Integral Limits: Infinity.

  • Thread starter PFStudent
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  • #1
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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
 

Answers and Replies

  • #2
179
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
170
0
Hey,

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
179
2
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
HallsofIvy
Science Advisor
Homework Helper
41,847
966

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
3
0
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"
 

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