# B Integrating any e^f(x) function

1. Mar 20, 2016

### Bounceback

Solve $f'(x)*a(x)+a'(x)=1$ For a(x)
(there's another less important question at the bottom)

Background behind equation (trying to find a function to integrate any e^f(x)):
$\int e^{f(x)}\,dx=e^{f(x)}*a(x)$
$e^{f(x)}=e^{f(x)}*{f'(x)}*a(x)+e^{f(x)}*a'(x)$
$1=f'(x)*a(x)+a'(x)$

A few of my attempts:

$df(x)*a(x)+da(x)=d(x)$
$\int a(x)\,df(x)+a(x)=x$
$a(x)=x-\int a(x)\,df(x)$
$a(x)=x-\int x-\int x-\int x ...\,df(x)\,df(x)\,df(x)$
$a(x)=x-\int x\,df(x)+\iint x\,d^2f(x)-\iiint x\,d^3f(x)...$
Note, this attempt only works if f(x) is an integrable function

This method doesn't work however, since $\int e^x\,dx=e^x$, You know the value of a and f(x) (1 and x respectively).
Substituting into the equation, the method stops working at line 2. Assuming that this is because you can only integrate a side of an equation with respect to a single function, I tried a different method.

$f'(x)*a(x)=1-a'(x)$
$\int f'(x)*a(x)\,dx=\int 1-a'(x)\,dx$
$\int f'(x)*a(x)\,dx=x-a(x)$
A rearranged version of line two...

Where have I gone wrong in my attempts?

Last edited: Mar 20, 2016
2. Mar 20, 2016

### SteamKing

Staff Emeritus
I'm not sure the above equation is true for any f(x).

$\int e^u\, du = e^u + C$

so if u = f(x), then du = f'(x) in order for substitution to work.

[In Latex, remember to use {} around any multi-character exponents or subscripts, to keep everything on the same line.]

3. Mar 20, 2016

### Bounceback

Just doing some testing of the two integrals I know off hand:

$\int e^{x^{1/3}}\,dx=3*e^{x^{1/3}}*(2-2*x^{1/3}+x^{2/3})$
$f'(x)*a(x)+a'(x)=1$
$(1/(x^{2/3}*3))*(3*(2-2*x^{1/3}+x^{2/3}))+(2*(x^1/3-1))/(x^{2/3})$
Which, as it happens, equals 1 (according to the calculator)

$\int e^x\,dx=e^x*1$
$f'(x)*a(x)+a'(x)=1$
$(1)*(1)+(0)=1$
Again, equals 1.

Am I misinterpreting what you are asking?

I'm not entirely sure what you mean here. Could you rephrase your question?

4. Mar 20, 2016

### SteamKing

Staff Emeritus
It's not a question. It's a basic statement of fact:

just because $\int e^x\,dx = e^x + C$, it does not logically follow that $\int e^{f(x)}\,dx = e^{f(x)} + C$

The latter integral is usually evaluated by u-substitution, where u = f(x) and du = f'(x) dx. If you don't have f'(x) dx somewhere in the original integral expression, you can't evaluate the integral.

For example, $\int e^{x^2}\, dx$ cannot be integrated using elementary functions, but $\int e^{x^2} ⋅ 2x dx = e^{x^2} + C$

By using u-substitution with u = x2 = f(x), then du = 2x dx = f'(x) dx and $\int e^{x^2} ⋅ 2x dx = \int e^u du = e^u + C = e^{x^2} + C$

5. Mar 20, 2016

### Bounceback

Hmm, I'm not claiming that $\int e^{f(x)}\,dx=e^{f(x)} + C$. I'm claiming that $\int e^{f(x)}\,dx=e^{f(x)}*a(x) + C$

My reasoning behind this is $derivative(e^{f(x)})=e^{f(x}*a(x)$ Therefore the derivative of any e^(f(x)) will have a e^(f(x)) as a multiplier, and the same in the reverse for integrals. So my reasoning in the original equation is that the integral of any e^f(x) will be e^f(x) times a function of x.

Is this what you were talking about?

6. Mar 20, 2016

### Samy_A

Yes, that is quite obvious.
If $\int e^{f(x)} dx=F(x)+c$, then clearly $F(x)=(F(x)e^{-f(x)})e^{f(x)}$.
So just set $a(x)=F(x)e^{-f(x)}$.

However, there is no reason to expect that $a$ will be an elementary function, even when $f$ is.

