# Existence of Integral

#### Petrus

##### Well-known member
Hello MHB,

Integrate $$\displaystyle \int_0^4 \frac{dx}{(x-2)^3}$$
We are suposed to integrate when x goes from zero to 4 but when x is 2 the integration does not exist so the integrate does not exist as well?

Regards,
$$\displaystyle |\pi\rangle$$

#### chisigma

##### Well-known member
Re: integrate

Hello MHB,

Integrate $$\displaystyle \int_0^4 \frac{dx}{(x-2)^3}$$
We are suposed to integrate when x goes from zero to 4 but when x is 2 the integration does not exist so the integrate does not exist as well?

Regards,
$$\displaystyle |\pi\rangle$$
A very interesting question!... for semplicity sake we set $x-2=u$ and the integral becomes $\displaystyle I= \int_{-2}^{2} \frac{d u}{u^{3}}$. Now, because the function has a singularity in u=0, we write...

$\displaystyle I= \int_{-2}^{2} \frac{d u}{u^{3}} = \int_{-2}^{0} \frac{d u}{u^{3}} + \int_{0}^{2} \frac{d u}{u^{3}}$ (1)

Very well!... now we are in front of a crossroads...

a) each integral in (1) diverges, so that the whole integral diverges...

b) the (1) can be written as...

$\displaystyle I = \lim_{\varepsilon \rightarrow 0} (\int_{-2}^{-\varepsilon} \frac{d u}{u^{3}} + \int_{\varepsilon}^{2} \frac{d u}{u^{3}}) = 0$ (2)

How to choose between a) and b)?... another interesting question!...

Kind regards

$\chi$ $\sigma$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: integrate

Suppose $f$ is continuous on an interval $I$ , then we can conclude that $f$is integrable that means $$\displaystyle \int_I f(x)dx$$ exists or finite . Now , suppose that $f$ is not continuous on $I$ , unfortunately we can not conclude the integral does not exist.

#### Petrus

##### Well-known member
Re: integrate

A very interesting question!... for semplicity sake we set $x-2=u$ and the integral becomes $\displaystyle I= \int_{-2}^{2} \frac{d u}{u^{3}}$. Now, because the function has a singularity in u=0, we write...

$\displaystyle I= \int_{-2}^{2} \frac{d u}{u^{3}} = \int_{-2}^{0} \frac{d u}{u^{3}} + \int_{0}^{2} \frac{d u}{u^{3}}$ (1)

Very well!... now we are in front of a crossroads...

a) each integral in (1) diverges, so that the whole integral diverges...

b) the (1) can be written as...

$\displaystyle I = \lim_{\varepsilon \rightarrow 0} (\int_{-2}^{-\varepsilon} \frac{d u}{u^{3}} + \int_{\varepsilon}^{2} \frac{d u}{u^{3}}) = 0$ (2)

How to choose between a) and b)?... another interesting question!...

Kind regards

$\chi$ $\sigma$
This was my thinking
integrate means that we calculate area, and if we 'imagine' the graph we se at x=2 the negative way will go to -infinity and positive way will go to positive infinity and I also made that substitute as you made and end with answer 0. Then I don't know what to choose and why?

Regards,
$$\displaystyle |\pi\rangle$$

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Re: integrate

Very well!... now we are in front of a crossroads...
There is no crossroads. If $f:[a,b]\to\mathbb{R}$ is continuos except at $c\in (a,b)$ with infinite discontinuity then, by definition $\int_a^bf$ is convergent if and only if $\int_a^cf$ and $\int_c^bf$ are both convergent. If $\int_a^bf$ is convergent, we define $$\int_a^bf=\int_a^cf+\int_a^cf$$

#### Opalg

##### MHB Oldtimer
Staff member
Re: integrate

A very interesting question!... for semplicity sake we set $x-2=u$ and the integral becomes $\displaystyle I= \int_{-2}^{2} \frac{d u}{u^{3}}$. Now, because the function has a singularity in u=0, we write...

$\displaystyle I= \int_{-2}^{2} \frac{d u}{u^{3}} = \int_{-2}^{0} \frac{d u}{u^{3}} + \int_{0}^{2} \frac{d u}{u^{3}}$ (1)

Very well!... now we are in front of a crossroads...

a) each integral in (1) diverges, so that the whole integral diverges...

b) the (1) can be written as...

$\displaystyle I = \lim_{\varepsilon \rightarrow 0} (\int_{-2}^{-\varepsilon} \frac{d u}{u^{3}} + \int_{\varepsilon}^{2} \frac{d u}{u^{3}}) = 0$ (2)

How to choose between a) and b)?... another interesting question!...
The problem with b) is that the same $\varepsilon$ is used for both limits. It would be better to write $\displaystyle I = \lim_{\varepsilon \rightarrow 0} \int_{-2}^{-\varepsilon} \frac{d u}{u^{3}} + \lim_{\delta \rightarrow 0} \int_{\delta}^{2} \frac{d u}{u^{3}}$. Both those integrals diverge, one to $-\infty$ and the other to $+\infty$, but it is not appropriate to think of the difference as cancelling out to $0$. So we are led back to conclusion a).

