# Compute the integral $\int_{-\infty}^\infty \frac{\sin(x) (1+\cos(x))}{x(2+\cos(x))} dx$

## Background

I recently found out about Lobachevsky's integral formula, so I tried to create a problem on my own for which I'd be able to apply this formula. The problem is presented below.

## Problem

Compute the integral $$\int_{-\infty}^\infty \frac{\sin(x) (1+\cos(x))}{x(2+\cos(x))} dx$$

## Attempt

If we define $$f(x) := \frac{1+\cos(x)}{2+\cos(x)}$$, which most definitely is $$\pi$$ - periodic.

The integral is, using our notation above, on the form

$$I = \int_{-\infty}^\infty \frac{\sin(x)}{x}f(x) dx.$$

The integrand is even, so we might as well compute

$$I = 2 \int_{0}^\infty \frac{\sin(x)}{x}f(x) dx.$$

We will now have to make use of a theorem.

Lobachevsky's integral formula states that if $$f(x)$$ is a continous $$\pi$$ - periodic function then we have that $$\int_0^\infty \frac{\sin(x)}{x}f(x) dx= \int_0^{\pi/2} f(x) dx.$$

Substituing our $$f(x)$$ yields us

$$\int_0^{\pi/2} \frac{1+\cos(x)}{2+\cos(x)} dx = \pi/2 - \int_0^{\pi/2}\frac{1}{2+\cos(x)}dx$$

where

$$I_2 = \int_0^{\pi/2}\frac{1}{2+\cos(x)}dx = \int_0^{\pi/2}\frac{\sec^2(x/2)}{3+\tan^2(x/2)}dx.$$

Letting $$u = \tan(x/2)/\sqrt{3}$$, for which $$du = \sec^2(x/2)/(2\sqrt{3})dx$$, therefore gives us:

$$I_2 = \int_0^{1/\sqrt{3}}\frac{2\sqrt{3}}{3u^2+3} = \frac{\pi}{3\sqrt{3}}.$$

Finally we can compute $$I$$ to

$$I = 2\left(\frac{\pi}{2} - \frac{\pi}{3\sqrt{3}}\right) = \frac{\pi(3\sqrt{3}-2)}{3\sqrt{3}}.$$

I've tried calculating this integral in Desmos where it gives me $$0$$ when I calculate the integrand on the interval $$(-\infty, \infty)$$, and something negative for $$(0,\infty)$$. This contradicts my answer.

I also tried typing it into Wolfram, without success. Can anyone confirm the validity of my result?

• Is not $f(x)$ $2\pi$-periodic? Commented Feb 23, 2023 at 19:58
• @MostafaAyaz There it is, should have changed it to $2x$ instead of $x$ as argumentin the cosines. Then my solution would be valid I guess, with some changes introduced to the values of the integrals of course. Thank you for spotting my mistake. Commented Feb 23, 2023 at 20:08
• @Tanamas No, there is a problem in that $\sin(2x)$ appears or $\sin(x/2)$ maybe, so the conditions of the formula wouldn't be satisfied Commented Feb 23, 2023 at 20:09
• But that would be changing the original problem! Commented Feb 23, 2023 at 21:04
• @FShrike Haha yup! Guess there'll be new things to learn instead. Either way, thank you! Commented Feb 24, 2023 at 7:39

Another approach:

Using the Fourier series $$1 + 2 \sum_{k=1}^{\infty} (-1)^{n} a^{n}\cos(nx) = \frac{1-a^{2}}{1+2a \cos (x) +a^{2}}, \quad |a| <1,$$ we have

\begin{align} \int_{-\infty}^{\infty} \frac{\sin (x)}{x}\frac{1-a^{2}}{1+2a \cos (x) +a^{2}} \, \mathrm dx &= \int_{-\infty}^{\infty} \frac{\sin (x)}{x}\left(1+2 \sum_{n=1}^{\infty} (-1)^{n} a^{n} \cos(nx) \right) \, \mathrm dx \\ &= \pi + 2 \sum_{n=1}^{\infty} (-1)^{n}a^{n}\int_{-\infty}^{\infty} \frac{\sin (x)}{x} \cos(nx) \, \mathrm dx \\ &= \pi - 2a \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \, \cos (x) \, \mathrm dx + 2 \sum_{n=2}^{\infty} (-1)^{n} a^{n} \int_{-\infty}^{\infty}\frac{\sin (x)}{x} \, \cos(nx) \mathrm dx \\ &= \pi - 2a \left(\frac{\pi}{2} \right)+2\sum_{n=2}^{\infty} (-1)^{n}a^{n}(0) \\ &= \pi \left(1-a \right) . \end{align}

