# Derivative of an integral and error functions

1. Oct 8, 2015

### TroyElliott

1. The problem statement, all variables and given/known data

differentiate ∫ e^(-x*t^4)dt from -x to x with respect to x.

2. Relevant equations

erf(x) = (2/sqrt(π)) ∫e^(-t^2)dt from 0 to x.

Leibniz rule.

I know that ∫t^2e^(-t^2)dt from 0 to x = (√π/4)*erf(x) - (1/2)*x*e^(-x^2)

3. The attempt at a solution

By using Leibniz rule, I get d/dx ∫ e^(-x*t^4)dt from -x to x with respect to x is equal to e^(-x^5) + e^(-x^5) - ∫ t^(4)*e^(-x*t^4)dt from -x to x.

I am stuck on this integral above. Trying to think of a nice substitution to write the integral in terms of an error
function.

Thanks for any tips!

Last edited: Oct 8, 2015
2. Oct 8, 2015

### BiGyElLoWhAt

Is this the problem?
$\frac{d}{dx}\int_{-x}^x e^{xt^4}dt$ ?

3. Oct 8, 2015

### TroyElliott

Yes, except for there should be a minus sign in the e function. e^(-xt^4)

4. Oct 8, 2015

### BiGyElLoWhAt

Hmmm... Can the solution be complex? You could try $u=\sqrt{-x}t^2$

5. Oct 8, 2015

### BiGyElLoWhAt

x is just a constant in the integral *

6. Oct 8, 2015

### TroyElliott

Yes, in the sense that the integral is being integrated with respect to t. But since we are taking a derivative of it with respect to x, we have to take that into consideration when using Leibniz's rule.

7. Oct 8, 2015

### BiGyElLoWhAt

Yea, but I was talking after you had applied it, however, there's a factor of t that I skipped over in my substitution when you substitute the differential.

8. Oct 8, 2015

### TroyElliott

Yes, it would be constant then.

9. Oct 8, 2015

### Ray Vickson

Why would you think the problem has anything to do with the error function? (It doesn't). Maple can evaluate the integral in terms of so-called "WhittakerM" functions (which can, in turn, be expressed in terms of hypergeometric functions), but NOT in terms of erf(..) or its relatives.

10. Oct 8, 2015

### LCKurtz

It looks to me that, after applying Leibnitz's rule, one of your terms is going to be
$$\int_{-x}^{x}-t^4e^{-xt^4}~dt$$which doesn't look to be an elementary integral, if that matters.

11. Oct 8, 2015

### TroyElliott

I am doing this problem out of a mathematical physics book. The only relevant section this problem could fall under would be an error function. It is similar to what I posted initially, namely ∫t^2e^(-t^2)dt from 0 to x = (√π/4)*erf(x) - (1/2)*x*e^(-x^2).

12. Oct 8, 2015

### TroyElliott

That will be one of the terms. You are correct, it isn't an elementary integral, but, like the following example shows ∫t^2e^(-t^2)dt from 0 to x = (√π/4)*erf(x) - (1/2)*x*e^(-x^2). It can still be written in terms of a special function, like the error function.

13. Oct 8, 2015

### Ray Vickson

But it's not of that form, so the error function does not apply. As I said in post #9, it involves the so-called WhittakerM function. That is, for sure, a "special" function, but it is not the error function or anything like it.

Also: why to you write in bold font? It looks like you are yelling at us.

14. Oct 8, 2015

### TroyElliott

I am not yelling, I did a copy and paste of the equation in the problem statement, which was bolded, and the text that followed it took on its font, sorry for any confusion. I am not doubting that you are correct when saying it involves the WhittakerM function. The reason I am not saying that must be the solution is because the problem comes from the first chapter of a book I am studying and in the first chapter, as well as the whole book, there is not a single reference to the WhittakerM function. The only special functions that are spoken of is the error function, gamma function, and the gaussian function. So, unless the author is playing a cruel joke on me in the first chapter of the text, I think there should be some clever way of writing it terms of one or more of these functions.

15. Oct 8, 2015

### Staff: Mentor

So, are you ever going to apply the Leibnitz Rule to the problem, or are you just going to keep talking about it forever?

16. Oct 8, 2015

### TroyElliott

You must not have read the very first part of the thread. That was the first thing I did under "The attempt at a solution" in the intro.

17. Oct 8, 2015

### BiGyElLoWhAt

Well, when I plugged it into wolfram, it gave me an answer in terms of either the error function or the gamma function, so I think the error function approach is appropriate. And it was erf (root(x)t), but I'm not sure how they got there.

18. Oct 8, 2015

### TroyElliott

Yeah, I see that. I am thinking I will need to either insert a parameter and differentiate with respect to that, then take the same integral and make a substitution somewhere wit hate same parameter and do another derivative with respect to that parameter, set them equal and hopefully something falls out from that mess.

Thanks for the response!

19. Oct 8, 2015

### Staff: Mentor

Sorry, I did miss that. Are you sure of that second term in your solution. I may be wrong, but I get an negative exponent also in the second term.

Chet

20. Oct 8, 2015

### TroyElliott

Yes, I didn't catch that, the second term should be a negative exponent as well.