#### chisigma

##### Well-known member
Re: Existance of Integral

The choice of alternative a) is quite 'obvious' but there are several cases of 'implicit choice' of the alernative b). In almost all the Fourier Transform tables is written that...

$\displaystyle \mathcal{F} \{\frac{1}{\pi\ t}\} =\begin{cases}-i &\text{if}\ \omega>0\\ 0 &\text{if}\ \omega=0\\ i &\text{if}\ \omega<0 \end{cases}$ (1)

The (1) is the basis of the so called 'Hilbert Transform', a powerful algorithm that I have used several times in pratical applications. If we write explicity the integral we find...

$\displaystyle \mathcal{F} \{\frac{1}{\pi\ t}\} = \int_{- \infty}^{+ \infty} \frac{e^{- i\ \omega\ t}}{\pi\ t}\ dt = \int_{- \infty}^{+ \infty} \frac{\cos \omega\ t}{\pi\ t}\ dt - i\ \int_{- \infty}^{+ \infty} \frac{\sin \omega\ t}{\pi\ t}\ dt$ (2)

Now if we observe (2) it is evident that the real part of the F.T. is an integral that, on the basis of alternative a) is divergent but in fact, according to alternative b), converges to 0...

Kind regards

$\chi$ $\sigma$

#### zzephod

##### Well-known member
Re: Existance of Integral

What you write comes down to the question of the convergence of:

$$\int_{- \infty}^{+ \infty} \frac{\cos \omega\ t}{\pi\ t}\ dt$$

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#### chisigma

##### Well-known member
Re: Existance of Integral

What you write comes down to the question of the convergence of:

$$\int_{- \infty}^{+ \infty} \frac{\cos \omega\ t}{\pi\ t}\ dt$$
Exactly!... The real part of $\mathcal{F}\ \{\frac{1}{\pi\ t}\}$ is...

$\displaystyle R = \int_{- \infty}^{+ \infty} \frac{\cos \omega t}{\pi\ t}\ dt$ (1)

... and, for the validity of the Hilbert Transform algorithm, it is 'requested' to be 0. Of course that's true only if we compute (1) as...

$\displaystyle R = \lim_{\varepsilon \rightarrow 0} (\int_{- \infty}^{- \varepsilon} \frac{\cos \omega t}{\pi\ t}\ dt + \int_{+ \varepsilon}^{+ \infty} \frac{\cos \omega t}{\pi\ t}\ dt) = 0$ (2)

I used the H.T. algorithm in several TC designs and one of them is protected by patent, so that I would be very 'unhappy' if the theoretical basis of the H.T. would 'founder' ...

Kind regards

$\chi$ $\sigma$

#### zzephod

##### Well-known member
Re: Existance of Integral

Exactly!... The real part of $\mathcal{F}\ \{\frac{1}{\pi\ t}\}$ is...

$\displaystyle R = \int_{- \infty}^{+ \infty} \frac{\cos \omega t}{\pi\ t}\ dt$ (1)

... and, for the validity of the Hilbert Transform algorithm, it is 'requested' to be 0. Of course that's true only if we compute (1) as...

$\displaystyle R = \lim_{\varepsilon \rightarrow 0} (\int_{- \infty}^{- \varepsilon} \frac{\cos \omega t}{\pi\ t}\ dt + \int_{+ \varepsilon}^{+ \infty} \frac{\cos \omega t}{\pi\ t}\ dt) = 0$ (2)

I used the H.T. algorithm in several TC designs and one of them is protected by patent, so that I would be very 'unhappy' if the theoretical basis of the H.T. would 'founder' ...

Kind regards

$\chi$ $\sigma$
I thought I had deleted that!

Singular FTs are defined rather differently and we only pretend (for the benefit of engineers and physicists) that the Fourier integral defines their transform.

.

#### chisigma

##### Well-known member
Re: Existance of Integral

Singular FTs are defined rather differently and we only pretend (for the benefit of engineers and physicists) that the Fourier integral defines their transform...
A suggestion I give to the people speacking in terms of 'benefit for engineers and physicists' is to read the introduction of...

Hilbert transform - Wikipedia, the free encyclopedia

In mathematics and in signal processing, the Hilbert transform is a linear operator which takes a function u(t) and produces a function H{u(t)} with the same domain. The Hilbert transform is named after David Hilbert who first introduced the operator in order to solve a special case of the Riemann-Hilbert for holomorphic functions...

David Hilbert... certainly not a second level mathematician!...

Kind regards

$\chi$ $\sigma$

#### Random Variable

##### Well-known member
MHB Math Helper
Re: integrate

By definition that's the Cauchy principal value of the integral. If an integral converges, it shouldn't matter how you take the limit.