Integral Limits: Infinity.

In summary: This can be solved by substitution. Let's say we want to find the derivative at x=-∞. We could solve it like this:\frac{{\partial \vec{F}}(r)}}{{\partial t}} = -1 And then we could use the chain rule to find the derivative at x=-∞:dF/dr = -1
  • #1
PFStudent
170
0

Homework Statement



Consider the function,

[tex]
{{\vec{F}}(r)} = {\int_{-\infty}^{\infty}}{{\vec{f}}(r)}{\times}{d{\vec{r}}}
[/tex]

Is the following true?,

[tex]
{\int_{-\infty}^{\infty}}{{\vec{f}}(r)}{\times}{d{\vec{r}}} = {2}{\int_{0}^{\infty}}{{\vec{f}}(r)}{\times}{d{\vec{r}}}
[/tex]

Homework Equations



Knowledge of infinite limits.

The Attempt at a Solution



I believe it is true, since zero is considered the mid-point between an infinite sum of numbers in one direction and an infinite sum of numbers in the opposite direction.

So is that right?

Any help is appreciated.

Thanks,

-PFStudent
 
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  • #2
Consider the following piece-wise function:

[tex]f(x) = \begin{cases}
x^2 & x \le 0,\\
\frac{1}{x^2} & x>0
\end{cases}[/tex]
 
  • #3
Hey,

rs1n said:
Consider the following piece-wise function:

[tex]f(x) = \begin{cases}
x^2 & x \le 0,\\
\frac{1}{x^2} & x>0
\end{cases}[/tex]

Thanks for the reply rs1n.

I noticed the function you mentioned approaches infinity for large negative values and approaches zero for large positive values. However, I am not sure how that relates to the question I proposed. What am I missing?

Thanks,

-PFStudent
 
  • #4
PFStudent said:
Hey,



Thanks for the reply rs1n.

I noticed the function you mentioned approaches infinity for large negative values and approaches zero for large positive values. However, I am not sure how that relates to the question I proposed. What am I missing?

Thanks,

-PFStudent

It could be that I'm missing something (yes, I notice that your integrand is a cross product). However, your function needs have some symmetry in order for your claim to be true. In the example I gave, 2 times the integral of just the "right" side is NOT equal to the overall integral (from negative infinity to positive infinity). You can see this geometrically (just consider the integral as area beneath the curve). The "right" side has finite area whereas the "left" side has infinite area. This can be proven rigorously using limits (see improper integrals in any calculus book).

For example, if you graph [tex]e^{-x^2}[/tex], you'll notice it's symmetric about the y-axis (i.e. it's an even function). In this case, you can say

[tex]\int_{-\infty}^\infty e^{-x^2}\ dx = 2\int_0^\infty e^{-x^2}\ dx[/tex]

Not all functions have such symmetry (e.g. the one I gave).
 
  • #5
PFStudent said:

Homework Statement



Consider the function,

[tex]
{{\vec{F}}(r)} = {\int_{-\infty}^{\infty}}{{\vec{f}}(r)}{\times}{d{\vec{r}}}
[/tex]
Here, you show [itex]\vec{r}[/itex] as a vector. How does a vector go from [itex]-\infty[/itex] to [itex]\infty[/itex]?

Is the following true?,

[tex]
{\int_{-\infty}^{\infty}}{{\vec{f}}(r)}{\times}{d{\vec{r}}} = {2}{\int_{0}^{\infty}}{{\vec{f}}(r)}{\times}{d{\vec{r}}}
[/tex]

Homework Equations



Knowledge of infinite limits.

The Attempt at a Solution



I believe it is true, since zero is considered the mid-point between an infinite sum of numbers in one direction and an infinite sum of numbers in the opposite direction.
As rs1 said, you need symmetry. Yes, you have "an infinite sum of numbers in one direction and an infinites sum of numbers in the opposite direction"- but not necessarily the same sum! The two sums are not necessarily the same.

So is that right?

Any help is appreciated.

Thanks,

-PFStudent
 
  • #6
how can I solve this integral ?

∫_(-∞)^(+∞)▒〖1/(16π^2 D^2 (t-t")(t" -t^')) e^((-〖(x-x")〗^2)/(4D(t-t"))-〖(x"-x')〗^2/(4D(t"-t'))) 〗dx"
 

1. What is an integral limit?

An integral limit is a boundary that is used to indicate the range over which an integral is being evaluated. It can be a finite number or it can be infinity.

2. What does it mean when an integral limit is infinity?

When an integral limit is infinity, it means that the integral is being evaluated over an unbounded range. This means that the function being integrated may continue indefinitely in one or both directions.

3. How do you handle infinity as an integral limit?

When infinity is used as an integral limit, it is typically handled by taking the limit of the integral as it approaches infinity. This is often done using techniques from calculus, such as L'Hôpital's rule or the comparison test.

4. What is the significance of using infinity as an integral limit?

Using infinity as an integral limit can be useful in many applications, particularly in physics and engineering. It allows for the evaluation of integrals over infinite ranges, which can model real-world situations more accurately.

5. Can an integral limit of infinity ever be a finite number?

Yes, it is possible for an integral limit of infinity to result in a finite number. This can happen when the function being integrated approaches zero at a certain point, or when the integral converges to a finite value despite having an infinite range.

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