Integration: int(exp(-2(s^2 - t^2)) ds, from 0 to t

[moy1989]

Hey guys, I am having difficulty finding out how to integrate the following:

y = int(exp(-2(s^2 - t^2)) ds, from 0 to t

can you please show me?
Thanks.

 

[tkhunny]

No need to "do" this one. You should just "know" it.

Hint: Your expression is the same as:

\(\displaystyle e^{2t^{2}}*\int_{0}^{t}e^{-2s^{2}}\;ds\)

 

[DR._Glockman]

TKhunny, if you had to do this one would you used a taylor expansion of the exponent? Just curious how you would attack this problem, even though we don't know the value of t , the factorial in the denominators would soon over whelm it.

 

[galactus]

\(\displaystyle \int_{0}^{t}e^{-2s^{2}+2t^{2}}ds\)

This one is a booger to integrate by elementary means.

We could use the error function if we played around awhile and get:

\(\displaystyle \frac{erf(\sqrt{2}t)e^{2t^{2}}\sqrt{2\pi}}{4}\)

But that is probably beyond the scope of your studies.

 

[tkhunny]

TKhunny, if you had to do this one would you used a taylor expansion of the exponent? Just curious how you would attack this problem, even though we don't know the value of t , the factorial in the denominators would soon over whelm it.

For elementary means, any numerical attack would be appropriate, I think. Really, though, galactus has what I had in mind. Let's recognize the erf() and be done with it.

 

[DR._Glockman]

The error funciton is the entire function; it has no singularities (except that at infinity) and its Taylor expansion always converges.If the integral cannot be evaluated in closed form in terms of elementary functions, by expanding the integrand in a Taylor series, one obtains the Taylor series for the error function . Galactus, this is what I ask tkhunny, however I would expand the Taylor series of e^(-2s^2) instead of using the error function, (Less protracted). I was just curious if you purist had a less protracted way of evaluating this obnoxious integral.

 

[tkhunny]

I've been accused rather often of being irritating, but only rarely a purist.

A year or two ago I had to come the grips with the meaning of "solve the integral" or "find the antiderivative". During the course of that discussion, the substance of which I really do not recall, I stepped out of my purist zone and added two valid attacks to my canon of methods worthy to be called appropriate, 1) Look it up in a table of integrals, and 2) Recall it from memory. I ruled out, "Get your brother to do it." That does solve the problem, but lacks in pedagogy.

 

[DR._Glockman]

Good show tkhunny, I concur wholeheartedly. There is (I seen one once) a tome of integrals from soup to nuts that other various purists laboriously compile for the pragmatist to avoid having (the pragmatist) to result in unfathomable grunt work. I remember the used book store I saw one in and the next time I am in that neighborhood I will purchase it if it is still available for my own self-edification.

Again tkhunny, thank you.

 

[moy1989]

Hey everyone, thanks to tkhunny, I figured I did not need to find the integral. My main concern was only to do a verification that a given function is the solution to the differential equation as shown by the following:

y' - 4ty = 1; y = int(exp(-2(s^2 - t^2))) ds from 0 to t

y = exp(2t^2) * int(exp(-2s^2)) ds from 0 to t

d/dt( (exp(2t^2)) * int(exp(-2s^2) ds) from 0 to t) - 4t( exp(2t^2) * int(exp(-2s^2) ds from 0 to t) ) = 1

4t * exp(2t^2) * int(exp(-2s^2) ds from 0 to t) + exp(2t^2) * exp(-2s^2) - 4t * exp(2t^2) * int(exp(-2s^2) ds from 0 to t) = 1

I simplified and ended with the answer

exp(2t^2) * exp(-2s^2) = 1 iff s = t

Did I do this right?

 

[tkhunny]

And thus, we see how important it is to provide the actual problem statement. Though this discussion was worthwhile, it simply was not to the point of the original problem statement.

Check that first derivative again. Assuming the antiderivative exists,

\(\displaystyle \frac{d}{dt}\left(\int_{0}^{t}f(s)\;ds\right)\;=\; \frac{d}{dt}\left(F(t) - F(0)\right)\;=\;f(t)\)

I'm not quite sure I'm seeing that in what you wrote.

 

[Deleted member 4993]

The error funciton is the entire function; it has no singularities (except that at infinity) and its Taylor expansion always converges.If the integral cannot be evaluated in closed form in terms of elementary functions, by expanding the integrand in a Taylor series, one obtains the Taylor series for the error function . Galactus, this is what I ask tkhunny, however I would expand the Taylor series of e^(-2s^2) instead of using the error function, (Less protracted). I was just curious if you purist had a less protracted way of evaluating this obnoxious integral.

This is a common function - comes into play:

Gaussian distribution (bell curve)

Diffusion of fluids in porous media

This function is a well-studied function - like exponential function.