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Determine the limit $$\lim_{k \to \infty} \int_{0}^1 x^{- \frac{1}{2}}\cos(x^k)e^{-\frac{x^2}{k}} dx.$$

I suppose the dominated convergence theorem would be in place here? If I denote $f_k(x)= x^{- \frac{1}{2}}\cos(x^k)e^{-\frac{x^2}{k}}$, then $$f_k(x) = x^{- \frac{1}{2}}\cos(x^k)e^{-\frac{x^2}{k}} \le x^{- \frac{1}{2}}e^{-\frac{x^2}{k}} \le e^{-\frac{1}{k}}$$

So now I would have that $$\lim_{k \to \infty} \int_{0}^1 x^{- \frac{1}{2}}\cos(x^k)e^{-\frac{x^2}{k}}dx = \int_0^1 e^{-\frac{1}{k}} dx = e^{-\frac{1}{k}} ?$$

Am I missing something here or is it really $e^{-\frac{1}{k}} $?

Jiming Le
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  • How did you come up with the second bound? $x\to 0$ will diverge. – MachineLearner Oct 01 '21 at 18:12
  • How can there be a $k$ on the RHS of the last line? It should have gone, right? Either way, if you want to apply DCT, you want to first know the pointwise limit $\lim_{k \to \infty} x^{-\frac 12} \cos(x^k)e^{\frac{-x^2}{k}}$. Try studying each part separately for large $k$, and then put it together. Use Taylor series to understand their behaviours as $x$ is close to $0$ or $1$. Then think about the bounding function (on your second last line, it's not clear how you went from $x^{-1/2}e^{-x^2/k}$ to $e^{-1/k}$). – Sarvesh Ravichandran Iyer Oct 01 '21 at 18:12
  • Now hold on - DCT lets you replace the integrand with it's pointwise limit, not with it's upper bound. – DMcMor Oct 01 '21 at 18:13
  • @TeresaLisbon I went from $x^{-1/2}e^{-x^2/k}$ to $e^{-x^2/k}$by assuming that $x \le 1$ which should be true by the bounds of the integral? – Jiming Le Oct 01 '21 at 18:24
  • @JimingLe Near $0$, $x^{-\frac 12}$ is near infinity. – Sarvesh Ravichandran Iyer Oct 01 '21 at 18:27

2 Answers2

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Mistakes in attempt

  • $x^{-\frac 12}e^{-\frac{x^2}{k}} \leq e^{-\frac 1k}$ is false, for example as $x$ approaches $0$ the LHS approaches infinity while the RHS stays fixed.

  • The limit is incorrect on the last line.

To apply the DCT , we want to make sure that we know whether the function $\lim_{k \to \infty} x^{-\frac 12}\cos(x^k)e^{-x^2/k}$ exists or not as an a.e. pointwise limit on $[0,1]$, and what it is.

To do this, we look at the parts.

  • $x^{-\frac 12}$ doesn't depend on $k$, so it stays as is.

  • What does $x^k$ look like for large $k$ and $x\in [0,1]$? What is this part going to converge a.e. to?

It would be the function $f(x) = 0$ for $x \neq 1$, and $f(1) =1$. So it's a.e. zero, in other words.

  • Hence, what does $\cos(x^k)$ converge a.e. to?

Since the cosine is continuous, $\cos(x^k)$ would converge to $1$ a.e.

  • Where does $-\frac{x^2}{k}$ converge to a.e. ?

Obviously it converges to the zero function.

  • Hence, where does $e^{-\frac{x^2}{k}}$ converge?

Since the exponential is continuous, close to $1$.

Thus, putting things together, we get by the product rule that $\lim_{k \to \infty} x^{-\frac 12}\cos(x^k)e^{-x^2/k}$ equals:

$x^{-\frac 12} \times 1 \times 1 = x^{-\frac 12}$ a.e.

and thus, providing the DCT held, the answer is:

$\int_0^1 x^{-\frac 12}dx = 2$.

How to think about the DCT? We obviously want to bound the function $|x^{-\frac 12}\cos(x^k)e^{-x^2/k}|$ uniformly in $k$ and get an integrable function of $x$ as the bound.

  • What is the obvious bound for $|\cos(x^k)|$?

It's $1$, isn't it?

  • What is an upper bound for $-\frac{x^2}{k}$ for any $k$ and $x \in [0,1]$?

Come on, they're all negative quantities, so $0$!

  • Thus, what is an upper bound for $e^{-\frac{x^2}{k}}$?

By monotonicity of the exponential, the required upper bound is $e^{0}=1$.

  • Leaving the first term as is, what does the eventual bound look like? Is it integrable?

$x^{-\frac 12} \times 1 \times 1 = x^{-\frac 12}$, obviously integrable!

Of course it turns out to be integrable, but it also turns out to be ... yeah, it's in the spoiler.

Here's the integral for $k=25$ computed via Wolfram Alpha, and you can check that it's already quite close to the actual answer we derived.

Sarvesh Ravichandran Iyer
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On the interval $(0,1)$ the integrands, which are positive, converge pointwise to $1/\sqrt x.$ Note that $1/\sqrt x\in L^1(0,1).$ Furthermore, each of these integrands is bounded above by $1/\sqrt x.$ We're done by the DCT: The limit is $\int_0^1 1/\sqrt x\, dx = 2.$

zhw.
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