We are searching data for your request:

**Forums and discussions:**

**Manuals and reference books:**

**Data from registers:**

**Wait the end of the search in all databases.**

Upon completion, a link will appear to access the found materials.

Upon completion, a link will appear to access the found materials.

## Improper integrals: unlimited integration intervals

In defining the definite integral as being on a finite interval $[\mathit{a},\mathit{b}]$ lying surface

- $$\mathit{A.}\phantom{\rule{.4em}{0ex}}=\phantom{\rule{.4em}{0ex}}\underset{\Delta \mathit{x}\to 0}{lim}\sum _{\mathit{i}=1}^{\mathit{n}}\mathit{f}({\mathit{x}}_{\mathit{i}})\phantom{\rule{.2em}{0ex}}\Delta \mathit{x}\phantom{\rule{.4em}{0ex}}=\phantom{\rule{.4em}{0ex}}\underset{\mathit{a}}{\overset{\mathit{b}}{\int}}\mathit{f}(\mathit{x})\phantom{\rule{.2em}{0ex}}\text{d}\mathit{x}\phantom{\rule{.4em}{0ex}},\phantom{\rule{.4em}{0ex}}\mathit{a}\le {\mathit{x}}_{\mathit{i}}\le \mathit{b}$$

the case of an infinite integration interval is not provided. What is meant are certain integrals of the form

- $$\underset{\mathit{a}}{\overset{\infty}{\int}}\mathit{f}(\mathit{x})\phantom{\rule{.2em}{0ex}}\text{d}\mathit{x},\phantom{\rule{1em}{0ex}}\underset{-\infty}{\overset{\mathit{b}}{\int}}\mathit{f}(\mathit{x})\phantom{\rule{.2em}{0ex}}\text{d}\mathit{x},\phantom{\rule{1em}{0ex}}\underset{-\infty}{\overset{\infty}{\int}}\mathit{f}(\mathit{x})\phantom{\rule{.2em}{0ex}}\text{d}\mathit{x}.$$

Since these so-called improper integrals are not initially defined, we proceed as follows. We restrict ourselves first to integrals of the type

- $$\underset{\mathit{a}}{\overset{\infty}{\int}}\mathit{f}(\mathit{x})\phantom{\rule{.2em}{0ex}}\text{d}\mathit{x}$$

and perform the integral function$\mathit{I.}(\lambda )$ over the finite interval $[\mathit{a},\lambda ]$ a:

- $$\mathit{I.}(\lambda )\phantom{\rule{.4em}{0ex}}=\phantom{\rule{.4em}{0ex}}\underset{\mathit{a}}{\overset{\lambda}{\int}}\mathit{f}(\mathit{x})\phantom{\rule{.2em}{0ex}}\text{d}\mathit{x}.$$

We then calculate the limit of $\mathit{I.}(\lambda )$ for $\lambda \to \infty $ and equate it, if it exists, with the improper integral:

- $$\underset{\mathit{a}}{\overset{\infty}{\int}}\mathit{f}(\mathit{x})\phantom{\rule{.2em}{0ex}}\text{d}\mathit{x}\phantom{\rule{.4em}{0ex}}=\phantom{\rule{.4em}{0ex}}\underset{\lambda \to \infty}{lim}\mathit{I.}(\lambda )\phantom{\rule{1em}{0ex}}\left\{\begin{array}{lcl}\text{exists}& \Rightarrow & \text{convergent}\\ \text{does not exist}& \Rightarrow & \text{divergent}\end{array}\right. $$

- Example 1: integral of$\mathit{y}={\mathit{x}}^{1/2}$ over the interval $[0,\infty )$$$\underset{0}{\overset{\infty}{\int}}{\mathit{x}}^{1/2}\phantom{\rule{.2em}{0ex}}\text{d}\mathit{x}$$
- Example 2: integral of$\mathit{y}=1/(1+\mathit{x}{)}^{2}$ over the interval $(-\infty ,\infty )$$$\underset{-\infty}{\overset{\infty}{\int}}\frac{1}{(1+\mathit{x}{)}^{2}}\phantom{\rule{.2em}{0ex}}\text{d}\mathit{x}$$

The improper integral over the interval $(-\infty ,\mathit{b}]$ is defined analogously:

- $$\underset{-\infty}{\overset{\mathit{b}}{\int}}\mathit{f}(\mathit{x})\phantom{\rule{.2em}{0ex}}\text{d}\mathit{x}\phantom{\rule{.4em}{0ex}}=\phantom{\rule{.4em}{0ex}}\underset{\lambda \to -\infty}{lim}\underset{\lambda}{\overset{\mathit{b}}{\int}}\mathit{f}(\mathit{x})\phantom{\rule{.2em}{0ex}}\text{d}\mathit{x}\phantom{\rule{1em}{0ex}}\left\{\begin{array}{lcl}\text{exists}& \Rightarrow & \text{convergent}\\ \text{does not exist}& \Rightarrow & \text{divergent}\end{array}\right. $$

if $\mathit{f}(\mathit{x})$ at every point $\mathit{x}$ is steady, you can then bet

- $$\begin{array}{rcl}\underset{-\infty}{\overset{\infty}{\int}}\mathit{f}(\mathit{x})\phantom{\rule{.2em}{0ex}}\text{d}\mathit{x}& \phantom{\rule{.4em}{0ex}}=\phantom{\rule{.4em}{0ex}}& \underset{-\infty}{\overset{0}{\int}}\mathit{f}(\mathit{x})\phantom{\rule{.2em}{0ex}}\text{d}\mathit{x}\phantom{\rule{.4em}{0ex}}+\phantom{\rule{.4em}{0ex}}\underset{0}{\overset{\infty}{\int}}\mathit{f}(\mathit{x})\phantom{\rule{.2em}{0ex}}\text{d}\mathit{x}\\ & \phantom{\rule{.4em}{0ex}}=\phantom{\rule{.4em}{0ex}}& \underset{\lambda \to -\infty}{lim}\underset{\lambda}{\overset{0}{\int}}\mathit{f}(\mathit{x})\phantom{\rule{.2em}{0ex}}\text{d}\mathit{x}\phantom{\rule{.4em}{0ex}}+\phantom{\rule{.4em}{0ex}}\underset{\lambda \to \infty}{lim}\underset{0}{\overset{\lambda}{\int}}\mathit{f}(\mathit{x})\phantom{\rule{.2em}{0ex}}\text{d}\mathit{x}.\end{array}$$

Only if both integrals converge on the right side is the integral on the left side convergent, otherwise it is divergent.

An exception is the trivial case in which both partial integrals divergent, but for each $\lambda $ are equal, so that the total integral is zero and thus convergent (example: $\mathit{f}(\mathit{x})=\mathit{x}$).

## Improper integral explanation with example

Integrals, which have a limit value when the upper or lower limit is variable and approaches infinity after they have been inserted into the antiderivative, are so-called improper integrals.

The calculation then comes down to examining the antiderivative for limit values or horizontal asymptotes.

With improper integrals, for example, areas can be calculated that extend infinitely, but still have a finite content because they are getting narrower and narrower.

The following notation is used for the area $ A $ between two converging functions $ f $ and $ g $, which extends from the straight line $ x = a $ to infinity: $ A = left | int_ ^ < infty>

The area is calculated that lies between the $ x $ axis and the exponential function $ f (x) = e ^ <-x> $, is limited to the left by the $ y $ axis and extends infinitely to the right . To do this, we first calculate the definite integral with the variable upper limit $ t $: $ int_ <0> ^

## Improper integrals

Oh, I actually understood that, but not how I continue when I let k run towards infinity?

Oh not at all seen that you already had that

Second I have to check if I did it right before I tell you the wrong thing

Do you basically know how the Limes works?

Did you realize everything Then please close the question

So if I let k (mostly yes zero or infinite) run against it, what happens?

Oh and how do I do that with the b when I have -infinite?

Understood everything or should I still calculate the b for you?

Would be great. Just for me as a control :))

I don't think you can see that well in my picture, but the integral goes from -infinite to -1

Will something change decisively then?

Ne the result is & # 431

Okay, I got that out too. Many thanks for your help:)

Please close the question

## Considerations on the e-function

Now I will try to calculate the area under such a function:

### Example:

Up until now the lower and upper limits of a certain integral were numbers.

So the area of integration was limited.

Now the area of integration is no longer limited.

That is why such an integral is called **Improper integral** with unlimited integration area.

These integrals can come in one of three forms.

For our area calculation it looks like this:

### Here is another example:

### Area under a compound function

We can put two functions together and calculate the area from them. Because this area is no longer infinite.

