Derivative of an integral and error functions

[TroyElliott]

Homework Statement

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

Homework 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)[/B]

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![/B]

 

[BiGyElLoWhAt]

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

 

[TroyElliott]

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

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

 

[BiGyElLoWhAt]

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

 

[BiGyElLoWhAt]

x is just a constant in the integral *

 

[TroyElliott]

x is just a constant in the integral *

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.

 

[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.

 

[TroyElliott]

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.

Yes, it would be constant then.

 

[Ray Vickson]

Homework Statement

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

Homework 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)[/B]

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![/B]

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.

 

[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.

 

[TroyElliott]

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.

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).

 

[TroyElliott]

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.

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.

 

[Ray Vickson]

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.

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.

 

[TroyElliott]

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.

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.

 

[Chestermiller]

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

 

[TroyElliott]

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

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.

 

[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.

 

[TroyElliott]

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.

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!

 

[Chestermiller]

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.

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

 

[TroyElliott]

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

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

 

[Ray Vickson]

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.

When you cut and paste you can go to the input ribbon at the top of the input page and "unclick" the B, to get rid of unwanted bolding. Or, you can manually input the command [/BOLD], which will turn off a previous [BOLD] command.

Anyway, if your initial statement of the problem was accurate, the person setting the problem did not seem to be asking for you to evaluate the integral; they just wanted you to use Leibnitz to write the derivative of the integral in another form. I don't think it is a cruel joke at all; it would have been cruel, indeed, if they had asked you to evaluate the integral in terms of the special functions they have introduced so far, because they would (very likely) have been asking for the impossible. You could spend a year trying, and get nowhere.

 

[TroyElliott]

When you cut and paste you can go to the input ribbon at the top of the input page and "unclick" the B, to get rid of unwanted bolding. Or, you can manually input the command [/BOLD], which will turn off a previous [BOLD] command.

Anyway, if your initial statement of the problem was accurate, the person setting the problem did not seem to be asking for you to evaluate the integral; they just wanted you to use Leibnitz to write the derivative of the integral in another form. I don't think it is a cruel joke at all; it would have been cruel, indeed, if they had asked you to evaluate the integral in terms of the special functions they have introduced so far, because they would (very likely) have been asking for the impossible. You could spend a year trying, and get nowhere.

That is very true, but a user above did say that he had mathematica spit the solution out in terms of error functions.

 

[TroyElliott]

I am getting a solution made up of incomplete gamma functions, so I think I will fiddle around with those for a little while and see what I come up with. Thanks for all the replys!

 

[Ray Vickson]

That is very true, but a user above did say that he had mathematica spit the solution out in terms of error functions.

I don't believe that claim; I mean, I think that user is mistaken. Certainly, you can get a solution in terms of incomplete Gamma functions, but I doubt that has much of a relationship to the error function, because the parameter is different from in the incomplete-Gamma representation of erf(.). I will stand corrected if I am shown an actual formula for the integral expressed in terms of the error function.

 

[BiGyElLoWhAt]

The solution I got in terms of the erf was for the substitution I suggested in post 4, not the original, so it may not apply.

 

[Ray Vickson]

The solution I got in terms of the erf was for the substitution I suggested in post 4, not the original, so it may not apply.

When you substitute $u = t^2 \sqrt{x}$ into
J = \int_{-x}^x e^{-x t^4} \, dt = 2 \int_0^x e^{-x t^4} \, dt
you will get
J = x^{-1/4} \int_0^{x^{5/2}} \frac{1}{\sqrt{u}} e^{-u^2} \, du
I don't think that can be evaluated in terms of the error function.

 

[BiGyElLoWhAt]

Yea, it was with an arbitrary constant a for sqrt x, I believe.

I'm having trouble getting the exact result back, so I'm not actually sure what I put in, but if you do this:
http://www.wolframalpha.com/input/?i=differentiate+∫+e^(-x*t^2)dt+from+-x+to+x+with+respect+to+x

You get an answer in terms of erf.

What I had gotten before, however, was in terms of erf(sqrt(x)), not erf(x^(3/2)), so I don't really remember what I had put into it.
I also tried this in getting back the same result, but was unsuccessful:
http://www.wolframalpha.com/input/?...sqrt(x)*t^2)dt+from+-x+to+x+with+respect+to+x

It's also worth noting that until this thread, I was unaware of the error function.

 

[Ray Vickson]

Yea, it was with an arbitrary constant a for sqrt x, I believe.

I'm having trouble getting the exact result back, so I'm not actually sure what I put in, but if you do this:
http://www.wolframalpha.com/input/?i=differentiate+∫+e^(-x*t^2)dt+from+-x+to+x+with+respect+to+x

You get an answer in terms of erf.

What I had gotten before, however, was in terms of erf(sqrt(x)), not erf(x^(3/2)), so I don't really remember what I had put into it.
I also tried this in getting back the same result, but was unsuccessful:
http://www.wolframalpha.com/input/?i=differentiate+∫+e^(-sqrt(x)*t^2)dt+from+-x+to+x+with+respect+to+x

It's also worth noting that until this thread, I was unaware of the error function.

Of course you get the error function when evaluating $\int e^{-xt^2}\, dt$. However, that was not what the OP asked about. His integral was $\int e^{-x t^4} \, dt$; note the $t^4$, not $t^2$.

 

[BiGyElLoWhAt]

I see that, the idea for that stemmed from the substitution idea, and also the fact that the title of this thread includes the error function.