# Limit of integral in the wrong variable

Evaluate and justify:

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

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 $\int_0^1 \frac{x}{\sqrt{1-x}}dx$ 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 $0^+$.)

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 $0$.

• You have to justify changing the limit with integration. What course is this? – science Mar 12 '15 at 4:54
• 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. – Tim Clark Mar 12 '15 at 5:02
• 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? – Mark Viola Mar 12 '15 at 5:08
• 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). – Mark Viola Mar 12 '15 at 5:26
• 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?". – science Mar 12 '15 at 5:35

$$\int_{0}^{1}\bigg|\frac{x\cos y}{{(1-x+y)^{1/3}}}\bigg| dx < \int_{0}^{1}\frac{x}{(1-x)^{1/3}} dx <\infty,$$
since $|\cos y| \leq 1$ and $(1-x+y) > 1-x$. Now you can exchange the order.
$$\int_{0}^{1}\frac{x}{(1-x)^{1/3}} dx = \frac{9}{10}.$$
just use the substitution $1-x=u$.