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## A Class of Trigonometric Integrals

### Re: A Class of Trigonometric Integrals

Sat Feb 08, 2014 3:34 am
Random Variable Integration Guru

Posts: 381
I'm going to evaluate integrals (5) and (6) using the identities I posted above, and two additional integrals.

For $\displaystyle |a| <1$,

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

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

$\displaystyle \frac{1}{1-a^{2}} \sum_{k=1}^{\infty} ka^{k-1} \sin(kx) = \frac{\sin x}{(1-2a \cos x +a^{2})^{2}}$ (Differentiate both sides of the first equation with respect to $\displaystyle x$.)

$\displaystyle \int_{0}^{\infty} \frac{dx}{(1+x^{2})(1-2a \cos x +a^{2})} = \frac{1}{1-a^{2}} \int_{0}^{\infty} \Big(1+2 \sum_{k=1}^{\infty} a^{k} \cos(kx) \Big) \ dx$

$\displaystyle = \frac{\pi}{2} \frac{1}{1-a^{2}} + \frac{2}{1-a^{2}} \sum_{k=1}^{\infty} a^{k} \int_{0}^{\infty} \frac{\cos(kx)}{1+x^{2}} \ dx$

$\displaystyle = \frac{\pi}{2} \frac{1}{1-a^{2}} + \frac{\pi}{1-a^{2}} \sum_{k=1}^{\infty} \Big(\frac{a}{e} \Big)^{k}$

$\displaystyle = \frac{\pi}{2} \frac{1}{1-a^{2}} + \frac{\pi}{1-a^{2}} \frac{a/e}{1-a/e}$

$\displaystyle = \frac{\pi}{2} \frac{1}{1-a^{2}} \Big(1+ \frac{2a}{e-a} \Big) = \frac{\pi}{2} \frac{1}{1-a^{2}} \frac{e+a}{e-a}$

$\displaystyle \int_{0}^{\infty} \frac{\cos x}{(1+x^{2})(1-2a \cos x+a^{2})} \ dx = \int_{0}^{\infty} \frac{1}{1+x^{2}} \Big( \frac{a}{1-a^{2}} + \frac{1+a^{2}}{1-a^{2}} \sum_{k=1}^{\infty} a^{k} \cos(kx) \Big) \ dx$

$\displaystyle = \frac{\pi}{2} \frac{a}{1-a^{2}} + \frac{1+a^{2}}{1-a^{2}} \sum_{k=1}^{\infty} a^{k} \int_{0}^{\infty} \frac{\cos (kx)}{1+x^{2}} \ dx$

$\displaystyle = \frac{\pi}{2} \frac{a}{1-a^{2}} + \frac{\pi}{2} \frac{1+a^{2}}{a(1-a^{2})} \sum_{k=1}^{\infty} \Big(\frac{a}{e} \Big)^{k}$

$\displaystyle = \frac{\pi}{2} \frac{a}{1-a^{2}} + \frac{\pi}{2} \frac{1+a^{2}}{a(1-a^{2})} \frac{a/e}{1-a/e}$

$\displaystyle = \frac{\pi}{2} \frac{1}{1-a^{2}} \Big(a + \frac{1+a^{2}}{e-a} \Big) = \frac{\pi}{2} \frac{1}{1-a^{2}} \frac{ae+1}{e-a}$

$\displaystyle \int_{0}^{\infty} \frac{x \sin x}{(1+x^{2})(1-2a \cos x +a^{2})^{2}} \ dx = \int_{0}^{\infty} \frac{x}{1+x^{2}} \frac{1}{1-a^{2}} \sum_{k=1}^{\infty} ka^{k-1} \sin(kx) \ dx$

$\displaystyle = \frac{1}{1-a^{2}} \sum_{k=1}^{\infty} k a^{k-1} \int_{0}^{\infty} \frac{x \sin (kx)}{1+x^{2}} \ dx$

$\displaystyle = \frac{\pi}{2} \frac{1}{a(1-a^{2})} \sum_{k=1}^{\infty} k \Big(\frac{a}{e} \Big)^{k}$

$\displaystyle = \frac{\pi}{2} \frac{1}{a(1-a^{2})} \frac{a/e}{(1-a/e)^{2}} = \frac{\pi}{2} \frac{1}{1-a^{2}} \frac{e}{(e-a)^{2}}$