Rewriting the integral as

$$\frac{1}{1+a^{2}}\int_{-\infty}^{\infty} \frac{\sin (x)}{x} \frac{1-a^{2}}{1+\frac{2a}{1+a^{2}}\cos(x)} \, \mathrm dx,$$ and letting $$a= 2-\sqrt{3}$$, we get $$\sqrt{3} \int_{-\infty}^{\infty}\frac{\sin (x)}{x} \frac{1}{2+\cos(x)} \, \mathrm dx = \pi \left(\sqrt{3}-1 \right).$$

Therefore, \begin{align} \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \frac{1+\cos (x)}{2+ \cos (x)} \, \mathrm dx &= \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \, \mathrm dx - \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \frac{1}{2+ \cos (x)} \, \mathrm dx \\ &= \pi - \frac{\pi}{\sqrt{3}} \left(\sqrt{3} - 1 \right) \\ &= \frac{\pi}{\sqrt{3}}. \end{align}

The one issue with this approach is justification for switching the order of integration and summation. Fubini's theorem is not satisfied.

Fortunately, we can use Sangchul Lee's result from the addendum of this answer to justify the switching.

• Nice solution! I guess you can apply the result in my answer to justify the switching (see the 'Addendum' part). Commented Feb 24, 2023 at 7:50
• @SangchulLee Since $$\sin(x) \cos(nx) = \frac{\sin \left((n+1)x \right) - \sin \left((n-1)x \right)}{2},$$ it indeed appears that your result can be used to justify the switch. Thanks. Commented Feb 24, 2023 at 15:31

There is a generalization of Lobachevsky's integral formula that is applicable.

The formula states that if $$f(x)$$ is an odd periodic function of period $$a$$, then $$\int_{0}^{\infty} \frac{f(x)}{x} \, \mathrm dx = \frac{\pi}{a} \int_{0}^{a/2} f(x) \cot \left(\frac{\pi x}{a} \right) \, \mathrm dx \tag{1}$$ if the integral on the left converges.

A proof can be found in the paper The integrals in Gradshteyn and Ryzhik. Part 16: Complete elliptic integrals by Boettner and Moll. It's Lemma 3.1.

The proof says to use the partial fraction expansion of $$\tan (z)$$. That's clearly a typo. It should be the partial fraction expansion of $$\cot (z)$$ since $$\sum_{k=0}^{\infty} \left( \frac{1}{x+ka}- \frac{1}{(k+1)a-x} \right) = \frac{1}{x} + \sum_{k=1}^{\infty} \frac{2x}{x^{2}-k^{2}a^{2}} = \frac{\pi}{a} \cot \left(\frac{\pi x}{a} \right).$$

Using $$(1)$$ and the tangent half-angle substitution, we get \begin{align} \int_{-\infty}^{\infty} \frac{\sin(x)}{x} \frac{1+ \cos x}{2+ \cos x} \, \mathrm dx &= 2 \int_{0}^{\infty} \frac{\sin(x)}{x} \frac{1+ \cos x}{2+ \cos x} \, \mathrm dx \\ &= \int_{0}^{\pi} \sin(x) \, \frac{1+ \cos x}{2+ \cos x} \, \cot\left(\frac{x}{2} \right) \, \mathrm dx \\ &= 8 \int_{0}^{\infty} \frac{\mathrm dt}{(t^{2}+1)^{2}(t^{2}+3)} \\&= 4 \int_{0}^{\infty} \frac{\mathrm dt}{(t^{2}+1)^{2}} -2 \int_{0}^{\infty} \frac{\mathrm dt}{t^{2}+1} + 2 \int_{0}^{\infty}\frac{\mathrm dt}{t^{2}+3} \\ &= 4 \left(\frac{\pi}{4} \right) - 2 \left( \frac{\pi}{2} \right) + 2 \left( \frac{\pi}{2\sqrt{3}} \right) \\ &= \frac{\pi}{\sqrt{3}}. \end{align}

