Proving $\int^{\infty}_{0}\frac{\sin x}{x} = \frac{\pi}{2}$

[alyafey22]

Prove the following :

$$\int^{\infty}_{0} \frac{\sin (x) }{ x} = \frac{\pi }{2}$$

There are three different methods to solve the integral .

 

[chisigma]

I'm affectionated to the Laplace Transform so that I remember it's following property...

$\displaystyle \mathcal{L} \{f(t)\} = F(s) \implies \mathcal{L} \{\frac{f(t)}{t}\} = \int_{s}^{\infty} F(u)\ du$ (1)

... so that is...

$\displaystyle \mathcal{L} \{\sin t\} = \frac{1}{1 + s^{2}} \implies \mathcal{L} \{ \frac{\sin t}{t}\} = \int_{s}^{\infty} \frac {d u}{1+ u^{2}} = \frac{\pi}{2} - \tan^{-1} s$ (2)

... and computing (2) for s=0 we find...

$\displaystyle \int_{0}^{\infty} \frac{\sin t}{t}\ dt = \frac{\pi}{2}$ (3)

Kind regards

$\chi$ $\sigma$

 

[alyafey22]

I'm affectionated to the Laplace Transform so that I remember it's following property...

$\displaystyle \mathcal{L} \{f(t)\} = F(s) \implies \mathcal{L} \{\frac{f(t)}{t}\} = \int_{s}^{\infty} F(u)\ du$ (1)

... so that is...

$\displaystyle \mathcal{L} \{\sin t\} = \frac{1}{1 + s^{2}} \implies \mathcal{L} \{ \frac{\sin t}{t}\} = \int_{s}^{\infty} \frac {d u}{1+ u^{2}} = \frac{\pi}{2} - \tan^{-1} s$ (2)

... and computing (2) for s=0 we find...

$\displaystyle \int_{0}^{\infty} \frac{\sin t}{t}\ dt = \frac{\pi}{2}$ (3)

Kind regards

$\chi$ $\sigma$

Excellent . Still two other ways to solve it .

 

[I like Serena]

I'm more of a Fourier Transform fan so I looked up (nr 202) its following property (picking one with a greek letter in it)...
$$\mathcal{F} \{\text{sinc }x\}(\xi) = \text{rect } \xi$$
$$\int_{-\infty}^{+\infty} \frac {\sin(\pi x)}{\pi x} e^{-2\pi i \xi x}dx = \text{rect } \xi$$

For $\xi = 0$ this is:
$$\int_{-\infty}^{+\infty} \frac {\sin(\pi x)}{\pi x} dx = 1$$

Substituting $\pi x = t$ gives:
$$\int_{-\infty}^{+\infty} \frac {\sin t}{t} dx = \pi$$

And since $\frac {\sin t}{t}$ is symmetric, we get:
$$\int_{0}^{\infty} \frac {\sin t}{t} dx = \frac \pi 2$$

 

[alyafey22]

I'm more of a Fourier Transform fan so I looked up (nr 202) its following property (picking one with a greek letter in it)...
$$\mathcal{F} \{\text{sinc }x\}(\xi) = \text{rect } \xi$$
$$\int_{-\infty}^{+\infty} \frac {\sin(\pi x)}{\pi x} e^{-2\pi i \xi x}dx = \text{rect } \xi$$

For $\xi = 0$ this is:
$$\int_{-\infty}^{+\infty} \frac {\sin(\pi x)}{\pi x} dx = 1$$

Substituting $\pi x = t$ gives:
$$\int_{-\infty}^{+\infty} \frac {\sin t}{t} dx = \pi$$

And since $\frac {\sin t}{t}$ is symmetric, we get:
$$\int_{0}^{\infty} \frac {\sin t}{t} dx = \frac \pi 2$$

I didn't know it can be solved this way . I had two other methods in mind :)

 

[chisigma]

A milestone of complex analysis is the use of Cauchy Integral Theorems for solving integrals that are 'resistent' to 'conventional' attaks. In particular the first of these theorems extablishes that if a complex variable function is analytic inside and along a closed path C is...

$\displaystyle \int_{C} f(z)\ d z = 0$ (1)

Now we computing (1) in the case $\displaystyle f(z)= \frac{e^{i\ z}}{z}$ when the path is represented in the figure...

