# Thread: Number of ways to evaluate an integral

1. ## Number of ways to evaluate an integral

This is a group challenge:

It's well known that $\displaystyle \int_0^\infty \frac{\sin x}{x}dx = \frac{\pi}{2}$.

I know of a few ways of showing this. My challenge to all of you is to come up with as many ways as you can to prove this result.

Let's see if there's any ways I'm not aware of yet!

2. There are different ways to compute the integral...

$\displaystyle \int_{0}^{\infty} \frac{\sin x}{x}\ dx$ (1)

... even if not all [in my opinion...] are 'fully convincing'... there is however also way to non compute it: if You consider (1) as a Lebesgue integral it diverges... in my opiniom this 'controversial' has to be, sooner or later, clarified...

Kind regards

$\chi$ $\sigma$

3. I only know of 2 and here they are.

$\displaystyle \int_{0}^{\infty}\frac{\sin(x)}{x}dx=\int_{0}^{\in fty}\int_{0}^{\infty}e^{-sx}\sin(x)dsdx$

Now switch the order of integration

$\displaystyle \int_{0}^{\infty}\int_{0}^{\infty}e^{-sx}\sin(x)dxds=\int_{0}^{\infty}\frac{1}{s^2+1}ds= \tan^{-1}(s) \bigg|_{0}^{\infty}=\frac{\pi}{2}$

The other is via the residue theorem

Let $C_1$ be the linear arc from $[-R,\-\rho]$ and $C_2: \rho e^{it};\pi < t < 2\pi$ $C_3:[\rho,R]$ and $C_4: Re^{it}; 0< t <\pi$

Now notice that $\frac{\sin(x)}{x}$ is an even function so

$\displaystyle \int_{0}^{\infty}\frac{\sin(x)}{x}dx=\frac{1}{2}\i nt_{-\infty}^{\infty}\frac{\sin(x)}{x}dx$

Now consider

$\displaystyle \oint \frac{e^{iz}}{z}dz=2\pi i(1)$

$f(z)=\frac{e^{iz}}{z}=\sum_{n=0}^{\infty}\frac{i^{ n}(z)^{n-1}}{n!}=\frac{1}{z}+g(z)$ where $g(z)$ is an analytic function.

Using this and the ML bound we get

$\displaystyle \lim_{\rho \to 0}\int_{C_2}\left( \frac{1}{z}+g(z)\right)dz=\pi i$

Now by Jordan's lemma on $C_4$

$\displaystyle \lim_{R \to \infty}\int_{C_4}\frac{e^{iz}}{z}dz =0$

This leaves

$\displaystyle \int_{-\infty}^{\infty}\frac{e^{iz}}{z}dz+\pi i=2\pi i \iff \int_{-\infty}^{\infty}\frac{\sin(z)}{z}dz=\pi i$

Since $e^{iz}=\cos(z)+i\sin)(z)$ we take the real and immaginary parts to get

$\int_{-\infty}^{\infty}\frac{\sin(x)}{x}dx=\pi$

This gives

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

4. Originally Posted by TheEmptySet
I only know of 2 and here they are.

$\displaystyle \int_{0}^{\infty}\frac{\sin(x)}{x}dx=\int_{0}^{\in fty}\int_{0}^{\infty}e^{-sx}\sin(x)dsdx$

Now switch the order of integration

$\displaystyle \int_{0}^{\infty}\int_{0}^{\infty}e^{-sx}\sin(x)dxds=\int_{0}^{\infty}\frac{1}{s^2+1}ds= \tan^{-1}(s) \bigg|_{0}^{\infty}=\frac{\pi}{2}$
Haven't seen this one!

Using this and the ML bound we get

$\displaystyle \lim_{\rho \to 0}\int_{C_2}\left( \frac{1}{z}+g(z)\right)dz=\pi i$
Also, I evaluated $\displaystyle \int_{C_2}$ differently. What exactly is "the ML bound"?

5. On the arc $\rho=e^{it};\pi < t < 2\pi$

We have

$\displaystyle \int_{\rho}\left(\frac{1}{z}+g(z) \right)dz$

So we get

$\int_{\pi}^{2\pi}idt+\int_{\pi}^{2\pi}g(\rho e^{it})dt$

The 2nd integral is bounded by it's maximum on the arc multiplied by its length so we get

$\pi i +\pi \rho M$ but as $\rho \to 0$ we get $\pi i$

6. Originally Posted by TheEmptySet
On the arc $\rho=e^{it};\pi < t < 2\pi$

We have

$\displaystyle \int_{\rho}\left(\frac{1}{z}+g(z) \right)dz$

So we get

$\int_{\pi}^{2\pi}idt+\int_{\pi}^{2\pi}g(\rho e^{it})dt$

The 2nd integral is bounded by it's maximum on the arc multiplied by its length so we get

$\pi i +\pi \rho M$ but as $\rho \to 0$ we get $\pi i$
Oh ok, that's what I did actually. I didn't understand what ML meant, but I do now.