## How do I find an improper integral?

You can easily determine the improper integral in three steps. We'll show you an example:

The area between the graph of the function f (x) = e ^ -x and the x-axis for x ≥ 0.

- Step: Imagine a right boundary and call it a variable z. Then set up a term A (z) for the area.

The area is exactly 1 FE.

### Improper integral: example exercise 1

Check whether the following functions include a finite area with the x-axis in the first quadrant. If this is the case, enter the area.

#### Solution to task 1:

The area is finite and amounts to:

If you proceed exactly as in a), you will get:

Therefore the enclosed area is not finite.

### Improper integral: example exercise 2

Check whether the following improper integral has a finite value:

#### Solution to task 2:

As you can see from the improper integral, this is an improper integral of the first kind with two critical integral limits. This means that the integral must be divided into two integrals, each with a critical limit:

We first determine the first improper integral:

Replace the critical interval limit -∞ with the variable ∝:

Depending on ∝ you now determine the integral:

Now you determine the limit value for ∝ → -∞:

That is, for the first integral:

Now we determine the value of the second improper integral:

First replace the critical interval limit ∞ with the variable β:

You now determine the integral as a function of:

Then determine the limit for β → ∞:

Finally you add up your results to get the value of the improper integral you are looking for:

## Integrals in physics

In high school physics dive frequently **Integrals** (sometimes even before they were discussed in depth in math class).

That **Work integral** (W = displaystyle int_a ^ b F (x) , text dx ) is that of a force (F !: x mapsto F (x), x in [ab] ), in the direction of travel along the path of *a* after *b* work done *W.*.

*Example:*

An electrical test charge *q* becomes the stationary charge *Q* from a great distance (the infinite) down to the distance *d* approximated. Here is the force required *F.*(*x*) equals the negative of the Coulomb force (F (x) = - F_ text

( begin

the **medium performance** ( overline

## Integration rules

Just as there are different rules when deriving, such as the product rule or the quotient rule, you also have to pay attention to a few things when integrating. After all, you are “reversing” the derivation rules, so to speak. In the next sections we will explain exactly what you have to do. We only formulate the integration rules for indefinite integrals, of course they apply analogously to certain integrals.

- The most important rule of integral calculus is that

**Power rule,**which is always used when the integral calculus contains power functions. She says

Obviously, deriving the right side again gives you .

- the

**Factor rule**is the simplest integration rule. You always use them when your function contains a constant factor. You can then “pull this in front of the integral”, you exclude it, so to speak. It applies

### Example

In order to carry out the following integral calculus, we need all of the above integration rules. Be sought

With the help of the sum and difference rule, we can “pull apart” the integral in the first step and get it

Now we apply, integrate and obtain the factor rule and the power rule, respectively

- Also for

**Sine and cosine**there are special rules in integral calculus.

In the **e function** integral calculus is not difficult:

It's a little more complicated with the **ln function** . Here you can determine and get the integral with the help of partial integration

Maybe you know that from the derivation is. So is of course the antiderivative of . This is a special case of **logarithmic integration rules**

### Example

We want the function integrate, so

If you take a closer look at the numerator and the denominator, you will see that the numerator almost contains the derivative of the denominator. In order to be able to apply the logarithmic integration rule, you have to transform the numerator a little. We cling to this and use the factor rule to pull it in front of the integral:

Now we can apply the rule of logarithmic integration and get as a result

**Note:** In most cases where you need to compute the antiderivative of a fraction, you will either use the logarithmic integration rule or rewrite the expression as a power function.

They also play an important role in quantum mechanics. The approach to derive quantum mechanics from path integrals is based on integrals of the form:

A practical formulation of the normalization constant $ mathcal

$ j $ is a whole natural number. For $ j = 0 $ is the integral

and is then called **Fresnel integral**. Integrals of this form appear in the Schrödinger equation derived from Feynman's path integrals.

A complex number results from the Fresnel integral, the real and imaginary parts of which are determined by

Both integrals converge. Due to the symmetry of the cosine, the cosine integral is invariant to a sign change of $ alpha $, the antisymmetric sine changes its sign. The addition results with $ sqrt*= e ^ <4>> $ and $ -1 = e ^ *

* *

*This also explains the normalization constant, which must be exactly the inverse of the integral solution so that the overall expression is 1. In quantum mechanics this is chosen for pragmatic reasons and from the idea that a wave function corresponds to a probability of location, so the integral over this function must be 1, since the particle described is finally somewhere.*