To emphasize what is stated in the paper, if $$g(x)$$ is an even $$\pi$$-periodic function, then $$g(x) \sin(x)$$ is an odd $$2 \pi$$-periodic function, and \begin{align} \int_{0}^{\infty} \frac{g(x) \sin (x)}{x} \, \mathrm dx &= \frac{1}{2} \int_{0}^{\pi} g(x) \sin(x) \cot \left(\frac{x}{2} \right) \, \mathrm dx \\ &= \frac{1}{2} \int_{0}^{\pi} g(x) \sin(x) \, \frac{1+ \cos(x)}{\sin(x)} \, \mathrm dx \\ &= \frac{1}{2} \int_{0}^{\pi} g(x) \, \mathrm dx + \frac{1}{2} \int_{0}^{\pi} g(x) \cos (x) \, \mathrm dx \\ &= \int_{0}^{\pi/2} g(x) \, \mathrm dx + \frac{1}{2} (0) \\ &= \int_{0}^{\pi/2} g(x) \, \mathrm dx \end{align} which is Lobachevsky's integral formula.

It wasn't immediately clear to me why $$\int_{0}^{\pi} g(x) \cos(x) \, \mathrm dx =0.$$

But since $$g(x) \cos(x)$$ is an even function, we have $$\int_{0}^{\pi} g(x) \cos(x) \, \mathrm dx = \int_{-\pi}^{0} g(x) \cos (x) \, \mathrm dx.$$

And making the substitution $$u = x- \pi$$, we have $$\int_{0}^{\pi} g(x) \cos(x) \, \mathrm dx = -\int_{-\pi}^{0} g(u) \cos (u) \, \mathrm du.$$

The only way both equations can be simultaneoulsy true is if $$\int_{0}^{\pi} g(x) \cos(x) \, \mathrm dx=0.$$

• Nice! Is this a straightforward case of interchanging integration and Mittag-Leffler expansion or is it a subtle theorem? Commented Feb 25, 2023 at 17:54
• @FShrike Unless I'm overlooking something subtle, the proof given in that paper seems straightforward. Commented Feb 25, 2023 at 17:59

Utilize the series $$\frac{p\sin x}{1+2p\cos x+p^2}=-\sum_{k=1}^\infty (-p)^{k}\sin kx$$ with $$p=2-\sqrt3$$ to express
\begin{align} \frac{\sin x}{2+\cos x}= -2\sum_{k=1}^\infty (\sqrt3-2)^k\sin kx \end{align} Then, integrate \begin{align} &\int_{-\infty}^\infty \frac{\sin x}x\frac{1+\cos x}{2+\cos x}dx\\ =& \ 2\int_{-\infty}^\infty \frac{\sin x}xdx- 2\int_{-\infty}^\infty \frac{1}x\frac{\sin x}{2+\cos x}dx\\ =&\ \pi +4\sum_{k=1}^\infty (\sqrt3-2)^k\int_0^\infty \frac{\sin kx}xdx\\ =&\ \pi + 2\pi \sum_{k=1}^\infty (\sqrt3-2)^k= \frac\pi{\sqrt3} \end{align} where the summation is geometric.

We start by $$\int_{-\infty}^\infty \frac{\sin x}{x} \frac{1+\cos x}{2+\cos x}dx {= \sum_{k\in \Bbb Z}\int_{2k\pi}^{2(k+1)\pi} \frac{\sin x}{x} \frac{1+\cos x}{2+\cos x}dx \\= \sum_{k\in \Bbb Z}\int_{0}^{2\pi} \frac{\sin (x+2k\pi)}{x+2k\pi} \frac{1+\cos x}{2+\cos x}dx \\= \int_{0}^{2\pi} \sum_{k\in \Bbb Z}\frac{\sin (x+2k\pi)}{x+2k\pi} \frac{1+\cos x}{2+\cos x}dx \\= \int_{0}^{2\pi} \sum_{k\in \Bbb Z}\frac{\sin (\omega+2k\pi)}{\omega+2k\pi} \frac{1+\cos \omega}{2+\cos \omega}d\omega. }$$ Note that $$F(\omega)= \sum_{k\in \Bbb Z}\frac{\sin (\omega+2k\pi)}{\omega+2k\pi}$$ and $$G(\omega)=\frac{1+\cos \omega}{2+\cos \omega}$$ are the Fourier transforms of $$x[n]$$ and $$y[n]$$, respectively, where $$x[n]=\frac{1}{2}\Pi(\frac{n}{2}) \ \ \ ,\ \ \ y[n]=\delta[n]-\frac{1}{\sqrt 3}(-2+\sqrt 3)^{|n|}.$$ Therefore, according to the Parseval's identity we find $$\int_{0}^{2\pi} \sum_{k\in \Bbb Z}\frac{\sin (\omega+2k\pi)}{\omega+2k\pi} \frac{1+\cos \omega}{2+\cos \omega}d\omega {= \int_{0}^{2\pi} F(\omega) G(\omega)d\omega = 2\pi\sum_{n\in \Bbb Z} x[n]y[-n] \\= 2\pi(x[-1]y[1]+x[0]y[0]+x[1]y[-1]) = \frac{\pi}{\sqrt 3}. }$$

