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## A Class of Trigonometric Integrals

### A Class of Trigonometric Integrals

Mon Sep 30, 2013 11:43 am
Shobhit Site Admin

Posts: 852
Location: Jaipur, India

$\displaystyle \;$

Lemma 1

For suitable $x$, we have
\begin{align*}\sum_{n=0}^\infty \frac{\cos(n\theta)}{x^{n+1}}&= \frac{x-\cos \theta}{x^2-2x\cos \theta+1} \tag{1}\\ \sum_{n=1}^\infty \frac{\sin(n\theta)}{x^{n+1}}&= \frac{\sin \theta}{x^2-2x\cos \theta+1}\tag{2}\end{align*}

Proof : (1) and (2) are the real and imaginary parts of the geometric series
$\sum_{n=0}^\infty \left( \frac{e^{i\theta}}{x}\right)^n = \frac{x}{x-e^{i\theta}}$

Theorem 1

For suitable $a$, we have
$\int_0^{\infty} \frac{a-\cos(x)}{(a^2-2a\cos x+1)(1+x^2)}dx = \frac{\pi e}{2(ae-1)}\tag{3}$

Proof:

By Lemma 1,
\begin{align*}\int_0^\infty \frac{a-\cos(x)}{(a^2-2a\cos x+1)(1+x^2)}dx &= \sum_{n=0}^\infty \frac{1}{a^{n+1}}\int_0^\infty \frac{\cos(n x)}{1+x^2} dx \\&= \frac{\pi}{2a}\sum_{n=0}^\infty \left( \frac{1}{ae}\right)^n \\ &=\frac{\pi}{2a}\left( \frac{1}{1-\frac{1}{ae}}\right) \\ &= \frac{\pi e}{2(ae-1)}\end{align*}

### Re: A Class of Trigonometric Integrals

Mon Sep 30, 2013 12:31 pm
Shobhit Site Admin

Posts: 852
Location: Jaipur, India

Theorem 2

We have

$\int_0^\infty \frac{a-\cos (x)}{a^2-2a \cos x +1}\cdot \frac{1}{1+x^4}dx=\frac{\pi}{2\sqrt{2}}\exp \left( \frac{1}{\sqrt{2}}\right)\frac{a \exp \left(\frac{1}{\sqrt{2}} \right)+\sin \left( \frac{1}{\sqrt{2}}\right)-\cos \left( \frac{1}{\sqrt{2}}\right)}{1-2a \exp \left(\frac{1}{\sqrt{2}}\right)\cos \left( \frac{1}{\sqrt{2}}\right) +a^2 \exp \left(\sqrt{2}\right)} \tag{4}$

Proof:

Again by Lemma 1,

\begin{align*}
\int_0^\infty \frac{a-\cos (x)}{a^2-2a \cos x +1}\cdot \frac{1}{1+x^4}dx &= \sum_{n=0}^\infty \frac{1}{a^{n+1}}\int_0^\infty \frac{\cos(nx)}{1+x^4}dx \\
&= \frac{\pi}{2\sqrt{2}}\sum_{n=0}^\infty \frac{1}{a^{n+1}}\left\{ e^{-n/\sqrt{2}} \left( \cos \frac{n}{\sqrt{2}}+\sin \frac{n}{\sqrt{2}}\right)\right\} \\
&= \frac{\pi}{2\sqrt{2}} \left[ \Re \sum_{n=0}^\infty \frac{e^{-n(1+i)/\sqrt{2}}}{a^{n+1}}+\Im \sum_{n=0}^\infty \frac{e^{-n(1+i)/\sqrt{2}}}{a^{n+1}}\right] \\
&= \frac{\pi}{2\sqrt{2}}\exp \left( \frac{1}{\sqrt{2}}\right)\frac{a \exp \left(\frac{1}{\sqrt{2}} \right)+\sin \left( \frac{1}{\sqrt{2}}\right)-\cos \left( \frac{1}{\sqrt{2}}\right)}{1-2a \exp \left(\frac{1}{\sqrt{2}}\right)\cos \left( \frac{1}{\sqrt{2}}\right) +a^2 \exp \left(\sqrt{2} \right)}
\end{align*}

