# limit $\lim_{k \to \infty} \int_{0}^{1} \frac{kx^k}{1+x} dx$

Consider the following sequence of functions and their integrals on $$[0,1]$$. Evaluate the limit of the following integral if possible

$$\lim_{k \to \infty} \int_{0}^{1} \frac{kx^k}{1+x} dx$$

At first glance, I do not know if the limit exists or not. Wolfram alpha will not take the limit for me, but it did evaluate the integral. The result is terms of a function I do not understand, but from looking at the series expansion:

$$x^k \left(\frac{kx}{k+1} - \frac{kx^2}{k+2} + \frac{kx^3}{k+3} - \dots\right)$$ it appears as though the limit is $$0$$ since $$x \in [0,1]$$. However, I have attempted the problem two different ways, and I am getting that the result is infinity. I am curious where I maybe going wrong.

Note: I am not looking for an answer. I just want to see where my logic is wrong (mainly in my 2nd attempt) so that I can try to make the correction.

Attempt 1: I can pull the $$k$$ outside the integral and get

$$\lim_{k \to \infty} k \int_{0}^{1} \frac{x^k}{1 + x} dx$$.

Now, $$|\frac{x^k}{1+x}| \leq \frac{1}{1+x} = g(x)$$ since we are on $$[0,1]$$. Since $$g(x)$$ is integrable, and dominates $$f(x)$$, the dominated convergence theorem tells me I can pass the limit inside.

$$\int_{0}^{1} \lim_{k \to \infty} \frac{kx^k}{1+x} dx = \infty$$ due to the factor of $$k$$.

Now, outside of the fact that I have not done measure theory for quite some time, I am uncertain about this for two reasons. One, DCT refers to Lebesgue integrable functions. I am always hesitant when I use it during a discussion of Riemann integration. Secondly, while I am definitely allowed to pull the $$k$$ outside of the integral, I am not sure I am allowed to apply DCT on the result without $$k$$, and then put the $$k$$ back inside along with the limit. If I had to dominate $$\frac{kx^k}{1 + x}$$, I am not sure it would work. These uncertainties made me want to try to brute force this with Riemann integration.

Attempt 2:

I will rewrite the integral as

$$\lim_{k \to \infty} k \int_{0}^{1} \frac{x^k}{(\sqrt{1 + x})^2}dx$$ so that I can apply Trigonometric substitution. I would get

$$\sec(\theta) = \sqrt{1 + x}$$

$$\tan(\theta) = \sqrt{x}$$

$$\tan^2(\theta) = x$$

$$2\tan(\theta)\sec^2(\theta)d\theta = dx$$

Substituting,

$$\lim_{k \to \infty} k \int \frac{\tan^{2k}(\theta)}{\sec^2(\theta)} 2 \tan(\theta) \sec^2(\theta) d\theta$$

Cleaning this up,

$$\lim_{k \to \infty} 2k \int \tan^{2k+1}(\theta) d\theta$$

Now, I can pull out 2 of the tangents.

$$\lim_{k \to \infty} 2k \int \tan^{2k - 1}(\theta) \tan^2(\theta)d\theta$$

Using the identity $$\tan^2(\theta) = \sec^2(\theta) - 1$$,

$$\lim_{k \to \infty} 2k \int \tan^{2k-1}(\theta) \sec^2(\theta) d\theta - \lim_{k \to \infty} 2k \int \tan^{2k-1}(\theta) d\theta$$

The first integral can be done with $$u$$-substitution. Let $$u = \tan(\theta)$$. Then, $$du = \sec^2(\theta) d\theta$$

$$\lim_{k \to \infty} 2k \left[\frac{u^{2k}}{2k}\right] - \lim_{k \to \infty} 2k \int \tan^{2k-1}(\theta) d\theta$$

Now, I can back substitute.

$$\lim_{k \to \infty} 2k \left[\frac{\tan^{2k}(\theta)}{2k}\right] - \lim_{k \to \infty} 2k \int \tan^{2k-1}(\theta) d\theta$$