$\displaystyle \int_{0}^{\infty} \frac{dx}{(x^{4}+4) (1-2 a \cos x +a^{2})} \ dx = \int_{0}^{\infty} \frac{1}{x^{4}+4} \frac{1}{1-a^{2}} \Big(1 + 2 \sum_{k=1}^{\infty} a^{k} \cos(kx) \Big) \ dx$

$\displaystyle = \frac{1}{1-a^{2}} \frac{\sqrt{2}}{4} \frac{\pi}{4} \csc \Big( \frac{\pi}{4} \Big) + \frac{2}{1-a^{2}} \sum_{k=1}^{\infty} a^{k} \int_{0}^{\infty} \frac{\cos(kx)}{x^{4}+4} \ dx$

$\displaystyle = \frac{\pi}{8} \frac{1}{1-a^{2}} + \frac{\pi}{4} \frac{1}{1-a^{2}} \sum_{k=1}^{\infty} a^{k}e^{-k} (\cos k + \sin k)$

$\displaystyle \sum_{k=0}^{\infty} (ae^{-1} e^{i})^{k} = \frac{1}{1-ae^{-1}e^{i}} = \frac{1-ae^{-1} \cos(1) + iae^{-1} \sin(1)}{1-2ae^{-1} \cos(1)+a^{2}e^{-2}}$

Therefore,

$\displaystyle \sum_{k=0}^{\infty} a^{k}e^{-k} \cos(k) = \frac{1-ae^{-1} \cos(1)}{1-2ae^{-1} \cos(1)+a^{2}e^{-2}} = \frac{e^{2}-ae \cos(1)}{e^{2}-2a e \cos(1)+a^{2}}$

$\displaystyle \implies \sum_{k=1}^{\infty} a^{k}e^{-k} \cos(k) = \frac{e^{2}-ae \cos(1)}{e^{2}-2a e \cos(1)+a^{2}} -1 = \frac{ae \cos(1)-a^{2}}{e^{2}-2ae\cos(1)+a^{2}}$

And

$\displaystyle \sum_{k=1}^{\infty} a^{k} e^{-k} \sin(k) = \sum_{k=0}^{\infty} a^{k} e^{-k} \sin(k) = \frac{ae \sin(1)}{e^{2}-2ae \cos(1)+a^{2}}$

So

$\displaystyle \int_{0}^{\infty} \frac{dx}{(x^{4}+4) (1-2 a \cos x +a^{2})} = \frac{\pi}{8} \frac{1}{1-a^{2}} + \frac{\pi}{4} \frac{1}{1-a^{2}} \frac{ae \cos(1)+ae \sin(1)-a^{2}}{e^{2}-2ae \cos(1)+a^{2}}$

$\displaystyle = \frac{\pi}{8} \frac{1}{1-a^{2}} \frac{e^{2}+2ae \sin(1)-a^{2}}{e^{2}-2ae \cos(1)+a^{2}}$

### Re: A Class of Trigonometric Integrals

Sun Feb 09, 2014 8:47 pm
Random Variable Integration Guru

Posts: 381
Differentiate the identity $\displaystyle \sum_{k=1}^{\infty} a^{k} \sin(kx) = \frac{a \sin x}{1-2a \cos x +a^{2}} \ (|a| <1 )$ with respect to $\displaystyle x$.

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

Now multiply both sides by $\displaystyle 2a(1-a^{2})$.

$\displaystyle 2(1-a^{2}) \sum_{k=1}^{\infty} k a^{k} \cos(kx) = \frac{2a(1-a^{4}) \cos(x) - 4a^{2}(1-a^{2})}{(1-2a \cos x +a^{2})^{2}}$ (1)

Next rewrite the identity $\displaystyle 1+ 2 \sum_{k=1}^{\infty} a^{k} \cos(kx) = \frac{1-a^{2}}{(1-2a \cos x +a^{2})^{2}} \ (|a|<1)$ in a different form.

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

And multiply both sides of that equation by $\displaystyle 1+a^{2}$.

$\displaystyle 1+ a^{2} + 2(1+a^{2}) \sum_{k=1}^{\infty} a^{k} \cos(kx) = \frac{(1-a^{4})(1-2a \cos x +a^{2})}{(1-2a \cos x +a^{2})^{2}}$ (2)

Finally add equations (1) and (2).

