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I come across a problem to find $f$ such that $f'(x)=e^{ae^{bx}}$ for constants $a$ and $b$.

I think $f(x)$ should have the form $c_1e^{c_2e^{c_3x}}$ (ignoring the indefinite constant), but it seems not working. Is there any other thought on it?

Connor
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    Where did you got the problem? The integral of this function is not any elementary function... – Marcos Apr 05 '22 at 14:16
  • Title and body are different. If you want the integral of $f'$ as in your title, I can handle that. – Randall Apr 05 '22 at 14:28
  • It comes from the problem that finding $y(x)$ such that $y'(x)=1-cy(x)e^{-dx}$ and $y(0)=0$ with positive constants $c,d$. After multiplying an integrating factor, it turns out to be this problem. – Connor Apr 05 '22 at 14:38
  • The problem is to find an explicit representation of $f(x)$ so that its derivative $f'(x)=\exp(a\exp(bx))$. – Connor Apr 05 '22 at 14:40

1 Answers1

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We can do this using the "Exponential integral" function $\operatorname{Ei}_1(z)$, which may be defined by: $$ \frac{d}{dz}\operatorname{Ei}_1(z) = -\frac{e^{-z}}{z},\qquad\lim_{z \to +\infty}\operatorname{Ei}_1(z) = 0 \tag1$$ or, equivalently for $z>0$, $$ \operatorname{Ei}_1(z) = \int_z^{+\infty}\frac{e^{-s}}{s}\;ds = \int_1^{+\infty} e^{-tz}\frac{dt}{t} . \tag2$$

Define $$ F(x) = -\frac{\operatorname{Ei}_1\left(-ae^{bx}\right)}{b} . \tag3$$ Differentiate using $(1)$ to get $$ F'(x) = \exp\left(a e^{bx}\right) $$ so $(3)$ is our answer.

It is known that $\operatorname{Ei}_1$ is not an elementary function.
For the OP we cannot expect an elementary function answer, such as $c_1e^{c_2e^{c_3x}}$.

GEdgar
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