Also for your first method not to be picky but I think you might need to justify switching the order of integration.

7. Here is a proof , first note that

$\tan(\frac{x}{2}) = \frac{x}{2} \displaystyle{ \prod_{k=1}^{\infty} \frac{ 1- \frac{x^2}{(2k\pi)^2 } }{ 1 - \frac{x^2}{((2k-1)\pi)^2} } }$

By taking log. differentiation , we obtain this equation :

$\frac{1}{\sin(x)} = \frac{1}{x} + 2x \displaystyle{ \sum_{k=1} \frac{(-1)^k }{ x^2 - k^2 \pi^2 } }$

I would like to stop here and back to the integral $\int_0^{\infty} \frac{\sin(x)}{x} ~dx$ .

We find that the integrand is even , i write it as

$\frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin(x)}{x} ~dx$

$= \frac{1}{2} \displaystyle{ \left( \int_0^{\pi} + \sum_{k=1}^{\infty} [ \int_{k\pi}^{(k+1)\pi} + \int_{-k\pi}^{(-k+1)\pi} ] \right) \frac{\sin(x)}{x} ~dx }$

Consider $\displaystyle{ \sum_{k=1}^{\infty} ( \int_{k\pi}^{(k+1)\pi} + \int_{-k\pi}^{(-k+1)\pi} ) \right) \frac{\sin(x)}{x} ~dx }$

Sub. $x = t + k\pi$ and $x = t - k\pi$ on the first and the second integral respectively ,we obtain :

$\displaystyle{ \sum_{k=1}^{\infty} (-1)^k \left( \int_{0}^{\pi} \frac{\sin(t)}{t+k\pi} ~dt + \int_{0}^{\pi} \frac{\sin(t)}{t-k\pi} ~dt \right) }$

$= \displaystyle{ \sum_{k=1}^{\infty} (-1)^k \int_{0}^{\pi} \frac{2t \sin(t) }{ t^2 -k^2\pi^2 }~ dt \right) }$

Therefore ,

$\int_0^{\infty} \frac{\sin(x)}{x} ~dx = \frac{1}{2} \displaystyle{ \int_0^{\pi} \left( \frac{1}{x} + 2x \sum_{k=1}^{\infty} \frac{ (-1)^k }{ x^2 -k^2\pi^2 } \right) ~\sin(x)~dx }$

$= \frac{1}{2} \int_0^{\pi} \frac{\sin(x)}{\sin(x)}~dx = \frac{\pi}{2}$

8. Originally Posted by TheEmptySet
I only know of 2 and here they are.

$\displaystyle \int_{0}^{\infty}\frac{\sin(x)}{x}dx=\int_{0}^{\in fty}\int_{0}^{\infty}e^{-sx}\sin(x)dsdx$

Now switch the order of integration
the problem here is that you don't have absolute convergence.

in order to justify the interchange, it's better to integrate by parts first by selecting $1-\cos x$ as an antiderivative for $\sin x,$ from there you introduce the parameter you said and the interchange is justified now by Tonelli's theorem since the function is no negative.

9. You could use the following laplace transform's property:

$\displaystyle L\left( \dfrac{f(t)}{t} \right) = \int_s^{\infty} F(\beta) \, d\beta$

10. Originally Posted by Krizalid
the problem here is that you don't have absolute convergence.

in order to justify the interchange, it's better to integrate by parts first by selecting $1-\cos x$ as an antiderivative for $\sin x,$ from there you introduce the parameter you said and the interchange is justified now by Tonelli's theorem since the function is no negative.

Something like this:

$\displaystyle = \int^{\infty}_{0} \frac{1- \cos x}{x^{2}} \ dx = \int^{\infty}_{0} \int^{\infty}_{0} (1-\cos x) t e^{-xt} \ dt \ dx$

so now the integrand is never negative

$\displaystyle = \int^{\infty}_{0} \int^{\infty}_{0} (te^{-xt} - t \cos x e^{-xt}) \ dx \ dt$

$\displaystyle = \int^{\infty}_{0} \big( 1- t \ \mathca{L} \{ \cos x \} (t) \big) \ dt$

$= \displaystyle \int^{\infty}_{0} \Big( 1 - \frac{t^{2}}{1+t^{2}} \Big) \ dt = \int^{\infty}_{0} \frac{1}{1+t^{2}} \ dt = \frac{\pi}{2}$

11. yes, that approach works fine and is justified.