Update

For connecting $$F(\omega)$$ to $$x[n]$$, I used the sampling theorem, which is:

Let the Fourier transform of $$x(t)$$ be $$X(f)$$. Then, the discrete-time Fourier transform (DTFT) of the sampled signal $$\hat x[n]=x(n/F_s)$$ is $$\hat X(f)=F_s\sum_{k\in\Bbb Z} X(F_s(f-k)).$$

I applied the aforementioned theorem to $$x(t)=\Pi(t)$$ and $$F_s=2$$.

• Why should it be clear that $F$ is the transform of (a shifted) Gamma function? That seems like a nasty integration, but I admit I haven't tried it yet Commented Feb 23, 2023 at 22:47
• In fact, in what sense do you mean Fourier Transform? $$\int\Pi(x)e^{-ix}$$Doesn’t converge over the real line Commented Feb 23, 2023 at 23:49
• @FShrike, the second integral converges, since the integral bounds over which the integrand is non-zero are bounded, i.e. $\int_{\Bbb R}\Pi(x)\exp(-ix)dx=\int_{-0.5}^0.5 \exp(-ix)$. About your first question, I used the sampling theorem. I will add it to my answer. Commented Feb 24, 2023 at 10:29
• Oh. Your use of Pi confused me! I now realise you’re using Pi for the window function whereas I had interpreted it as the Euler Pi function ie $x\mapsto\Gamma(x+1)$ Commented Feb 24, 2023 at 12:09

$$\newcommand{\d}{\,\mathrm{d}}$$I can tell you from long personal experience that Desmos is really really really really bad at doing trigonometric integrals over large domains. For example, we all know: $$\int_0^\infty\frac{\sin x}{x}\,\mathrm{d}x=\frac{\pi}{2}$$But:

If you play around with the size of the upper bound, Desmos' answer wavers between $$0$$ and $$3$$, roughly, and the differences don't at all converge. For reference, $$\pi/2\approx1.570796$$.

Let's try some others. For example: $$\int_0^\infty\sin(x^2)\d x=\frac{1}{2}\sqrt{\frac{\pi}{2}}$$

Yet:

Which is atrocious.

I could give more examples, but I hope I've made my point: if your integrand features trigonometric quantities that don't really rapidly decay, don't use Desmos, use Wolfram or some other computer.

As commented, the period of your $$f$$ is not $$\pi$$ but rather $$2\pi$$. This breaks the formula, so we need another method.

How else can we compute this integral? Complex analysis!

Rewrite the integrand: $$\frac{\sin x}{x}\left(1-\frac{1}{2+\cos x}\right)$$So that our integral is just: $$\pi-\int_{-\infty}^\infty\frac{\sin x}{x}(2+\cos x)^{-1}\d x=\pi-2\alpha\cdot\Im\int_{-\infty}^\infty\frac{1}{\alpha+\beta\cdot e^{-ix}}\cdot\frac{\d x}{x}$$Where: $$\alpha=\frac{\sqrt{3}+1}{2},\,\beta=\frac{\sqrt{3}-1}{2}$$

