I need a help with my appeal. My task was to find an expected value of some random variable $Y$, or to prove it doesn't exist.
All ended up to $$ \mathbb{E}[Y]= \int_0^{\infty} \frac{1}{ue^u}du \tag{1} $$ And know I need to check it for convergence.
Here is my solution:
Firstly, the integrand is always positive.
I showed that $$ \lim_\limits{u \to 1^-} \frac{\frac{1}{ue^u}}{\frac{1}{u}}=e^{-1} $$ which is finite and positive. Hence, by a Limit Comparison Test the integral: $$ \int_0^1 \frac{1}{ue^u}du \tag{2} $$ diverges since the integral: $$ \int_0^1 \frac{1}{u}du $$ diverges. And then I said that the divergence of the integral $(2)$ implies the divergence of the integral $(1)$ and we're done.
Feedback from examiners:
The divergence of the integral $(2)$ DOES NOT implies the divergence of the asked integral $(1)$.
What? Why? Are they wrong? I believe they are! This is my argument: $$ \int_0^\infty \frac{1}{ue^u}du = \int_0^1 \frac{1}{ue^u}du + \int_1^{\infty} \frac{1}{ue^u}du $$
We have $$\int_0^1 \frac{1}{ue^u}du=\infty$$
All we should know about the integral $$\int_1^{\infty} \frac{1}{ue^u}du>0$$ is that it is positive(does not depend convergent or divergent it is). And then $$\int_0^{\infty} \frac{1}{ue^u}du=\infty + something>0=\infty$$
And that's it! Is this argument good enough?