12. I asked a professor, and he said something about that because the issue is at a single point (the origin), that what the TheEmptySet did was OK. I think the absolute convergence requirement is too restrictive. I don't know.

13. perhaps is restrictive, but it works, but you'd need a good justification when reversing the integration order by not having absolute convergence.

14. Although $\displaystyle \int_{(0, \infty)} \frac{sin(x)}{x}d\lambda$ diverges, one can show $\displaystyle \lim_{a \to \infty} \int_{(0,a)} \frac{sin(x)}{x} = \frac{\pi}{2}$ in the following way:

First, note that since $0 \leq |sin(x)e^{-xy}|$ is measurable, we have that $\displaystyle \int_{(0,a) \times (0,\infty)}|sin(x)e^{-xy}| = \int_{0}^{a}|sin(x)| (\int_{0}^{\infty}|e^{-xy}|dy)dx = \int_{0}^{a} \frac{|sin(x)|}{|x|}dx < \infty$ and so $sin(x)e^{-xy} \in L^1((0,a)\times(0,\infty))$

Now note that using Fubini on $f(x,y)=sin(x)e^{-xy}$ gives $\displaystyle \int_{0}^{a}\frac{sin(x)}{x}dx$ and changing the order of integration the other way gives, after using integration by parts twice, $\displaystyle \int_{0}^{a}\frac{sin(x)}{x}dx = \frac{\pi}{2} - cos(a)\int_{0}^{\infty}\frac{e^{-ay}}{1+y^2}dy$ $\displaystyle -sin(a)\int_{0}^{\infty} \frac{ye^{-y}}{1+y^2}dy$

Taking the limit as $a \to \infty$ and using DCT finished off the argument.

15. Here's a way using minimum machinery, but it requires the simple lemma that if $f\in\mathcal{C}[a,b]$ then $\displaystyle \lim_{n\to\infty}\int_a^b f(x)\sin((2n+1)x)\text{ }dx=0$ and the fact that $\displaystyle \int_0^{\infty}\frac{\sin(x)}{x}$ exists, although this follows immediately from Dirichlet's test for integrals.

Ok, so note that since $\displaystyle \int_0^{\infty}\frac{\sin(x)}{x}\text{ }dx$ exists we have that

$\displaystyle \int_0^{\infty}\frac{\sin(x)}{x}\text{ }dx=\lim_{\lambda\to\infty}\int_0^\lambda \frac{\sin(x)}{x}\text{ }dx=\lim_{\substack{n\to\infty\\ n\in\mathbb{N}}}\int_0^{\frac{\pi n}{2}}\frac{\sin(x)}{x}\text{ }dx=\lim_{n\to\infty}\int_0^{\frac{\pi}{2}}\frac{\ sin(nx)}{x}\text{ }dx$

and thus clearly $\displaystyle \int_0^\infty \frac{\sin(x)}{x}\text{ }dx=\lim_{n\to\infty}\int_0^{\frac{\pi}{2}}\frac{\ sin\left((2n+1)x\right)}{x}\text{ }dx$. We next claim that $\displaystyle \int_0^{\frac{\pi}{2}}\frac{\sin\left((2n+1)x\righ t)}{\sin(x)}\text{ }dx=\frac{\pi}{2}$ for every $n\in\mathbb{N}$. But, this follows immediately from the common trigonometric identity associated with the Dirichlet kernel. We then note that

\displaystyle \begin{aligned}\lim_{n\to\infty}\left(\int_0^{\fra c{\pi}{2}}\frac{\sin\left((2n+1)x\right)}{x}\text{ }dx-\int_0^{\frac{\pi}{2}}\frac{\sin\left((2n+1)x\righ t)}{\sin(x)}\text{ }dx\right) &=\lim_{n\to\infty}\int_0^{\frac{\pi}{2}}\frac{\si n(x)-x}{\sin(x)}\sin\left((2n+1)x\right)\text{ }dx\\ &=0\end{aligned}

where we've made use of the lemma mentioned at the beginning of the post. Thus, it follows that

$\displaystyle \int_0^{\infty}\frac{\sin(x)}{x}\text{ }dx=\lim_{n\to\infty}\int_0^{\frac{\pi}{2}}\frac{\ sin\left((2n+1)x\right)}{x}\text{ }dx=\lim_{n\to\infty}\int_0^{\frac{\pi}{2}}\frac{\ sin\left((2n+1)x\right)}{\sin(x)}\text{ }dx=\frac{\pi}{2}$

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