1/2}{\left(x+\dfrac12\right)^2 + \left(\dfrac{\sqrt3}2 \right)^2}$$

I trust you can take it from here.

For the second one, from Taylor series of $\dfrac1{(b+ax)}$, we have $$\dfrac1{(b+a \cos(t))} = \sum_{k=0}^{\infty} \dfrac{(-a)^k}{b^{k+1}} \cos^{k}(t)$$ Now $$\int_0^{\pi} \cos^k(t) dt = 0 \text{ if $k$ is odd}$$ We also have that $$\color{red}{\int_0^{\pi}\cos^{2k}(t) dt = \dfrac{(2k-1)!!}{(2k)!!} \times \pi = \pi \dfrac{\dbinom{2k}k}{4^k}}$$ Hence, $$I=\int_0^{\pi}\dfrac{dt}{(b+a \cos(t))} = \sum_{k=0}^{\infty} \dfrac{a^{2k}}{b^{2k+1}} \int_0^{\pi}\cos^{2k}(t) dt = \dfrac{\pi}{b} \sum_{k=0}^{\infty}\left(\dfrac{a}{2b}\right)^{2k} \dbinom{2k}k$$ Now from Taylor series, we have $$\color{blue}{\sum_{k=0}^{\infty} x^{2k} \dbinom{2k}k = (1-4x^2)^{-1/2}}$$ Hence, $$\color{green}{I = \dfrac{\pi}{b} \cdot \left(1-\left(\dfrac{a}b\right)^2 \right)^{-1/2} = \dfrac{\pi}{\sqrt{b^2-a^2}}}$$

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Although the solution above for the second integral works, it's nice to see a rather general method for turning such trignonometric integrals into contour integrals. The following method I learned from Alfhors. We have $$ I = \frac{1}{2} \int_{0}^{2 \pi} \frac{d \theta}{a \cos \theta + b} $$ For the right hand integral, consider the contour $C = \{|z| = 1\}$ in the complex plane. For $z = e^{i \theta} \in C$ we have $$ \cos \theta = \frac{1}{2} (z + \frac{1}{z}) $$ and $$ dz = i e^{i \theta} d \theta = i z d \theta $$ This leads to $$ I = \frac{1}{2} \int_{C} \frac{i dz}{z(\frac{a}{2}(z + \frac{1}{z}) + b)} $$ You can now apply the usual residue calculus to this integral.

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Here are the complex variables techniques you have asked for. The first one, consider the complex integral

$$ \int\limits_{C} \frac{z^2}{1+z^2+z^4}dz, $$

where $C$ is the upper half of the circle $z=Re^{i\theta}$ centered at the origin. Check Jordan' lemma. The second one, write it in the form

$$ \int\limits_{0}^\pi \frac{d\theta}{a\cos\theta+ b}= \frac{1}{2}\int\limits_{-\pi}^\pi\frac{d\theta}{a\cos\theta+ b}, $$

since the integrand is an even function, then use the change of variables $ \theta=\pi-\theta$, to write the integral in the form

$$ \frac{1}{2} \int\limits_{-\pi}^\pi \frac{d\theta}{a\cos\theta+ b}= \frac{1}{2}\int\limits_{0}^{2\pi} \frac{d\theta}{a\cos\theta+ b}.$$

Now, use complex variables techniques $z=e^{i\theta}$ where the $C:|z|=1$. For techniques to find the residue see here.

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