If we take the contours $$-R\to-1/R$$ (straight line), $$-1/R\to1/R$$ (semicircle of radius $$1/R$$), $$1/R\to R$$ (straight line) and then $$R\to-R$$ (semicircle of size $$R$$) for large $$R$$ which are uniformly bounded away from the moduli of the poles (see below), $$n$$ an integer, and limit as $$R\to+\infty$$, our remaining integral shall equal: $$\Im\left[2\pi i\cdot\sum\operatorname{Res}_{\Im z>0}+\frac{\pi i}{\alpha+\beta}\right]=\frac{\pi}{\alpha+\beta}+2\pi\cdot\Re\sum\operatorname{Res}_{\Im z>0}$$ Since it's not too hard to see that the integrals over the larger semicircles vanish as $$R\to\infty$$, and the integrals smaller semicircles near zero can be found with the half-residue principle. This would yield a genuine evaluation of the integral of the imaginary part (which converges) and a principal value (of zero) for the non-convergent real part. I here have only focused on the desired imaginary part.

When does our integrand have poles in the upper half plane? If $$\alpha+\beta\cdot e^{-i\sigma+\tau}=0$$ for some $$z=\sigma+i\tau$$ then we know $$e^{-i\sigma+\tau}=-\frac{\alpha}{\beta}=-2\alpha^2$$ which is purely real. Hence $$\sigma$$ must $$\pi n$$ for some integer $$n$$, and then $$(-1)^n\cdot e^{\tau}=-2\alpha^2<0$$ shows $$n$$ is necessarily an odd integer. Then $$\tau=\ln(2\alpha^2)$$ is forced: let's define $$\zeta:=\ln(2\alpha^2)$$. The poles are precisely $$z=\pi(2n+1)+i\zeta$$ for integer $$n$$. The residues are $$\frac{1}{z}\cdot\frac{1}{-i\beta e^{-iz}}=\frac{i}{-2\alpha^2\beta}\cdot\frac{1}{\pi(2n+1)+i\zeta}=-2i\beta\cdot\frac{\pi(2n+1)-i\zeta}{\zeta^2+\pi^2(2n+1)^2}$$. Thus the integral is: $$\frac{\pi}{\sqrt{3}}-4\pi\beta\zeta\cdot\sum_{n=-\infty}^\infty\frac{1}{\zeta^2+\pi^2(2n+1)^2}$$The other part of the sum, with summands $$-2\beta i\cdot\frac{\pi(2n+1)}{\zeta^2+\pi^2(2n+1)}$$, I have ignored since the residue sums arise symmetrically, and these summands have (almost) an odd symmetry in $$n$$ and the sums vanish. This confirms the clear fact that the principal value of the non-convergent real part is zero.

Our final answer should then be: $$\int_{-\infty}^\infty\frac{\sin x}{x}\cdot\frac{1+\cos x}{2+\cos x}\d x=4\pi\zeta\sum_{n=-\infty}^\infty\frac{1}{\zeta^2+\pi^2(2n+1)^2}-\frac{\pi}{\sqrt{3}}=\frac{\pi}{\sqrt{3}}$$

This matches the Wolfram Alpha numerical data. The final series computation was done using well known formulae: $$\forall a,b\in\Bbb R\setminus\{0\},\quad\quad\sum_{n=-\infty}^\infty\frac{1}{a^2+b^2(2n+1)^2}=\frac{\pi}{2ab}\tanh\left(\frac{\pi a}{2b}\right)$$Which can itself be derived from complex analysis. There is a master theorem:

Let $$f:\Bbb C\to\hat{\Bbb C}$$ be a meromorphic function with a finite set of poles $$P$$, none of which are integers, such that $$\lim_{z\to\infty}zf(z)=0$$. Then: $$\lim_{N\to\infty}\sum_{|n|\le N}f(n)=-\pi\sum_{p\in P}\operatorname{Res}_{s=p}(\cot(\pi s)f(s))$$

This can be strengthened a little bit. You need to apply it to the meromorphic map $$s\mapsto(a^2+b^2(2s+1)^2)^{-1}$$, with poles: $$s=-\frac{1}{2}\pm\frac{a}{2b}i$$

• Late answer back but thank you for your thorough explanation. Also, think I might be keeping myself distant from Desmos :) at least for calculating infinite sums. Commented Feb 27, 2023 at 20:31
• @Tanamas You’re welcome Commented Feb 27, 2023 at 20:37