Again,

$$\lim_{k \to \infty} 2k \left[\frac{x^k}{2k}\right]_{0}^{1} - \lim_{k \to \infty} 2k \int \tan^{2k-1}(\theta) d\theta$$

Evaluating,

$$\lim_{k \to \infty} 2k \frac{1}{2k} - \lim_{k \to \infty} 2k \int \tan^{2k-1}(\theta) d\theta$$

At this point, I can say the first limit is $$1$$. Now, I still have an odd power of tangent in the second integral. So, I can repeat this process again

$$1 - \lim_{k \to \infty} 2k \int \tan^{2k - 3}(\theta) \sec^2(\theta) d\theta + \lim_{k \to \infty} 2k \int \tan^{2k - 3}(\theta) d\theta$$

Then,

$$1 - \lim_{k \to \infty} 2k \left[\frac{u^{2k-2}}{2k-2}\right] + \lim_{k \to \infty} \int \tan^{2k - 3}(\theta)$$

Again,

$$1 - \lim_{k \to \infty} 2k \left[\frac{\tan^{2k-2}(\theta)}{2k-2}\right] + \lim_{k \to \infty} \int \tan^{2k - 3}(\theta)$$

Then,

$$1 - \lim_{k \to \infty} 2k \left[\frac{x^{k-1}}{2k-2}\right]_{0}^{1} + \lim_{k \to \infty} \int \tan^{2k - 3}(\theta)$$

Evaluating,

$$1 - \lim_{k \to \infty} 2k \frac{1}{2k-2} + \lim_{k \to \infty} \int \tan^{2k-3}(\theta) d\theta$$

Then, this limit is also $$1$$

$$1 - 1 + \lim_{k \to \infty} 2k \int \tan^{2k-3}(\theta) d\theta)$$

So,

$$\int \tan^{2k-3}(\theta) d\theta)$$

Well, this is very nice. I have ended up with a single integral with an odd power of tangent. So, I can repeat these two steps over and over until I am left with.

$$\lim_{k \to \infty} 2k \int \tan(\theta) d\theta$$

after going through all of the powers. Well, this integral I know

$$\lim_{k \to \infty} 2k \left[- \ln|\cos(\theta)| \right]$$

Going back to our trig sub, we can see that

$$\cos(\theta) = \frac{1}{\sqrt{1 + x}}$$. Therefore,

$$\lim_{k \to \infty} 2k \left[ -\ln|\frac{1}{\sqrt{1+x}}|\right]_{0}^{1}$$

Evaluating,

$$\lim_{k \to \infty} -2k\ln|\frac{1}{\sqrt{2}}| = \infty$$

This matched the result I got before, and I will much better about this because I just used straight up integration rather than a sophisticated tool that I don't know very well, the dominated convergence theorem.

My uncertainty lies in that the answer does not appear to be $$\infty$$ based on the power series wolfram alpha gave. Does anyone see where I am going wrong?

• First deal with $\int_0^1 x^kf(x) \,dx$ and show that if $M$ is a bound for $f(x)$ on $[0,1]$ then the integral doesn't exceed $M\int_0^1 x^k\,dx$ and hence tends to $0$. Next use integration by parts for your integral. Commented Aug 3, 2022 at 2:39

\begin{align} \lim_{k\to\infty}\int_0^1\frac{kx^k}{1+x}\,\mathrm{d}x &=\lim_{k\to\infty}\int_0^1\frac{x}{1+x}\,\color{#C00}{kx^{k-1}\,\mathrm{d}x}\tag{1a}\\ &=\lim_{k\to\infty}\int_0^1\frac{u^{1/k}}{1+u^{1/k}}\,\color{#C00}{\mathrm{d}u}\tag{1b}\\ &=\int_0^1\lim_{k\to\infty}\frac{u^{1/k}}{1+u^{1/k}}\,\mathrm{d}u\tag{1c}\\ &=\int_0^1\frac12\,\mathrm{d}u\tag{1d}\\ &=\frac12\tag{1e} \end{align} Explanation:
$$\text{(1a):}$$ $$kx^k=x\cdot kx^{k-1}$$
$$\text{(1b):}$$ substitute $$u=x^k$$ so that $$\mathrm{d}u=kx^{k-1}\,\mathrm{d}x$$
$$\text{(1c):}$$ Monotone Convergence or Dominated Convergence (dominated by $$\frac12$$)
$$\text{(1d):}$$ evaluate the limit
$$\text{(1e):}$$ evaluate the integral