View attachment 805

... and when $R \rightarrow \infty$ and $r \rightarrow 0$. The Jordan's lemma extablishes that the integral along the 'big circle' tends to 0 if $ R \rightarrow \infty$ , so that, tacking into account (1), we conclude that is...

$\displaystyle \int_{- \infty}^{+ \infty} \frac{\sin x}{x}\ dx = \lim_{r \rightarrow 0} \int_{0}^{\pi} e^{i\ r\ e^{i \theta}}\ d \theta = \pi$ (2)

Details have been omitted but it's clear that this approach is more complex and less elegant that the Laplace Transform approach...

Kind regards

$\chi$ $\sigma$

 

[alyafey22]

Details are been omitted but it's clear that this approach is more complex and less elegant that the Laplace Transform approach...

A complex analysis approach is necessarily complex ;) , but I don't I agree it is less elegant .

 

[alyafey22]

The last method

See hint

Spoiler

Differentiation under the integral sign

 

[polygamma]

You can also use contour integration to evaluate $\displaystyle \int_{0}^{\infty} \frac{\sin^{n} x}{x^{n}} \ dx \ (n = 2, 3, \ldots) $.$\displaystyle \int_{0}^{\infty} \frac{\sin^{n} x}{x^{n}} \ dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin^{n} x}{x^{n}} \ dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{1}{x^{n}} \Big( \frac{e^{ix}-e^{-ix}}{2i} \Big)^{n} \ dx$

$ \displaystyle = \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{1}{(x-i \epsilon)^{n}} \Big( e^{ix}-e^{-ix} \Big)^{n} \ dx $The last line is justified for $n >1$ by the dominated convergence theorem.$ \displaystyle = \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{1}{(x-i \epsilon)^{n}} \sum_{k=0}^{n} \binom{n}{k} (e^{ix})^{n-k} (-e^{-ix})^{k} \ dx $ (binomial theorem)

$ \displaystyle = \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{1}{(x-i \epsilon)^{n}} \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} e^{ix(n-2k)} \ dx $

$ \displaystyle = \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \int_{-\infty}^{\infty} \frac{e^{ix(n-2k)}}{(x- i \epsilon)^{n}} \ dx$Now let $\displaystyle f(z) = \frac{e^{iz(n-2k)}}{(z- i \epsilon)^{n}}$ and integrate around a large closed half-circle in the upper-half complex plane if $n-2k \ge 0$, and around a large closed half-circle in the lower-half plane if $n - 2k < 0$.

There will be no contribution from the integrals around the half-circle in the lower-half plane since the pole is in the upper half plane. So we'll be summing from $k=0$ to the largest integer such that $k \le \frac{n}{2}$.$ \displaystyle = \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{k} (-1)^{k} \ 2 \pi i \ \text{Res}[f,i \epsilon]$The pole at $z=i \epsilon$ is of order $n$.So $\displaystyle \text{Res} [f, i \epsilon] = \frac{1}{(n-1)!} \lim_{z \to i \epsilon} \frac{d^{n-1}}{dx^{n-1}} \ e^{iz(n-2k)} = \frac{1}{(n-1)!} \lim_{z \to i \epsilon} i^{n-1} (n-2k)^{n-1} e^{iz(n-k)} $

$ \displaystyle = \frac{1}{(n-1)!} i^{n-1} (n-2k)^{n-1} e^{-\epsilon(n-k)} $And $\displaystyle \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{k} (-1)^{k} 2 \pi i \ \text{Res}[f,i \epsilon] $

$ \displaystyle = \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{k} (-1)^{k} 2 \pi i \ \frac{1}{(n-1)!} i^{n-1} (n-2k)^{n-1} e^{-\epsilon(n-k)}$

$ \displaystyle = \frac{\pi}{2^{n}(n-1)!} \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{k} (-1)^{k} (n-2k)^{n-1} $So, for example, $\displaystyle \int_{0}^{\infty} \frac{\sin^{5} x}{x^{5}} \ dx = \frac{\pi}{2^{5}4!} \sum_{k=0}^{2} \binom{5}{k} (-1)^{k}(5-2k)^{4}= \frac{\pi}{768} \Big(5^4-5(3)^{4} +10(1)^4 \Big) = \frac{115 \pi}{384}$

