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I need to integrate a function from either 0- or 0+ to infinity. While there is no problem entering the upper limit, how is it possible to enter either of the lower limits?
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Use limits
Since 0+ and 0- are not actual numbers, you will need to go back to what this notation actually represents: a limit.
assume( s>0 ); f := t; t F := int( f*exp(-s*t), t=a..infinity ); (1 + s a) exp(-s a) ------------------- 2 s limit( F, a=0, right ); 1 -- 2 sThis gives the limit from the right (0+). Replace the last argument of the limit command with left to take the limit from the left (0-).
Of course, in this case the same result can be obtained without limits.
I hope this is helpful.
Doug
Limits and FTOC
By default, the fundamental theorem of calculus definite integrator will take the limit from the right of the the lower limit (0+), so Doug's instructions should only be necessary if you want the integral from 0-.
My appologies, but I did not
My appologies, but I did not state the problem very clearly! The difficulty is not with the maths, but with getting Maple to display the lower limit of the integral as 0 subscript + or -
Displaying the lower limit
Perhaps something like this?
> Zeroplus:= `#msub(mn("0"),mo("+"))`; Int(f(x), x=Zeroplus .. infinity);And similarly, replacing plus with minus.
Also of course integrating
Also of course integrating the Delta function from 0- to infinity should yield 1, whereas integrating the Delta function from 0+ to infinity should yield 0.
use inert limit (and int)
To have Maple setup but not evaluate the integrals, you can use the intert version of int and limit: Int and Limit, respectively.
f := t; t F := Int( f*exp(-s*t), t=a..infinity ); /infinity | | t exp(-s t) dt | /a Limit( F, a=0, right ); / /infinity \ | | | lim | | t exp(-s t) dt| a -> 0+ | | | \/a /The eval command can be used to evaluate an inert expression.
If you are not interested in the evaluation, then you might be able to use the inert integral to create something that looks more like what you are requesting. For example:
Int( f*exp(-s*t), t=0^`+`..infinity ); /infinity | | t exp(-s t) dt | / + 0Doug
Integration between limits
Further to this discussion, am I correct in thinking that the integral of any continuous function between 0- and 0+ (or more generally a- to a+) is always equal to zero?
yes, but continuity is essential
Further to this discussion, am I correct in thinking that the integral of any continuous function between 0- and 0+ (or more generally a- to a+) is always equal to zero?
You should be careful to indicate exactly where the function is continuous: at 0 or at a for the two statements in your post.
The result is still zero even for some discontinuous functions. If the integrand has a simple jump discontinuity at a (different left and right limits), then there is still a continuous antiderivative and the definite integral from a- to a+ will be zero.
Things get interesting when the integrand is something like a impulse function, such as a Dirac distribution. The Dirac distribution at x=a is the "function", f, with the property that f(x)=0 for all x<>a and int( f(x), x=-infinity..infinity ) = 1. Thus, int( f(x), x=-infinity..a- ) = 0, int( f(x), x=a+..infinity ) = 0, and int( f(x), x=a-..a+ ) = 1.
For more information, take a look at the Wikipedia page on Dirac delta functions.
At some point in this discussion you will have to take some time to learn about the different types of integral. Most people start with the Riemann integral, but that is only the tip of the proverbial mathematical iceberg. (Hey, it's cold here today.)
For starters, read the Wikipedia page on Lebesgue integration.
Continuity Again
Can you please give one or more examples of these functions, that are discontinuous but have a continuous antiderivative. As the problem that I am studying, is concerned with the functions that the two Laplace transforms return different values. For example, the Laplace transform of the delta function is either 1 or 0, depending upon which definition of the transform is used, that is
int(f(t)*exp(-s*t),t = 0- to infinity) = 1
or
int(f(t)*exp(-s*t),t = 0+ to infinity) = 0
Whereas the transform of sin(wt) is always w/(s^2 + w^2). Hence the concern with functions that are continuous or discontinuous in the region [0-,0+]
distributions and discontinuous derivatives
Did you look at the Wikipedia references I gave in an earlier post? These will give you some background on the Dirac distributions (they are NOT "functions" - they are not defined at 0). In particular, you will learn that
int( f(t)*delta(t-tau), t=I ) = f(tau) when tau is in I,
otherwise this integral is 0.
This is what you are reporting with the different values of the integral. 0 is in (0-,infinity) but not in (0+,infinity).
As for your question about discontinuous functions (f) with continuous antiderivatives (F). Let F(x)=abs(x) and define f=F'. More generally, form F(x) = piecewise( x<a, F1(x), x>a, F2(x) ) where F1(a-)=F2(a+) (to force continuity).
Then f(x) = F'(x) = piecewise( x<a, F1'(x), x>a, F2'(x) )will be discontinuous at x=a unless F1'(a-)=F2'(a+).
Doug
Many thanks all that was
Many thanks all that was very helpful
Laplace Transform
I am currently compiling a manuscript about the Laplace Transform, and am trying to locate some detailed information about its history. While it has been easy to find information about its development by the two well known players, that is Laplace and Heaviside, I have been unable to obtain very much about the development prior to Laplace, by mathematicians such as Euler? As a result of this I would be grateful of any input as to possible sources of information.
Euler
Euler's integral for the Gamma function is the source of many Laplace transforms.
According to www-history.mcs.st-andrews.ac.uk/Biographies/Euler.html these were studied in Institutiones calculi integralis (1768-70). You might try
P J Davis, Leonhard Euler's integral : A historical profile of the gamma function, Amer. Math. Monthly 66 (1959), 849-869.