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This is a detailed question from Introduction to Calculus and Analysis, by Courant & John.

1.8b(a) Exercise 2

Let $$F(y) = \int_0^1 \frac{(x-1)x^y}{\log x} dx \quad \text{for} \; y>1$$ Consider $\epsilon \in (0,1/100).$

(a) Prove that there exists two constants $c_1, c_2$ such that $$\left|\int_{1-\epsilon}^1 \frac{(x-1)x^y}{\log x} dx - \frac{1}{y+1}\right| \le \frac{1}{y+1}e^{-c_1(y+1)\epsilon} + c_2\epsilon^2 .$$ for any $\epsilon \in (0,1/100)$ and $y > 1$.

(Hint given by professor: $\log(1-t) = -t + O(t^2)$ for t small, and >$\frac{x-1}{\log x} = 1 + O(\vert x-1 \vert)$ for $\vert x-1 \vert$ small.)

(b) Prove that there exists a constant $c_3$ such that $$\left|\int_0^{1-\epsilon} \frac{(x-1)x^y}{\log x} dx\right| \le e^{-c_3y\epsilon} .$$ for any $\epsilon \in (0,1/100)$ and $y > 1$.

(Hint given by professor: The value of $\frac{(x-1)x}{\log x}$ is between $0$ and $1$ for $x \in [0,1]$)

Some attempts to solve this problem:

$$\int_0^1 x^y\,dx = \frac{1}{y+1}$$ $$\int_{1-\epsilon}^1 x^y\,dx < \frac{1}{y+1}$$ for $\epsilon > 0.$ So, $$\left|\int_{1-\epsilon}^1 \frac{(x-1)x^y}{\log (x)}\,dx - \frac{1}{y+1}\right| > \left|\int_{1-\epsilon}^1 \left(\frac{(x-1)x^y}{\log (x)} - x^y\right)\,dx \right|$$

By applying the hint,

$$ \left|\int_{1-\epsilon}^1 x^y\left(\frac{x-1}{\log (x)}-1\right)\,dx\right| = \left|\int_{1-\epsilon}^1 x^y(1-x)\,dx\right|$$ $$= \frac{1}{y+1}(1-(1-\epsilon)^{y+1}) - \frac{1}{y+2}(1-(1-\epsilon)^{y+2}).$$

macton
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  • Welcome to StackExchange, it is appreciated, especially when posting homework or class questions (as you have sort of admitted this is by adding hints from professors) to show what work you have already done towards solving the question – lioness99a Mar 20 '17 at 15:16
  • Attempts have added. – macton Mar 20 '17 at 15:35

1 Answers1

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Using the "hint" that $\log(1-t)=-t+O(t^2)$ for $t\to 0$, we see that

$$\begin{align} \int_{1-\epsilon}^1\frac{(x-1)}{\log(x)}x^y\,dx&=\int_{\epsilon}^0\frac{t}{\log(1-t)}(1-t)^y\,dt\\\\ &=\int_0^\epsilon \left(1+O(t)\right)(1-t)^y\,dt\\\\ &=\frac{1-(1-\epsilon)^{y+1}}{y+1}+\int_0^\epsilon O(t)(1-t)^y\,dt\\\\ &=\frac{1}{y+1}-\frac{e^{(y+1)\log(1-\epsilon)}}{y+1}+\int_0^\epsilon O(t)e^{y\log(1-t)}\,dt \end{align}$$

Next, we use the mean value theorem to assert that there exists a number $\xi\in (0,\epsilon)$ such that $\log(1-\epsilon)=-\frac{1}{1-\xi}\epsilon=-c_1\epsilon$.

Furthermore, we have $e^{y\log(1-t)}=e^{-yt+O(t^2)}=1-yt+O(t^2)$ so that $\int_0^\epsilon O(t)\,dt=O(\epsilon^2)=-c_2\epsilon^2$ for some $c_2$.

Putting it all together reveals

$$\left|\int_{1-\epsilon}^1\frac{(x-1)}{\log(x)}x^y\,dx-\frac{1}{y+1}\right|=\frac{e^{-c_1(y+1)\epsilon}}{y+1}+c_2\epsilon^2$$

Mark Viola
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  • Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark – Mark Viola Mar 21 '17 at 03:36
  • Sorry, but I can't understand the part when you integrate $\int_0^\epsilon (1-yt+O(t^2)),dt$. – macton Mar 21 '17 at 03:48
  • Well, you're missing an extra $O(t)$ term. We are integrating $$\int_0^\epsilon O(t)e^{y\log(1-t)},dt=\int_0^\epsilon O(t)\left(1-yt+O(t^2) \right),dt=O(\epsilon^2)=-c_2\epsilon^2$$for some $c_2$ – Mark Viola Mar 21 '17 at 03:52
  • But how do we integrate this? By parts or something? – macton Mar 21 '17 at 03:58
  • @Tsai Note that if $f(x)=O(g(x))$ as $x\to x_0$, then there exists a number $C$ such that $f(x)\le Cg(x)$. Then, $\int f(x),dx=C\int g(x),dx$. Can you integrate $x$, $x^2$, and $x^3$? For example, $\int O(t),dt=O(t^2)$. – Mark Viola Mar 21 '17 at 04:06
  • So, by integration, I get $O(\epsilon^2)+O(\epsilon^3)+O(\epsilon^4). $ But why can we omit $ \epsilon^3 & \epsilon^4 $? – macton Mar 21 '17 at 04:27
  • Good question. The higher order terms are much smaller. That is, as $\epsilon\to 0$, $O(\epsilon^3)/\epsilon^2=C\epsilon \to 0$. – Mark Viola Mar 21 '17 at 04:30
  • @Tsai Since you're new to the site, I wanted to let you know that when you have enough reputation points, you can also up vote answers. – Mark Viola Mar 21 '17 at 15:40