Calculate a hard limit involving logs and exponentials For practice I gave myself some limits to compute. I gave myself hard limits so that the test might be easier.
Limit #1.
Evaluate the limit:

$$\lim_{n \to \infty} \log (n) \int_0^1 \bigg(\exp\bigg(\frac{1}{\log(x)}\bigg)\log(x)x\bigg)^n ~dx.$$

My attempt:
I saw the expression and boosted $n$ in my mind immediately to gain conceptual ground on the problem. In parallel I placed a small amount of my energetic resources into thinking analytically. I was able to get to a $0~\cdot \infty$ form for the limit fairly quickly. This is because I knew that the integrand would go to $0$  if I ignored the pre-multiplication of $\log(n).$ 
I figured out that I could manipulate the indeterminate form and then hit it with L'Hopital's rule. 
So I rewrote the big expression above as:
$$ \lim_{n \to \infty} \frac{\log (n)} {\int_0^1 \bigg(\exp\bigg(\frac{1}{\log(x)}\bigg)\log(x)x\bigg)^n ~dx}.$$
And then I unfortunately realized that this is a form $\frac{\infty}{0}.$ I realized that I could not use L'Hopitals rule...
Then I decided to try again, this time with more urgency and purpose. So I rewrote it a different way and realized I could in fact now use the rule!
$$\lim_{n \to \infty} \frac{1}{\frac{1}{\log(n)}} \int_0^1 \bigg(\exp\bigg(\frac{1}{\log(x)}\bigg)\log(x)x\bigg)^n ~dx$$ because we have the form $\frac{0}{0}.$
L'Hopitals rule goes like: "differentiate the numerator and differentiate the denominator.Then take the limit." So that's what I did:
$$ \frac{\lim_{n\to \infty} \frac{d}{dx} A_n(x)}{\lim_{n\to \infty} \frac{d}{dx} B_n(x)}$$
where $A_n(x)\equiv \int_0^1 \bigg(\exp\bigg(\frac{1}{\log(x)}\bigg)\log(x)x\bigg)^n dx  $
and $ B_n(x)\equiv \frac{1}{\log(n)}. $
Then I got confused and wasn't sure if I had defined everything correctly...I went back and checked my work.
I noticed that I should have put, $\frac{d}{dn} B(n),$ in the numerator. I should have used $n$ instead of using the variable $x.$ This is because we're taking the limit as $n$ goes to infinity not the limit as $x$ approaches infinity.
So here I was with $10$ minutes left, and I had not even finished the first question. So I took a deep breath and continued.
I played the end game and asked myself what the answer was. Immediately I came up with 3 options: $0,1,\infty.$ But I needed to verify the correct answer still. This was just a guess.
At this point I'm fairly tired and I just want to give up on the whole thing and come back to it the next day, but I decided to just solve this one problem and forget about the others.
But then I actually did call it quits and save it for the next day. 

How do you find the limit? I think it's $0,$ but of course that's no proof.

 A: I agree with Tavish that the limit only makes sense for integer $n$.
Claim: for $n\geq 1$, we have
$$
 \int _0^1 \left( \exp\left(\frac{1}{\log(x)}\right) x\log(x)\right)^n\,dx= 2 (-1)^n \left(\frac{n}{n+1}\right)^{\frac{n+1}{2}}
   K_{n+1}\left(\sqrt{4n(n+1)}\right),
$$where $K_{\alpha}(x)$ is the modified Bessel function of the second kind:
$$
K_{\alpha}(x) = \frac{1}{2}\left(\frac{x}{2}\right)^{\alpha}\int_{0}^{\infty} \exp\left(-t - \frac{x^2}{4t}\right) t^{-(\alpha+1)}\,dt
$$Since $K_{\alpha}(x)\sim1/2 \Gamma(\alpha) (2/x)^{\alpha}$, the limit in question is indeed $0$ as the $\log(n)$ term gets rapidly outpaced. 
So how to prove the claim? Start with the substitution $y=\log(x)$ or $x=e^y$, with $dx=e^ydy$:
$$
 \int _0^1 \left( \exp\left(\frac{1}{\log(x)}+\log(x)\right) \log(x)\right)^n\,dx
$$
$$
\Rightarrow  \int _{-\infty}^{0} \exp\left(\frac{n}{y}+(n+1)y\right) y^n\,dy
$$
Now put $y\mapsto - ny$ (this could have been done in the previous step)
$$
=(-1)^n n^{n+1}\int _0^{\infty} \exp\left(-\frac{1}{y}-n(n+1)y\right) y^n\,dy
$$
Now put $y=1/t$
$$
\Rightarrow (-1)^n n^{n+1}\int _0^{\infty} \exp\left(-t-\frac{n(n+1)}{t}\right) t^{-(n+2)}\,dt
$$
$$
=(-1)^n n^{n+1}\int _0^{\infty} \exp\left(-t-\frac{(\sqrt{4n(n+1)})^2}{4t}\right) t^{-(n+2)}\,dt
$$
$$
= 2 (-1)^n \left(\frac{n}{n+1}\right)^{\frac{n+1}{2}}
   K_{n+1}\left(\sqrt{4n(n+1)}\right)
$$
Sources:


*

*https://en.wikipedia.org/wiki/Bessel_function

*https://mathworld.wolfram.com/ModifiedBesselFunctionoftheSecondKind.html

*https://dlmf.nist.gov/10.32
