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A Class of Trigonometric Integrals


Post Mon Sep 30, 2013 11:43 am
Shobhit Site Admin
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\(\displaystyle \;\)

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

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

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

Post Mon Sep 30, 2013 1:48 pm
Shobhit Site Admin
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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
$$\frac{\pi (1+e)^2}{2(e-a)(ae-1)}\tanh\left( \frac{1}{2}\right) \left\{\frac{a}{a^2-1}-\frac{e}{e^2-1}\right\}$$

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

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

Post Wed Oct 02, 2013 12:36 pm
galactus User avatar
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This is fun stuff, S. Interesting identities/results.

Post Wed Oct 02, 2013 9:46 pm

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

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

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

Post Sun Jan 19, 2014 3:54 am
Shobhit Site Admin
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Posts: 850
Location: Jaipur, India

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

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