# Comparison of integrals

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
I know we have the following

$$\displaystyle \big | \int^{b}_{a} f(t) \, dt \big | \leq \int^{b}_{a} |f(t)|\, dt$$

1- How to prove the inequality ,what are the conditions ?
2- Does it work for improper integrals ?

#### chisigma

##### Well-known member
I know we have the following

$$\displaystyle \big | \int^{b}_{a} f(t) \, dt \big | \leq \int^{b}_{a} |f(t)|\, dt$$

1- How to prove the inequality ,what are the conditions ?
If f(*) is Riemann integrable on (a,b) then the integral is...

$$\int_{a}^{b} f(t)\ dt = \lim_{n \rightarrow \infty\ , \text{max} \Delta t_{i} \rightarrow 0} \sum_{i=0}^{n-1} f(t_{i})\ \Delta t_{i}\ (1)$$

For any finite sum is...

$$|\sum_{i=0}^{n-1} a_{i}| \le \sum_{i=0}^{n-1} |a_{i}|\ (2)$$

... and that proves the item 1...

Kind regards

$\chi$ $\sigma$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
$$|\sum_{i=0}^{n-1} a_{i}| \le \sum_{i=0}^{n-1} |a_{i}|\ (2)$$
That is the triangle inequality on the elements of the sequence , right ?

#### chisigma

##### Well-known member
That is the triangle inequality on the elements of the sequence , right ?
The so called 'triangle inequality' holds in general for vectors or complex numbers and extablishes that...

$$|\sum_{i=1}^{n} a_{i}| \le \sum_{i=1}^{n} |a_{i}|\ (1)$$

See here...

Triangle Inequality -- from Wolfram MathWorld

Kind regards

$\chi$ $\sigma$

#### Ackbach

##### Indicium Physicus
Staff member
And since the metric $d(x,y)=|x-y|$ is uniformly continuous, you are justified in passing to the limit for the integral.

#### Random Variable

##### Well-known member
MHB Math Helper
Here's how I remember the outline of the proof.

If $f(x)$ and $g(x)$ are integrable on $[a,b]$ (by which I mean Riemann integrable) and $f(x) \le g(x)$ for all $x \in [a,b]$, then $\int_{a}^{b} f(x) \ dx \le \int_{a}^{b} g(x) \ dx$.

Now if $f(x)$ is integrable on $[a,b]$, so is $|f(x)|$.

And $-|f(x)| \le f(x) \le |f(x)|$ for all $x \in [a,b]$.

So $- \int_{a}^{b} |f(x)| \ dx \le \int_{a}^{b} f(x) \ dx \le \int_{a}^{b} |f(x)| \ dx$.

#### chisigma

##### Well-known member
I know we have the following

$$\displaystyle \big | \int^{b}_{a} f(t) \, dt \big | \leq \int^{b}_{a} |f(t)|\, dt$$

1- How to prove the inequality ,what are the conditions ?
2- Does it work for improper integrals ?
The answer to point 2 is slighly more complex. If we consider an improper integral in (a,b) where a is a singularity of f(*), then we intend...

$$\int_{a}^{b} f(t)\ dt = \lim_{x \rightarrow a+} \int_{x}^{b} f(t)\ dt\ (1)$$

The problem in such a case is that in can be that $\int_{a}^{b} f(t)\ dt$ converges and $\int_{a}^{b} |f(t)|\ dt$ diverges. An interesting example of such a case is...

$$\int_{0}^{1} \frac{\sin \frac {1}{t}}{t}\ dt\ (2)$$

Kind regards

$\chi$ $\sigma$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
The answer to point 2 is slighly more complex. If we consider an improper integral in (a,b) where a is a singularity of f(*), then we intend...

$$\int_{a}^{b} f(t)\ dt = \lim_{x \rightarrow a+} \int_{x}^{b} f(t)\ dt\ (1)$$

The problem in such a case is that in can be that $\int_{a}^{b} f(t)\ dt$ converges and $\int_{a}^{b} |f(t)|\ dt$ diverges. An interesting example of such a case is...

$$\int_{0}^{1} \frac{\sin \frac {1}{t}}{t}\ dt\ (2)$$

Kind regards

$\chi$ $\sigma$
This is just like absolute convergence in series . If the integral absolutely convergent then it is convergent . If it is ''absolutely divergent" then the integral may or may not converge.