7. Mar 20, 2016

### HallsofIvy

But that does not help if you cannot find that "function of x"!
Starting from $e^{f(x)}a(x)+ C$ and differentiating, we get $e^{f(x)}f'(x)a(x)+ e^{f(x)}a'(x)$ and we want that equal to $e^{f(x)}$. That is, we want $e^{f(x)}(f'(x)a(x)+ a'(x))= e^{f(x)}$ so that $f'(x)a(x)+ a'(x)= 1$. That is, we need to solve the differential $a'(x)+ f'(x)a(x)= 1$ for some known but general function f(x). There is no generic way to solve that equation.

Last edited by a moderator: Mar 20, 2016
8. Mar 20, 2016

### SteamKing

Staff Emeritus
I cite the case of $\int e^{x^2}\,dx = e^{x^2} * a(x) + C$, using your thesis.

As far as I know, there is no a(x) composed of elementary functions which can be found to make this statement true.

https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions

While it is true that if $\frac{d}{dx} (e^{x^2}) = 2x ⋅ e^{x^2}$ and you integrate both sides of this equation, you will obtain $e^{x^2}$, but that's not the same as saying $\int e^{x^2}\,dx = 2x ⋅ e^{x^2}$, which is not true.

9. Mar 20, 2016

### Bounceback

If I'm correct, an elementary function doesn't encompass an infinite series. I was intending to find something along this line for a(x).

Ah, this answers the original question, thank you. Could you explain why this can't be solved (based on differential principles)?

My attempts (in the original question) ended up not working, could someone explain where it went wrong?

Last edited: Mar 20, 2016
10. Mar 20, 2016

### HallsofIvy

I didn't say it couldn't- that depends upon what f is! I just said there was no method that could be used for all f.

11. Mar 20, 2016

### Bounceback

Upon imputing some more values into $\int a(x)\,df(x)+a(x)=x$ I realised that the equations are always differing by a constant -_-

Modifying the process in order to involve a constant results in:

$df(x)*a(x)+da(x)=d(x)$
$\int a(x)\,df(x)+a(x)=x+C$
$a(x)=x+C-\int a(x)\,df(x)$
$a(x)=x+C-\int x+C-\int x+C-\int x+C ...\,df(x)\,df(x)\,df(x)$
$a(x)=x+C-\int x+C\,df(x)+\iint x+C\,d^2f(x)-\iiint x+C\,d^3f(x)...$

This method works for $\int e^x\,dx=e^x$; I'm still stuck on getting it to work for any other integral of e^f(x)

12. Mar 21, 2016

### Anama Skout

If you could find a closed form for $a(x)$ then you'd be able to find any closed form formula for any primitive of any function, by putting $f(x)=\ln g(x)$ in your formula, but that's impossible by Louisville's theorem.

13. Mar 24, 2016

### Bounceback

There is an integral of $e^{x^2}$

$\int e^x*(a(x)+a'(x))\,dx=e^x*a(x)$
$S=a(x)+a'(x)$
$S-S'=a(x)-a''(x)$
$S-S'+S''-S'''...=a(x)\pm S'''^{...}$
Note, $S'''^{...}$ ends up as 0 for all S=x^constant

$\int e^{\sqrt{x}}\,dx=e^{\sqrt{x}}*a(x)$
$x=\sqrt{b}$
$dx=\frac{1}{2*\sqrt{b}}db$
$\int \frac{e^b}{2*\sqrt{b}}\,db$
$S=\frac{1}{2*\sqrt{b}}$
$S-S'+S''-S'''...=\frac{1}{2*\sqrt{b}}-\frac{-1}{4*\sqrt{b^3}}+\frac{3}{8*\sqrt{b^5}}...=a(x)$
$S-S'+S''-S'''...=\frac{1}{2*x^1}-\frac{-1}{4*x^3}+\frac{3}{8*x^5}...=a(x)$
Therefor, $\int e^{x^2}\,dx=e^{x^2}*(\frac{1}{2*x^1}-\frac{-1}{4*x^3}+\frac{3}{8*x^5}...)$
By using the equation in my first post, I'm trying to find a formula that generates a series, like the one above. The above series is diverging however, that's the reason why a(x) isn't an elementary function for e^(x^2).

My current problem is trying to figure out why the series below doesn't work for any f(x) other than x (the same series in my last post).