 

[alyafey22]

Here is the third , using differentiation under the integral sign F(a)=\int^{\infty}_0 \frac{\sin(x) e^{-ax}}{x}\, dxF&#039;(a)=-\int^{\infty}_0 \sin(x) e^{-ax}\, dxF&#039;(a)=-\int^{\infty}_0 \sin(x) e^{-ax}\, dx=\frac{-1}{a^2+1}F(a)=-\arctan(a)+CC=\lim_{a\to \infty }F(a) +\arctan(a)= \frac{\pi}{2}F(a)=-\arctan(a)+ \frac{\pi}{2}F(0)=\int^{\infty}_0 \frac{\sin(x) }{x}=\frac{\pi}{2}

 

[polygamma]

$\int_{0}^{\infty} \sin (x) e^{-ax} = \frac{1}{1+a^{2}} $ for $a >0$

So really what you're saying is that

$ \lim_{a \to 0^{+}} \int_{0}^{\infty} \frac{\sin (x) e^{-ax}}{x} \ dx = \int_{0}^{\infty} \lim_{a \to 0} \frac{\sin (x) e^{-ax}}{x} \ dx = \int_{0}^{\infty} \frac{\sin x}{x} \ dx = \lim_{a \to 0} \Big( \arctan(a) + \frac{\pi}{2} \Big) = \frac{\pi}{2}$

Bringing the limit inside of the integral is not easy to justify since $\int_{0}^{\infty} \frac{\sin x}{x} \ dx$ does not converge absolutely.The following is an example where things go horribly wrong.

$\lim_{a \to 0^{+}} \int_{0}^{\infty} \sin(x) e^{-ax} \ dx = \int_{0}^{\infty} \lim_{a \to 0^{+}} \sin(x) e^{-ax} \ dx =\int_{0}^{\infty} \sin x \ dx = \lim_{a \to 0^{+}} \frac{1}{1+a^{2}}= 1 $

 

[alyafey22]

$$ \lim_{a \to 0^{+}} \int_{0}^{\infty} \frac{\sin (x) e^{-ax}}{x} \ dx = \int_{0}^{\infty} \lim_{a \to 0} \frac{\sin (x) e^{-ax}}{x} \ dx = \int_{0}^{\infty} \frac{\sin x}{x} \ dx = \lim_{a \to 0} \Big( \arctan(a) + \frac{\pi}{2} \Big) = \frac{\pi}{2}$$

If We take $$F(a)=\lim_{a \to 0^{+}} \int_{0}^{\infty} \frac{\sin (x) e^{-ax}}{x} \ dx $$

since F(a) exists for a=0 then we can pass the limit inside .

The following is an example where things go horribly wrong.

$$\lim_{a \to 0^{+}} \int_{0}^{\infty} \sin(x) e^{-ax} \ dx = \int_{0}^{\infty} \lim_{a \to 0^{+}} \sin(x) e^{-ax} \ dx =\int_{0}^{\infty} \sin x \ dx = \lim_{a \to 0^{+}} \frac{1}{1+a^{2}}= 1 $$

Since at a=0 the integral doesn't converge , we cannot pass the limit inside .

 

[polygamma]

since F(a) exists for a=0 then we can pass the limit inside

There is no theorem that says that's true in general.

 

[alyafey22]

There is no theorem that says that's true in general.

It always works for me , may be I need to prove it. Can you give an example that if we swap we get a different convergent value ?

