# 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.

• The integrand is negative in $(0,1)$ so $n$ cannot be a non-integer. – Tavish Apr 29 at 19:52
• what do you mean? could you explain that better – geocalc33 Apr 29 at 20:06
• The function you have inside the integral is negative, but a negative number to a non-integral power is not defined. – Tavish Apr 29 at 20:42
• @Tavish oh I did not know that. Thank you for pointing it out to me. I'm sure $n$ can run over the naturals though – geocalc33 Apr 29 at 20:52

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 $$y=1/t$$ $$\Rightarrow (-1)^n n^{n+1}\int _0^{\infty} \exp\left(-t-\frac{n(n+1)}{t}\right) t^{-(n+2)}\,dt$$ $$=(-1)^n n^{n+1}\int _0^{\infty} \exp\left(-t-\frac{(\sqrt{4n(n+1)})^2}{4t}\right) t^{-(n+2)}\,dt$$ $$= 2 (-1)^n \left(\frac{n}{n+1}\right)^{\frac{n+1}{2}} K_{n+1}\left(\sqrt{4n(n+1)}\right)$$