### Re: A Class of Trigonometric Integrals

Mon Sep 30, 2013 12:37 pm
Shobhit Site Admin

Posts: 852
Location: Jaipur, India

Example

Putting $a=2$ in theorem 2, we arrive at

$\displaystyle \int_0^\infty \frac{2-\cos x}{5-4\cos x}\times \frac{1}{1+x^4}dx=\frac{\pi}{2\sqrt{2}}\exp \left( \frac{1}{\sqrt{2}}\right)\frac{2 \exp \left(\frac{1}{\sqrt{2}} \right)+\sin \left( \frac{1}{\sqrt{2}}\right)-\cos \left( \frac{1}{\sqrt{2}}\right)}{1-4 \exp \left(\frac{1}{\sqrt{2}}\right)\cos \left( \frac{1}{\sqrt{2}}\right) +4 \exp \left(\sqrt{2} \right)}$

Note that Mathematica can't evaluate this integral.

### Re: A Class of Trigonometric Integrals

Mon Sep 30, 2013 1:48 pm
Shobhit Site Admin

Posts: 852
Location: Jaipur, India

Lemma 2

\begin{align*} \sum_{n=1}^\infty \frac{n \cos(n \theta)}{x^{n+1}} &= \frac{\cos \theta (1+x^2)-2x}{\left(x^2-2x\cos \theta+1 \right)^2} \end{align*}

Theorem 3

$\displaystyle \int_0^\infty \frac{1}{(1-2a\cos x+a^2)(1+x^2)}dx =\frac{\pi (e^2-1)}{2(e-a)(ae-1)} \left\{\frac{a}{a^2-1}-\frac{e}{e^2-1}\right\} \tag{5}$

Proof:

\begin{align*} \int_0^\infty \frac{1}{(1-2a\cos x+a^2)(1+x^2)}dx &= \frac{1}{2}\sum_{n=-\infty}^{+\infty} \int_{2\pi n}^{2\pi (n+1)} \frac{1}{(1-2a\cos x+a^2)(1+x^2)}dx \\ &= \frac{1}{2} \sum_{n=-\infty}^{+\infty} \int_0^{2\pi}\frac{1}{(1-2a\cos x+a^2)(1+(x+2\pi n)^2)}dx \\ &= \frac{1}{2} \int_0^{2\pi} \frac{1}{1-2a\cos x+a^2}\sum_{n=-\infty}^{+\infty} \frac{1}{1+(x+2\pi n)^2}dx \\ &= \frac{1}{4}\tanh\left( \frac{1}{2}\right) \int_0^{2\pi} \frac{\sec^2 \frac{x}{2}}{(a^2-2a\cos x+1)\left( \tan^2 \frac{x}{2}+\tanh^2 \frac{1}{2}\right)}dx \end{align*}

This integral can be evaluated by elementary techniques of integration. The final answer is
$$\frac{\pi (1+e)^2}{2(e-a)(ae-1)}\tanh\left( \frac{1}{2}\right) \left\{\frac{a}{a^2-1}-\frac{e}{e^2-1}\right\}$$

### Re: A Class of Trigonometric Integrals

Mon Sep 30, 2013 4:36 pm
Shobhit Site Admin

Posts: 852
Location: Jaipur, India

Theorem 4

$\displaystyle \int_0^\infty \frac{\cos x}{1-2a \cos x+a^2} \times \frac{1}{1+x^2}dx = \frac{a\pi (e^2-1)}{2(e-a)(ae-1)} \left\{\frac{a}{a^2-1}-\frac{e}{e^2-1}\right\}-\frac{\pi e}{2(ae-1)} \tag{6}$

Using this theorem we obtain the mysterious result

$\displaystyle \int_0^\infty \frac{\cos x}{5-4\cos x} \times \frac{1}{1+x^2}dx = \frac{\pi }{2e-1}\left[ \frac{e^2-1}{e-2}\left\{ \frac{2}{3}-\frac{e}{e^2-1}\right\}-\frac{e}{2} \right]=0.5568468705 \cdots$

### Re: A Class of Trigonometric Integrals

Wed Oct 02, 2013 12:36 pm
galactus
Global Moderator

Posts: 902
This is fun stuff, S. Interesting identities/results.

