can someone tell me a way of integrating functions like $e^{f(x)}$
I have a specific case: $\int e^{-3x}\,\mathrm{d}x$
PS: I'm not looking for the answer of this, but the way of doing it.
Thanks for your help
can someone tell me a way of integrating functions like $e^{f(x)}$
I have a specific case: $\int e^{-3x}\,\mathrm{d}x$
PS: I'm not looking for the answer of this, but the way of doing it.
Thanks for your help
Interesting, I fired my old symbolic algebra program and typed the following
$ \int{\exp\left(C_{0}+C_{1}x-C_{2}x^{2}\right)}\,\mathrm{d}x= $
and it gave the answer
$ =\sqrt{\frac{\pi}{4C_{2}}}\exp\left(C_{0}+\frac{C_{1}^{2}}{4C_{2}}\right)\mathrm{erf}\left(\frac{C_{1}}{2\sqrt{C_{2}}}-\sqrt{C_{2}}\, x\right) $
or more simply
$ \int{\exp\left(-x^2\right)}\,\mathrm{d}x = \frac{\sqrt{\pi}}{2}\,\mathrm{erf}(x) $
with $\mathrm{erf}(x)$ the error-function which is not an elementary function.
So although a general solution may not exist, I am sure there are special cases where as solution is feasible.
If you mean a way to obtain an anti-derivative in terms of elementary functions, there is no such general algorithm: it is known that for $f(x) = -x^2$, $\int e^{f(x)}$ cannot be written in terms of elementary functions.
There are some general algorithms for computing anti-derivatives though, for instance: Risch's algorithm.
Your specific case is much easier than what you have generalized your problem too.
Hint: What is the derivative of $e^{-3x}$ ?
For the case where $f(x)$ is linear, a nice $u$-substitution works. I assume you know how to integrate $\int e^xdx$? So in order to integrate a function of the form $e^{f(x)}$, let $u=f(x)$, and thus $du=f'(x)dx$, which allows you to 'solve' for $dx$ in terms of $du$. Then your original integral goes from:
$ \int e^{f(x)}dx $ to $ \int \frac{e^u}{f'(x)}du. $ Of course, this is not always so easy to integrate, as Moron points out. When $f(x)$ is linear, you have a nice situation, because $f'(x)$ is just a constant. Other situations may not be so easily handled, as far as I'm aware.