$\sum_{n=1}^\infty (-1)^{n-1}*\frac{d^{1-n}}{df(x)^{1-n}}*(x+C)$
Note, currently you can only work out C using $\int f'(x)*a(x)\,dx+a(x)-x=C$; Otherwise you have to guess. (for f(x)=x, c=1; f(x)=x^(1/2), c=-1; f(x)=x^(1/3), c=6)

Last edited: Mar 24, 2016
14. Mar 24, 2016

### bigfooted

To understand this, you need much more mathematical tools. Specifcally, you need some abstract algebra and group theory.
Like Anama Skout mentioned, Liouville's theorem explains why it is not possible what you want to do. Today, this falls under differential Galois theory. Some reading to get you started, for real understanding you have to read the references of the wikipedia article:
https://en.wikipedia.org/wiki/Liouville's_theorem_(differential_algebra)
https://www.math.bgu.ac.il/~kamenskm/lectures/diffgalois.pdfhttps://www.math.bgu.ac.il/~kamenskm/lectures/diffgalois.pdf [Broken]

Last edited by a moderator: May 7, 2017
15. Mar 24, 2016

### Bounceback

To which statement of mine are you referring to?

If you're referring to finding a(x) for $\int e^{f(x)}\,dx=e^{f(x)}*a(x)$:
I'm not looking for an elementary function for a(x), I'm looking for an infinite series. The series will converge if a(x) is an elementary function, otherwise the series will diverge. I've given the integral of e^(x^2), despite it not being elementary, as an example.

If you're referring to fixing the series $\sum_{n=1}^\infty (-1)^{n-1}*\frac{d^{1-n}}{df(x)^{1-n}}*(x+C)$:
Although this series is an infinite sum (not an elementary function); the wikipedia article says the theorem does apply for some infinite sums (presumably converging sums), so I'll continue reading if it is this that you are referring to.

16. Mar 31, 2016

### Bounceback

As an update to this, using a different method, I've a formula for any e^f(x) (assuming that the infinith derivative of any f(x) is 0, and that f(x) is differentiable)

$\int e^{f(x)}\,dx=e^{f(x)}*a(x)$
$f(x)=b$
$x=f^i(b)\ Note, f^i(f(x))=x, for\ example, if\ f(x)=x^2\ then\ f^i(x)=x^{1/2}$ (Essentially the solve function)
$dx=f^i{'(b)}db$
$\int e^b*f^i{'(b)}\,db=e^b*a(x)$
$e^b*f^i{'(b)}=e^b*a(x)+e^b*a'(x)$
$f^i{'(b)}=a(x)+a'(x)$
$f^i{'(b)}-f^i{''(b)}=a(x)-a''(x)$
$f^i{'(b)}-f^i{''(b)}+f^i{'''(b)}...=a(x)\pm f^i{'''^{...}(b)}=a(x)$
$\int e^{f(x)}\,dx=e^{f(x)}*(f^i{'(b)}-f^i{''(b)}+f^i{'''(b)}...)$
$\int e^{f(x)}\,dx=e^{f(x)}*(\frac{df^i(b)}{db}-\frac{d^2f^i(b)}{db^2}+\frac{d^3f^i(b)}{db^3}...)$
$\int e^{f(x)}\,dx=e^{f(x)}*(\frac{dx}{df(x)}-\frac{d^2x}{df(x)^2}+\frac{d^3x}{df(x)^3}...)$
$\int e^{f(x)}\,dx=e^{f(x)}*(A-B+C-D...)$
This method only works however if f(x)=b can be solved for x (same thing as saying 'only works if f^i(b) can be found'). By taking the derivative of the last equation, a work-around can be found.
$e^{f(x)}=e^{f(x)}*f'(x)*(\frac{dx}{df(x)}-\frac{d^2x}{df(x)^2}+\frac{d^3x}{df(x)^3}...)+e^{f(x)}*\frac{d}{dx}(\frac{dx}{df(x)}-\frac{d^2x}{df(x)^2}+\frac{d^3x}{df(x)^3}...)$
$1=1-f'(x)(\frac{d^2x}{df(x)^2}-\frac{d^3x}{df(x)^3}...)+\frac{d}{dx}(\frac{dx}{df(x)}-\frac{d^2x}{df(x)^2}+\frac{d^3x}{df(x)^3}...)$
$\frac{df(x)}{dx}(\frac{d^2x}{df(x)^2}-\frac{d^3x}{df(x)^3}...)=\frac{d}{dx}(\frac{dx}{df(x)}-\frac{d^2x}{df(x)^2}+\frac{d^3x}{df(x)^3}...)$
$f'(x)*(B-C+D...)=A'-B'+C'...$
Assuming that {term number} of series 1 cancels with {term number} of series 2:
$B=\frac{A'}{f'(x)},C=\frac{B'}{f'(x)},D=\frac{C'}{f'(x)}...$, also, since $f'(x)*A=1, A=\frac{1}{f'(x)}$

Last edited: Mar 31, 2016