$\displaystyle 1+a^{2} + 2 (1-a^{2}) \sum_{k=1}^{\infty} ka^{k} \cos(kx) + 2 (1+a^{2}) \sum_{k=1}^{\infty} a^{k} \cos(kx) = \frac{1-3a^{2}+3a^{4}-a^{6}}{(1-2a \cos x +a^{2})^{2}} = \frac{(1-a^{2})^{3}}{(1-2a \cos x +a^{2})^{2}}$

That identity can then be used to evaluate something like $\displaystyle \int_{0}^{\infty} \frac{dx}{(1+x^{2})(1-2a \cos x +a^{2})^{2}}$.

### Re: A Class of Trigonometric Integrals

Thu Jun 19, 2014 9:22 am

Posts: 852
Location: Jaipur, India

\displaystyle \begin{align*} \tan^{-1}\left( \frac{p\sin qx}{1+p\cos qx}\right) &= \Im \; \log \left(1+i \frac{p\sin qx}{1+p\cos qx}\right) \\ &= \Im \; \log\left(1+p \cos qx+i p \sin qx \right) \\ &= \Im \; \log\left(1+p e^{iqx} \right) \\ &= \Im \; \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}p^n e^{iq nx} \\ &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}p^n \sin(qnx) \end{align*}

Using this series we can create some more crazy results. We will assume $1>p>-e^q$ from now on.

1.

\displaystyle \begin{align*} \int_0^\infty \frac{x}{1+x^2}\tan^{-1}\left( \frac{p\sin qx}{1+p\cos qx}\right) \; dx &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}p^n \int_0^\infty \frac{x \sin(qn x)}{1+x^2}dx \\ &= \frac{\pi}{2}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}p^n e^{-qn} \\ &= \frac{\pi}{2} \log(1+p e^{-q}) \end{align*}

2.

\displaystyle \begin{align*} \int_0^\infty \frac{x}{(1+x^2)^2}\tan^{-1}\left( \frac{p\sin qx}{1+p\cos qx}\right) \; dx &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}p^n \int_0^\infty \frac{x \sin(qn x)}{(1+x^2)^2}dx \\ &= \frac{q\pi }{4}\sum_{n=1}^\infty (-1)^{n+1}\; p^n e^{-qn} \\ &= \frac{\pi }{4} \frac{pq}{e^q+p} \end{align*}

3.

\displaystyle \begin{align*} \int_0^\infty \frac{x}{1+x^4}\tan^{-1}\left( \frac{p\sin qx}{1+p\cos qx}\right) \; dx &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}p^n \int_0^\infty \frac{x \sin(qn x)}{1+x^4}dx \\ &= \frac{\pi}{2}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}p^n e^{-qn/\sqrt{2}}\sin\left(\frac{qn}{\sqrt{2}} \right) \\ &= \frac{\pi}{2}\Im \log \left( 1+p e^{-q/\sqrt{2}+iq/\sqrt{2}}\right) \\ &= \frac{\pi}{2}\tan^{-1}\left(\frac{p \sin \left( \frac{q}{\sqrt{2}}\right)}{e^{q/\sqrt{2}}+p \cos \left( \frac{q}{\sqrt{2}}\right)} \right) \end{align*}

### Re: A Class of Trigonometric Integrals

Thu Jun 19, 2014 9:47 am

Posts: 852
Location: Jaipur, India

Differentiating 1,2,3 of the previous post respectively w.r.t $p$, we obtain

1. $\displaystyle \int_0^\infty \frac{x}{1+x^2}\frac{\sin qx}{1+2p \cos(qx)+p^2}dx =\frac{\pi}{2}\frac{1}{p+e^q}$

2. $\displaystyle \int_0^\infty \frac{x}{(1+x^2)^2}\frac{\sin qx}{1+2p \cos(qx)+p^2}dx =\frac{\pi}{4}\left(\frac{q}{e^q+p}-\frac{pq}{(e^q+p)^2}\right)$

3. $\displaystyle \int_0^\infty \frac{x}{1+x^4}\frac{\sin qx}{1+2p \cos(qx)+p^2}dx =\frac{\pi}{2} \frac{\sin\left(\frac{q}{\sqrt{2}} \right)}{e^{q/\sqrt{2}}+2p\cos\left(\frac{q}{\sqrt{2}} \right) +e^{-q/\sqrt{2}}p^2}$

These are similar to what RV derived.

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