### Re: A Class of Trigonometric Integrals

Wed Oct 02, 2013 9:46 pm

Posts: 138
Location: North Londinium, UK
Really, really... Really nice, Shobhit! Thanks for sharing!

### Re: A Class of Trigonometric Integrals

Sat Oct 05, 2013 11:11 am
Shobhit Site Admin

Posts: 852
Location: Jaipur, India

I am currently working on $\displaystyle \int_0^\infty \frac{1}{1-2a\cos x+a^2} \times \frac{1}{1+x^4}dx$. If I find something, I will post it.

### Re: A Class of Trigonometric Integrals

Sat Jan 18, 2014 6:37 pm
Random Variable Integration Guru

Posts: 381
I hope you don't mind me resurrecting an old thread, but I found a way to simplify the calculations.

For $\displaystyle |a|<1$, $\displaystyle \sum_{k=0}^{\infty} a^{k} \cos (kx) = \frac{1- a \cos x}{1- 2a \cos x+a^{2}}$.

If one wants a formula for $\displaystyle a >1$, then just replace $\displaystyle a$ with $\displaystyle \frac{1}{a}$.

But notice that $\displaystyle 1 + 2 \sum_{k=1}^{\infty} a^{k} \cos(kx) = 1 + 2 \Big( \frac{1-a \cos x}{1- 2a \cos x+a^{2}} - 1\Big) = \frac{1-a^{2}}{1- 2 a \cos x+a^{2}}$

Using the above identity would simply the evaluation of (5).

And again if you want a formula for $\displaystyle |a|>1$, just replace $\displaystyle a$ with $\displaystyle \frac{1}{a}$.

$\displaystyle 1 + 2 \sum_{k=1}^{\infty} \left(\frac{1}{a} \right)^{k} \cos (kx) = \frac{a^{2}-1}{a^{2}-2a \cos x +1}$

$\displaystyle \implies -1 - 2 \sum_{k=1}^{\infty} \left(\frac{1}{a} \right)^{k} \cos(kx) = \frac{1-a^{2}}{1-2 a \cos x + a^{2}}$

Also notice that $\displaystyle \frac{a}{1-a^{2}} + \frac{1+a^{2}}{a(1-a^{2})} \sum_{k=1}^{\infty} a^{k} \cos (kx) = \frac{a}{1-a^{2}} + \frac{1+a^{2}}{a(1-a^{2})} \Big( \frac{1- a \cos x}{1-2 a \cos x +a^{2}} -1 \Big) = \frac{\cos x}{1-2 a \cos x +a^{2}}$

Or if $\displaystyle |a| >1$, $\displaystyle \frac{a}{a^{2}-1} + \frac{a(a^{2}+1)}{a^{2}-1} \sum_{k=1} \Big(\frac{1}{a} \Big)^{k} \cos kx = \frac{a^{2} \cos x}{a^{2}- 2a \cos x+1}$

$\displaystyle \implies \frac{1}{a(a^{2}-1)} + \frac{a^{2}+1}{a(a^{2}-1)} \sum_{n=1}^{\infty} \Big(\frac{1}{a} \Big)^{k} \cos kx = \frac{\cos x}{1- 2a \cos x +a^{2}}$

which would be very useful for the evaluation of (6)

### Re: A Class of Trigonometric Integrals

Sun Jan 19, 2014 3:54 am
Shobhit Site Admin

Posts: 852
Location: Jaipur, India

Hey, thanks for the suggestion! I will definitely try it.

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