• Note that $kx^{k-1}\,\mathrm{d}x$ is often denoted as $\mathrm{d}x^k=\mathrm{d}\!\left(x^k\right)$
– robjohn
Commented Aug 3, 2022 at 9:05
• Do you have a source where that is used? I have literally never seen that before, and I would like to read some more about it for when I teach Calculus in case another student sees this. Commented Aug 19, 2022 at 20:25
• It is most often used with Riemann-Stieltjes Integration, though it is also used in Integration by Parts, among other places.
– robjohn
Commented Aug 20, 2022 at 8:05

1st attempt: You can't apply the dominated convergence theorem to the set of functions $$\frac{x^k}{1+x}$$ and then bring in the $$k$$ after you've evaluated the limit. For dominated convergence, you need a sequence of integrable functions that are bounded by an integrable function and converge pointwise to an integrable function.

So, dominated convergence does say $$\lim_{k \to \infty} \frac{x^k}{1+x}dx = 0$$ because $$\frac{x^k}{1+x}$$ converges pointwise to the function $$f(x) = 0$$ on $$(0,1)$$, and indeed these functions are bounded by $$\frac{1}{1+x}.$$ Once you bring in the $$k$$ here, you have to try to apply dominated convergence to the functions $$\frac{kx^k}{1+x},$$ which is an entirely different problem because when you look at $$x = 1$$, we see that these functions diverge to $$+\infty.$$ In general, you can't separate out limits unless you know that each of them converge.

Also, your measurability concerns are trivial: these functions are continuous, and if a function is Riemann integrable on a compact set, they are lebesgue integrable on said compact set. (The issue arises when you integrate over the real numbers.)

2nd attempt: you need to be more careful with how you apply limits. You can't just take a limit of one part of an equation and take the limit of another part of the equation years later. The terms you were throwing away make up a nontrivial part of the limit.

So, let's take your method of eventually getting to $$\int \tan(\theta) d\theta.$$ Let's fix $$k$$, apply your reductions, and see what we get.

\begin{align} \int 2k \tan^{2k+1}(\theta)d\theta &= 2k \frac{x^k}{2k(1+x)} \bigg|_{x=0}^1 - 2k \int \tan^{2k-1}(\theta)d\theta \\ &= 2k \frac{x^k}{2k(1+x)} - 2k\frac{x^{k-1}}{(2k-2)(1+x)} + 2k\int \tan^{2k-3}(\theta)d\theta... \\ &= 2k \bigg( \frac{x^k}{2k(1+x)} - \frac{x^{k-1}}{(2k-2)(1+x)} + ... \pm \frac{x^1}{2(1+x)} \bigg)\bigg|_{x=0}^1 \pm 2k\int \tan(\theta)d\theta. \end{align} You see that we can't just take the limit as $$k \to \infty$$ because of the tail terms (i.e., $$\frac{x}{1+x}$$,) so you have to find some other way of obtaining this limit.

