An ACTUAL post: Integrating exp() over certain range

1. May 11, 2008

Baggio

An ACTUAL urgent post: Integrating exp() over certain range

Hi,

Simplified problem:

Suppose I have two exponentials

$$$e^{ - (x + a - b)} \forall x + a -b> 0$ $e^{ - (x + b)} \forall x + b> 0$$$

Then suppose I wanted to integrate:

$$$\int\limits_{ - \infty }^\infty {e^{ - (x + b)} e^{ - (x + a - b)} dx}$$$

How would I do this? I'm guessing I need to break the integral up and integrate over a certain range but what are the limits???

Thanks

Last edited: May 11, 2008
2. May 11, 2008

Gib Z

Those 2 first inequalities can not be true over the whole interval of integration..

Regardless, do you know how to simplify e^a * e^b? Use that identity, use a substitution and it should be evident from there.

3. May 11, 2008

Baggio

Yes I know they can't be BOTH true, which is why I need to find the condition where they are both satisfied. a and b are real variables I don't understand your 2nd point

4. May 11, 2008

Gib Z

Neither of them can be true for all x in the interval of integration...And My second point helps with the actual integration.

5. May 11, 2008

Baggio

...Why is that? Maybe I should have mentioned that each exp() is 0 otherwise.. integration isn't a problem it's the limits of integration that I need to find.

6. May 11, 2008

Baggio

Essentially what this is isa simplified integral I'm trying to solve which describes two photons overlapping, each with an exponential wavefunction in the time domain. a an b describe offsets in time between the two. Regardless

If you plot out those two functions you can see that there should obviously be some range where they both coincide, the upper limit would be infinity and the lower limit is what I'm trying to find.

7. May 11, 2008

Gib Z

So you want to find out the interval where the both are not equal to zero. The condition for the first is x> b-a, the seconds is x > -b. Just find the intersection of these intervals.

8. May 11, 2008

Baggio

I know, but this lower limit I'm trying to find would vary depending on whether -b>b-a if this is true -b would be the lower limit, otherwise b-a would be.

9. May 11, 2008

HallsofIvy

What conditions do you have on a and b? Since they are constants, you should be able to break this into cases.

For example, if a and b are both positive, b> a, then x+ a- b> 0 implies x> b- a> 0 while x+ b> 0 implies x> -b< 0. Those are both satisified if x> b-a.

If a and b are both positie and a> b, then x+ a-b> 0 implies x> b-a< 0 but b-a> -b. Again both are satisfied if x> b-a.

10. May 11, 2008

Hurkyl

Staff Emeritus
Hrm... allow me to speculate upon what you really meant.

The thing you are integrating isn't an exponential at all -- instead, it's the product of two piecewise-defined functions, where each function is exponential on one part and zero on the other part.

i.e. if we set

$$f(x) = \begin{cases} e^x & x > 0 \\ 0 & x < 0 \end{cases}$$

then you are trying to simplify the definite integral

$$\int_{-\infty}^{+\infty} f(x + a - b) f(x + b) \, dx.$$

Is my speculation correct?

11. May 11, 2008

Baggio

Hurkyl yes that's correct, as for a and b they can take any real value and are not dependent on each other. The function that I posted at the beginning is part of a larger function that I want to eventually plot as a function of a and b. x in this case is like a dummy variable over which I need to integrate. I've only included the real part of the functions I'm trying to integrate as this is the only part that determines the limits of integration. In the end I want to integrate:

$$\int_{ - \infty }^{ + \infty } {\left| {f(x + a - b)f(x + b)} \right|^2 \,dx} \]$$

Last edited: May 11, 2008
12. May 11, 2008

Hurkyl

Staff Emeritus
When integrating a piecewise-defined function by breaking into pieces, it should be clear what pieces to use: the individual pieces in the piecewise definition!

13. May 11, 2008

Baggio

I understand that but how do I obtain a general expression for the limits when a and b are independent of each other?

14. May 11, 2008

Redbelly98

Staff Emeritus
Why not solve it for each case, separately? Then list the two solutions, according to which condition holds.

15. May 11, 2008

Baggio

Because I know what the final answer should be and it is just a single expression so it should be possible to have a single expression for the simplified form in post #1

Last edited: May 11, 2008
16. May 11, 2008

Redbelly98

Staff Emeritus
Perhaps if you worked out the separate solutions, you would then find they are equivalent.

17. May 11, 2008

Baggio

yes i think i need to do that , i'll try that and get back to you