Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

According to the fundamental theorem of calculus, if $f$ is continuous and $F$ is defined as $F(x)=\int_a^x f(t) dt$ then $F'=f$. But what happens if $x$ appears inside the integral? I'm trying to find the derivattive of

$$f(x)=\int_0^x \frac{e^{xy}}{y}dy$$

I read about the Libniz integral rule, but when I try to use it I get

$$\frac{df(x)}{x}=\int_0^x e^{xy} dy + \frac{e^{x^2}}{x}-\frac{1}{0}\cdot0$$

Is this because $\frac{e^{xy}}{y}$ is not defined when $y=0$? Also, is the Leibniz rule the only way to solve problems like this one?

share|cite|improve this question
Do you mean $\frac{df(x)}{dx}$? – Calvin Lin Jan 4 '13 at 2:12
up vote 1 down vote accepted

The main hypotheses for applying the Leibniz integral rule is that the derivative of $e^{xy}/y$ in $x$ must exist and be continuous on a rectangle whose base contains the domain of integration. Clearly $e^{xy}/y$ cannot be extended to continuously to any point at which $y=0$ due to the division by 0, so it certainly cannot be differentiable. Thus it fails the main hypothesis.

If you wanted to calculate $f'(x)$ in this case you would have to do it by hand at $x=0$, but the Leibniz formula you derived should work fine for $x\neq 0$ as a $\int_0^{x+h}-\int_0^{x} = \int_x^{x+h}$ stays away from the bad point when $x\neq 0$ (so long as you first prove these integrals exist).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.