# Integral [0,PI] sinx^2/ sqrt(x)

## Homework Statement

I have to use the comparison test although I'm bad at coming up with what to compare it to

## Homework Equations

What if I compare it to 1/sqrt(x)

## The Attempt at a Solution

I take the integral of 1/sqrt(x) from [0,PI] I get -2sqrt(x) and then I use limits and evaluate
I get 2sqrt(PI)

Can I do this? So since this integral is larger then the one in question my integral converges?

Yes you have the general idea. In a sense, we're sandwiching the integral $\displaystyle\int ^{\pi}_0 \frac{\sin ^2x}{\sqrt{x}}$ between two integrals with finite value, showing that it converges. You have the upper bound right; since $\sin ^2x\le 1$, then $\displaystyle\int ^{\pi}_0 \frac{\sin ^2x}{\sqrt{x}}\le \displaystyle\int ^{\pi}_0 \frac{1}{\sqrt{x}}$. By $p$-test you can show that this larger integral converges. You had this part. However, you do need some "protection" against your integral diverging to negative infinity. This should be pretty obvious though.

What do you mean protection?

Mark44
Mentor
By "protection" Arkuski means that you also need to show that your integral is bounded below, not just bounded above. We know that sin2(x) ≤ 1 for all real x. Is there a lower bound for this as well?

I'm kind of confused. Guess: lower bound is -1? For all real x?

Mark44
Mentor
A better one is 0.

Why would it be 0. Because you are squaring?

Yes you have the general idea. In a sense, we're sandwiching the integral $\displaystyle\int ^{\pi}_0 \frac{\sin ^2x}{\sqrt{x}}$ between two integrals with finite value, showing that it converges. You have the upper bound right; since $\sin ^2x\le 1$, then $\displaystyle\int ^{\pi}_0 \frac{\sin ^2x}{\sqrt{x}}\le \displaystyle\int ^{\pi}_0 \frac{1}{\sqrt{x}}$. By $p$-test you can show that this larger integral converges. You had this part. However, you do need some "protection" against your integral diverging to negative infinity. This should be pretty obvious though.

Doesn't p test say that convergent if p > 1 and divergent if p<=1 ? So how come this integral that I'm comparing to is convergent? I don't get it.

Doesn't p test say that convergent if p > 1 and divergent if p<=1 ? So how come this integral that I'm comparing to is convergent? I don't get it.

It depends on the bounds that the integral has. Let $c\in ℝ^+$. The integral $\displaystyle\int_c^∞\frac{1}{x^p}dx$ is convergent if $p>1$ and divergent otherwise. However, the integral $\displaystyle\int_0^c\frac{1}{x^p}dx$ is convergent if $p<1$ and divergent otherwise.

Please note that in the following argument I'm throwing precision into the wind. I like to think of this in terms of inverses. Suppose we start with $f(x)=\frac{1}{x}$. This will be our base. Now let's think about $g(x)=\frac{1}{x^2}$. This function gets comparitively closer to the x-axis, but farther away from the y-axis. We note that $\displaystyle\int_c^∞\frac{1}{x^2}dx$ converges, and intuitively so since it hugs the x-axis so tight. Now let us imagine its inverse, $g^{-1}(x)=\frac{1}{\sqrt{x}}$. Since the inverse essentially reflects a function over the diagonal line $y=x$, we now see that $g^{-1}(x)$ hugs the y-axis very closely, and an integral that accounts for that area, namely $\displaystyle\int_0^c\frac{1}{\sqrt{x}}dx$, converges.

Mark44
Mentor
I think that Arkuski misspoke. The p-test doesn't apply here - it's used for infinite series, and that's not what you have here.

This integral is improper, but it converges:
$$\int_0^{\pi}\frac{dx}{\sqrt{x}}$$

Ah clear. Thanks.

What about if the lower bound was 1 and the upper bound infinity? In that case, the larger function or our f(x) (1/sqrt(x)) is divergent, and you can't prove that g(x) (sin^2(x)/sqrt(x)) is divergent based on f(x) being divergent.

I'm guessing you'd have to pick a different integral for comparison, but I'm stuck as to which one to pick.

I know sin^2(x)/x is less than or equal to sin^2(x)/sqrt(x) for all x>1, so I thought maybe 1/x would work for comparison, but I graphed them and it doesn't.

Ray Vickson