Evaluate and justify:

$\lim_{y\to 0^+} \int_0^1 \frac{x\cos{y}}{\sqrt[3]{1-x+y}}dx$

Can I just apply the limit to the $y$ directly, since in regards to the variable being integrated it's just a constant? Then I'd get $\int_0^1 \frac{x}{\sqrt[3]{1-x}}dx$ and integrate via u-substitution $u=1-x$ , $du = -x $.

That's gut instinct. And I'm forgetting a certain theorem about limits that I'm supposed to apply? How should I solve this integral? (I'm sorta suspicious that I'm missing something because of the $0^+$.)

EDIT: After thinking about the comments:

I can pull $\cos{y}$ out of the integral because it's a (relative) constant factor. Then set my $u=1-x+y$ and get the same $du=-x$. Then integrate $$-\int u^{{-1/3}}du = \frac{3}{2}u^{2/3} = [\frac{3}{2}(1-x+y)^{2/3}\Big|_0^1 = cy^a - \frac{3}{2}(1+y)^{2/3} $$

whereby I can apply the limit and get $\frac{-3}{2}$.

Does that look better? I'm no longer exchanging any orders (I think).

Oops! Then $du= -dx$ which doesn't help me.

I still don't get why the limit is $0^+$ instead of just $0$.

  • 1
    $\begingroup$ You have to justify changing the limit with integration. What course is this? $\endgroup$ – science Mar 12 '15 at 4:54
  • $\begingroup$ You have to integrate first, then apply the limit. For example, if the integral was $\int_0^1 sin(xy) dx$ you clearly couldn't switch the order. Applying the limit first would just give 0. Applying it after would lead to $-cos(xy)/y$, which would mean division by 0. Switching the order gives a different result. $\endgroup$ – Tim Clark Mar 12 '15 at 5:02
  • $\begingroup$ You can set $y=0$ in the cosine term, but need to justify interchanging the limit with the integral. Uniform convergence of $f$ is a sufficient condition to justify the interchange. But do you you have that here? $\endgroup$ – Mark Viola Mar 12 '15 at 5:08
  • $\begingroup$ Not quite. The result has to be positive, does it not? You're on the right track. Just be careful. Remember if $u=1+y-x$, the $x = 1+y-u$, $du =-dx$, and the limits switch (which after you absorb the minus sign, the limits go back from 0 to 1). $\endgroup$ – Mark Viola Mar 12 '15 at 5:26
  • $\begingroup$ Your method of solution depends on what's the goal of this exercise? So that's why I asked you in my first comment "what course is this?". $\endgroup$ – science Mar 12 '15 at 5:35

You need to prove the uniform convergence of the integral. One way to do this is

$$ \int_{0}^{1}\bigg|\frac{x\cos y}{{(1-x+y)^{1/3}}}\bigg| dx < \int_{0}^{1}\frac{x}{(1-x)^{1/3}} dx <\infty, $$

since $|\cos y| \leq 1$ and $ (1-x+y) > 1-x $. Now you can exchange the order.

Added: To evaluate the integral

$$ \int_{0}^{1}\frac{x}{(1-x)^{1/3}} dx = \frac{9}{10}. $$

just use the substitution $1-x=u$.


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