\int_0^{\infty}\frac{\sin x}{x}\ dx

**Also sprach Zarathustra** · August 17th 2011, 08:33 AM

Okay, I'm trying to compute the following integral:

$I=\int_0^{\infty}\frac{\sin x}{x}\ dx$

But first, I will try to prove that $I$ converge.

$I=\int_0^{\infty}\frac{\sin x}{x}\ dx =\int_0^{1}\frac{\sin x}{x}\ dx +\int_1^{\infty}\frac{\sin x}{x}\ dx$ .

The integral $\int_1^{\infty}\frac{\sin x}{x}\ dx <\infty$ (Dirichlet's test)

But what about $\int_0^{1}\frac{\sin x}{x}\ dx$ ?

I know that $x=0$ is our problematic point and also I know that $\lim_{x\to 0} \frac{\sin x}{x}=1$ .

Can I say that because that limit exists the function is integrable?

Now to the the second part-computation.

It's from a book by G. M. Fikhtengoltz, The Fundamentals of Mathematical Analysis.

(I'm translating this from Russian into Hebrew and then into English.)

We will represent I in form of sum:

$I=\sum_{\nu =0}^{\infty}\int^{(\nu+1)\frac{\pi}{2}}_{\nu\frac{ \pi}{2}}$ .

Putting $\nu=2\mu$ or $\nu=2\mu-1$ and the he says the following:

we substituting $x=\mu\pi+t$ or $x=\mu\pi-t$ , and we get:

$\int_{2\mu\frac{\pi}{2}}^{(2\mu+1)\frac{\pi}{2}} = (-1)^{\mu}\int_{0}^{\frac{\pi}{2}}\frac{\sin t}{\mu\pi+t}$

and

$\int_{(2\mu-1)\frac{\pi}{2}}^{2\mu\frac{\pi}{2}} =(-1)^{\mu - 1}\int_{0}^{\frac{\pi}{2}}\frac{\sin t}{\mu\pi-t}$

Question:

Why we need the both substitutions, $x=\mu\pi+t$ or $x=\mu\pi-t$ ?

Thank you.

To be continued...

**General** · August 17th 2011, 10:48 AM

Your integral is just the laplace transformation of $f(t)=\frac{sin(t)}{t}$ evaluated at s=0.

**Also sprach Zarathustra** · August 17th 2011, 10:54 AM

Originally Posted by General

Your integral is just the laplace transformation of $f(t)=\frac{sin(t)}{t}$ evaluated at s=0.

Without Laplace transformation please.

**chisigma** · August 17th 2011, 12:44 PM

A more general form of the indefinite integral is...

$\int_{0}^{\infty} \frac{\sin p x}{x}\ d x =\begin{cases}\frac{\pi}{2}&p>0\\0&p=0\\-\frac{\pi}{2}&p<0\end{cases}$ (1)

This integral has great importance in several applcations, so that almost nobody remembers the 'theoretical bag' hidden in it: the definite integral (1) converges if it is processed as Riemann integral and doesn't converge if it is processed as Lebesgue integral, as for example in...

Lebesgue integration - Wikipedia, the free encyclopedia

All that is well known of course but, proper considering this example, I wonder how to justify motivations regarding the 'superiority' of the Lebesgue integral like that

...

... the Riemann integral is not defined for a wide range of functions and situations of importance in applications (and of interest in theory). For example, the Riemann integral can easily integrate density to find the mass of a steel beam, but cannot accommodate a steel ball resting on it. This motivates other definitions, under which a broader assortment of functions are integrable. The Lebesgue integral, in particular, achieves great flexibility by directing attention to the weights in the weighted sum...

Kind regards

$\chi$ $\sigma$

**Krizalid** · August 20th 2011, 08:40 PM

Also sprach Zarathustra, have you tried double integration? It's a neat approach and the justification is easy.

**Also sprach Zarathustra** · August 21st 2011, 12:16 AM

Originally Posted by Krizalid

Also sprach Zarathustra, have you tried double integration? It's a neat approach and the justification is easy.

No, I didn't.

I just want to understand that interesting method that the author presenting.

By the way, he using in the computation, the interesting formula for $\frac{1}{\sin t}$ ,

$\frac{1}{\sin t}=\frac{1}{t}+\sum^{\infty}_{1}(-1)^\mu \left (\frac{1}{t+\mu\pi}+\frac{1}{t-\mu\pi} \right )$ .

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