• Do you have a suggestion for a different way to do the limit? Commented Aug 19, 2022 at 20:25
• The other answers give good methods, so I'd just default to those. Personally I'd do what robjohn did. You could also try integrating by parts, $f_k' = kx^k, g_k = 1+x$ and see that you get $$\lim_{k \to \infty} \frac{k}{k+1}x^{k+1} \frac{1}{1+x} \bigg|_{x=0}^1 - \int_0^1 \frac{k}{k+1}x^{k+1}\frac{-1}{(1+x)^2}$$ and look at each of those limits. Commented Aug 19, 2022 at 23:18

Note that $$\lim_{k \to \infty} {kx^k}=0$$ for $$x\in(0,1)$$, which allows the integration below $$\lim_{k \to \infty} \int_{0}^{1} \frac{kx^k}{1+x} dx = \lim_{k \to \infty,\epsilon\to 1} \int_{\epsilon}^{1} \frac{kx^k}{1+x} dx =\lim_{k \to \infty} \int_{\epsilon}^{1} \frac{kx^k}{2} dx=\frac12$$

• I'm not follow the step where you involved the epsilon. Why is it going to 1? And how did it remove the 1 + x in the denominator? Commented Aug 19, 2022 at 20:26
• I'm not following that. What does that mean and what does it have to do with epsilon going to 1? Commented Aug 19, 2022 at 20:39
• @NolanP - because $\lim_{k \to \infty} {kx^k}=0$, except when $x$ is close to 1. So, the integration is only non-vanishing near $x=1$. You only need to integrate $\int_{\epsilon\to 1}^1 (\cdots )dx$ Commented Aug 19, 2022 at 20:42
• Ok I see that now. Wow, that's the second answer on this problem that has used notation I've never seen before. I wonder if this problem is above my level of comprehension. Two of the answers I don't understand. Now what about the next step? Where did the 1 + x go? Commented Aug 19, 2022 at 20:45
• @NolanP - since you only integrate $\int_{\epsilon\to 1}^1 (\cdots )dx$, you replace $1+x$ with $1+1=2$. Commented Aug 19, 2022 at 20:49

Now, $$|\frac{x^k}{1+x}| \leq \frac{1}{1+x} = g(x)$$ since we are on $$[0,1]$$.

You miss the factor $$k$$, if you want to use dominated convergence theorem, you have to show $$|\frac{kx^k}{1+x}| \leq g(x)$$ on $$[0,1]$$, where $$g(x)$$ is independent of $$k$$.

$$\lim_{k \to \infty} \int_{0}^{1} \frac{kx^k}{1+x} dx$$

$$I_k=\int_{0}^{1} \frac{kx^k}{1+x} dx=\frac{k}{k+1}\int_{0}^{1} \frac{1}{1+x} dx^{k+1}=\frac{1}{2}\cdot\frac{k}{k+1}+\frac{k}{k+1}\int_{0}^{1} \frac{x^{k+1}}{(1+x)^2} dx$$

$$\lim_{k\to\infty}I_k=\frac{1}2+\lim_{k\to\infty}\int_0^1\frac{x^{k+1}}{(1+x)^2} dx$$

Since $$\frac{x^{k+1}}{(1+x)^2} \le\frac{1}{(1+x)^2}~~~\text{for}~~x\in[0,1]$$ and $$\frac{1}{(1+x)^2}$$ is integrable on $$[0,1]$$. So by dominated convergence theorem, we can interchange the limit and integration. Namely,

$$\lim_{k\to\infty}I_k=\frac{1}2+\int_0^1\lim_{k\to\infty}\frac{x^{k+1}}{(1+x)^2} dx=\frac{1}2$$

• I'm not sure what you did when you wrote $dx^{k+1}$. I have never seen that before. Commented Aug 2, 2022 at 23:29
• That's integration by part. @NolanP Commented Aug 2, 2022 at 23:30
• I don't understand that. I have never seen that notation before. Can you explain what it is? Maybe if you explain it I can try to decipher what it is? Commented Aug 2, 2022 at 23:40
• For example, $\frac{dx^2}{dx}=2x\Leftrightarrow xdx=\frac{1}2dx^2$, did you see this notation before? @NolanP Commented Aug 2, 2022 at 23:47
• Unfortunately, I have not seen that either. I have never seen a power on the differential $dx$ in this context. I have only seen that with higher order derivatives. Commented Aug 2, 2022 at 23:59