 

[polygamma]

$\displaystyle \lim_{n \to \infty} \int_{0}^{\infty} \frac{n \arctan x}{n^2+x^{2}} \ dx = \int_{0}^{\infty} \lim_{n \to \infty} \frac{n \arctan x}{n^2+x^{2}} \ dx = \int_{0}^{\infty} 0 \ dx = 0 $

But $ \displaystyle \lim_{n \to \infty} \int_{0}^{\infty} \frac{n \arctan x}{n^2+x^{2}} \ dx$ tends to $\frac{\pi^{2}}{4}$

 

[I like Serena]

Here is the third , using differentiation under the integral sign F(a)=\int^{\infty}_0 \frac{\sin(x) e^{-ax}}{x}\, dxF&#039;(a)=-\int^{\infty}_0 \sin(x) e^{-ax}\, dxF&#039;(a)=-\int^{\infty}_0 \sin(x) e^{-ax}\, dx=\frac{-1}{a^2+1}F(a)=-\arctan(a)+CC=\lim_{a\to \infty }F(a) +\arctan(a)= \frac{\pi}{2}F(a)=-\arctan(a)+ \frac{\pi}{2}F(0)=\int^{\infty}_0 \frac{\sin(x) }{x}=\frac{\pi}{2}

Isn't this the same as chisigma[/color]'s solution?

Note that $$\mathcal L\{\frac {\sin x} x\}(a) = \int^{\infty}_0 \frac{\sin(x) e^{-ax}}{x}\, dx$$.
Combine with the Laplace formula for taking a derivative and there you go...

 

[alyafey22]

Isn't this the same as chisigma[/color]'s solution?

Note that $$\mathcal L\{\frac {\sin x} x\}(a) = \int^{\infty}_0 \frac{\sin(x) e^{-ax}}{x}\, dx$$.
Combine with the Laplace formula for taking a derivative and there you go...

Lots of things look different while they are actually the same. Finding the connection is not so easy, though .

 

[polygamma]

My preferred method

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

Integrate by parts by letting $u= \frac{1}{x}$ and $dv = \sin x \ dx$.

For $v$ choose the antiderivative $1- \cos x$.

$ \displaystyle = \frac{1- \cos x}{x} \big|^{\infty}_{0} + \int_{0}^{\infty} \frac{1- \cos x}{x^{2}} \ dx = \int_{0}^{\infty} \frac{1- \cos x}{x^{2}} \ dx$

$ \displaystyle = \int_{0}^{\infty} \int_{0}^{\infty} (1- \cos x) te^{-xt} \ dt \ dx $

Since the integrand is always nonnegative, Fubini's/Tonelli's theorem says that we can switch the order of integration. That's why I initially integrated by parts.

$ \displaystyle = \int_{0}^{\infty} \int_{0}^{\infty} t (1- \cos x) e^{-tx} \ dx \ dt = \int_{0}^{\infty} t \Big( \frac{1}{t} - \frac{t}{1+t^{2}} \Big) \ dt$

$ \displaystyle = \int_{0}^{\infty} \frac{1}{1+t^{2}} \ dt = \frac{\pi}{2}$

 

[I like Serena]

Lots of things look different while they are actually the same. Finding the connection is not so easy, though .

I guess the trick is in understanding how the Laplace transform works.
Once you learn how to apply it in its integral form, your derivation pops out.
As for the Fourier transform, I have to admit that it's not so nice to look up a transform.
On the other hand, it's really easy to deduce the inverse transform of the rectangle function:
$$\mathcal F^{-1}\{\text{rect } \xi\}(x) = \int_{-\infty}^\infty \text{rect } \xi \ e^{2\pi i \xi x} d\xi = \int_{-1/2}^{1/2} e^{2\pi i \xi x} d\xi = \left. \frac 1 {2\pi i x} e^{2\pi i \xi x} \right|_{-1/2}^{1/2} = \frac 1 {2i \pi x}(e^{i \pi x} - e^{-i \pi x}) = \text{sinc x}$$

So that makes the Fourier transform solution still pretty easy and really different (that is, no derivative involved).

- - - Updated - - -

$ \displaystyle = \int_{0}^{\infty} \int_{0}^{\infty} (1- \cos x) te^{-xt} \ dt \ dx $

Where did that new integral come from?
It looks suspiciously like a Laplace transform.

 

[polygamma]

Where did that new integral come from?
It looks suspiciously like a Laplace transform.

I just expressed the integral as an iterated integral by using the Laplace transform $ \displaystyle \int_{0}^{\infty} t e^{-xt} \ dt = \frac{1}{x^{2}}$.

In general, $\displaystyle \int_{0}^{\infty} t^{n} e^{-xt} \ dt = \frac{n!}{x^{n+1}}$.

A lot of integrals can be evaluated in this manner.