Impossible Integrand?

S

Skelly4444

Junior Member
Joined
Apr 4, 2019
Messages
63
I would like to know why is it that we cannot integrate the following?

ex2

The online integral calculator states that this is an imaginery error function erfi(x)

What exactly does this mean in simple terms?
 
Skelly4444 said:
I would like to know why is it that we cannot integrate the following?

ex2

The online integral calculator states that this is an imaginery error function erfi(x)

What exactly does this mean in simple terms?
Click to expand...
It means that although the function is integrable in theory, the result can't be expressed in terms of the elementary functions like exponentials and sines. So if you want to state the result of the integral, you have to invent a new function, whose value can be evaluated more or less the same way you can directly evaluate (approximate) the value of the definite integral.

Why does this happen? Just because the set of elementary functions we teach can't cover everything. None of them happens to have [imath]e^{x^2}[/imath] as its derivative. If we added more functions to the list, we'd find something else that we still couldn't integrate.
 
Dr.Peterson said:
It means that although the function is integrable in theory, the result can't be expressed in terms of the elementary functions like exponentials and sines. So if you want to state the result of the integral, you have to invent a new function, whose value can be evaluated more or less the same way you can directly evaluate (approximate) the value of the definite integral.

Why does this happen? Just because the set of elementary functions we teach can't cover everything. None of them happens to have [imath]e^{x^2}[/imath] as its derivative. If we added more functions to the list, we'd find something else that we still couldn't integrate.
Click to expand...
Many thanks for clearing that up. Can it be successfully integrated if we combine it into a product with another function such as:

x3ex2

According to my maths book, this one is 100% solvable but I'm not sure why?
 
Skelly4444 said:
Many thanks for clearing that up. Can it be successfully integrated if we combine it into a product with another function such as:

x3ex2

According to my maths book, this one is 100% solvable but I'm not sure why?
Click to expand...
If you know how to use parts in integration use [imath]u=x^2~\&~dv=x e^{x^2}dx[/imath].
 
Assume that neither limit of the integration is infinite. (Why?)

For this specific case, [imath]\int x^3 e^{x^2} ~ dx = \int x^2 e^{x^2} ~ (x dx)[/imath], we let [imath]u = x^2[/imath] so that [imath]\int x^2 e^{x^2} ~ (x dx) = \dfrac{1}{2} \int u e^u ~ du[/imath] which can be done with integration by parts. We can do any integral [imath]\int u^n e^u ~ du[/imath] if n is a non-negative integer, and we can convert [imath]\int x^{2p +1} e^{x^2}~ dx[/imath] into this form (for any non-negative integer p.)

-Dan
 
Hi Dan,
I can use integration by parts and obtain an answer that agrees with my maths book. What I'm still puzzled about is why the function ex2 isn't defined as a legal integral when we can successfully graph it and find an area under it?

The graph is obviously a steep symmetrical curve that passes through the point (0,1). If integration is concerned with finding areas under curves etc, why can't we find an area under this particular graph? We could certainly use numerical integration on ex2 to find an area using the trapezium rule, so why not integration?
 
Skelly4444 said:
Hi Dan,
I can use integration by parts and obtain an answer that agrees with my maths book. What I'm still puzzled about is why the function ex2 isn't defined as a legal integral when we can successfully graph it and find an area under it?

The graph is obviously a steep symmetrical curve that passes through the point (0,1). If integration is concerned with finding areas under curves etc, why can't we find an area under this particular graph? We could certainly use numerical integration on ex2 to find an area using the trapezium rule, so why not integration?
Click to expand...
To use numerical integration - you need to have "numerical limits" (definite integral).
 
The answer to that is basically what Dr Peterson said, "although the function is integrable in theory, the result can't be expressed in terms of the elementary functions". Yes there is an area under the graph that can be measured (approximately as all measurement is) but we cannot assign a precise numerical value to it.

The fact is that the great majority of "integrable functions' cannot be integrated in terms of functions that we already know!
 
Last edited:
Shiloh said:
The answer to that is basically what Dr Peterson said, "although the function is integrable in theory, the result can't be expressed in terms of the elementary functions". Yes there is an area under the graph that can be measured (approximately as all measurement is) but we cannot assign a precise numerical value to it.

The fact is that the great majority of "integrable functions' cannot be integrated in terms of functions that we already know!
Click to expand...
Answer to what?

Please use the reply button while responding - so that we know the correct reference!
 
There is something puzzling me about this. I can successfully integrate this on my calculator and plug in upper and lower limits. For instance, using the limits of -1 and +1, it gives me an area of 2.925 sq units.

How is the calculator able to do this calculation if the integrand isn't able to be integrated into anything?
 
Skelly4444 said:
There is something puzzling me about this. I can successfully integrate this on my calculator and plug in upper and lower limits. For instance, using the limits of -1 and +1, it gives me an area of 2.925 sq units.

How is the calculator able to do this calculation if the integrand isn't able to be integrated into anything?
Click to expand...
First: As I said in #2, and Shiloh restated in #8, this function does have an integral (both a definite integral = area under the curve, and an indefinite integral = antiderivative); it just can't be expressed in terms of elementary functions.

Second: As we said, definite integrals with numerical limits can be approximated by numerical methods, and that is what a calculator does. (Look in the manual to see what it says this button does.)

Unfortunately, I suspect that most calculus courses don't emphasize enough what an integral, in both senses, really is, and that it exists regardless of whether we have a method to calculate it. Not all functions have formulas. Not all math is methods.
 
Thanks for explaining this, I now have a clearer picture.

You're quite right in what you say about integration courses. We were just taught it was a tool for finding areas and volumes and can be thought of as the sum of infinite parts that are just added up automatically, similar to the trapezium rule but with infinite strips. I know this is not how it really works though.
 
The fact is that most integrable functions do not have "anti-derivatives" in terms of elementary functions.
 
Skelly4444 said:
You're quite right in what you say about integration courses. We were just taught it was a tool for finding areas and volumes and can be thought of as the sum of infinite parts that are just added up automatically, similar to the trapezium rule but with infinite strips. I know this is not how it really works though.
Click to expand...
Steven G said:
What do you mean why you say I know this is not how it really works though?
Click to expand...
If one reads a journal article on some aspect of ignoration integration theory you will see that it assumes that the integral is a number,
That is [imath]\displaystyle\int_0^\pi {\cos (2x)dx}[/imath] is an integral while [imath]\int {\cos (2x)dx} [/imath] denotes an ante-derivative.
There just some things, especially mathematical terminology, we must learn to live with no matter how distasteful.
 
You must log in or register to reply here.
Top