1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The integrand is negative in $(0,1)$ so $n$ cannot be a non-integer. $\endgroup$ – Tavish Apr 29 at 19:52
  • $\begingroup$ what do you mean? could you explain that better $\endgroup$ – geocalc33 Apr 29 at 20:06
  • $\begingroup$ The function you have inside the integral is negative, but a negative number to a non-integral power is not defined. $\endgroup$ – Tavish Apr 29 at 20:42
  • $\begingroup$ @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 $\endgroup$ – geocalc33 Apr 29 at 20:52
0
$\begingroup$

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:

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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