# Test for Convergence

1. Mar 14, 2013

### whatlifeforme

1. The problem statement, all variables and given/known data
Test for convergence or divergence.

2. Relevant equations
$\displaystyle\int_0^∏ {\frac{sin∅}{\sqrt{pi-∅}} d∅}$

3. The attempt at a solution
the solution manual does the following:

pi -∅ = x

$\displaystyle\int_0^pi {\frac{sinx}{\sqrt{x}} d∅}$

$0 <= \frac{sinx}{\sqrt{x}} <= \frac{1}{\sqrt{x}}$

but: sqrt( ∏ - ∅) > sqrt(∅) when ∅< ∏

2. Mar 14, 2013

### SammyS

Staff Emeritus
(Use "\theta" in LaTeX for θ and "\pi" for pi.)

Yes, but dx = -dθ , and $\displaystyle \ \ -\int_{a}^{b}f(x)\,dx=\int_{b}^{a}f(x)\,dx$

3. Mar 14, 2013

### whatlifeforme

i'm sorry what do you mean?

4. Mar 15, 2013

### SammyS

Staff Emeritus
I thought you were having difficulty understanding why the limits of integration for the integral in x don't go from π to 0 .

OK, then what did you mean by the following?
the solution manual does the following:

pi -∅ = x

$\displaystyle\int_0^pi {\frac{sinx}{\sqrt{x}} d∅}$

$0 <= \frac{sinx}{\sqrt{x}} <= \frac{1}{\sqrt{x}}$

but: sqrt( ∏ - ∅) > sqrt(∅) when ∅< ∏​
Especially the statement following the word "but" .

5. Mar 15, 2013

### whatlifeforme

they are replacing (∏ - ∅) by x (which is greater than which is essentially a replacement for ∅.

in other words:

sqrt( ∏ - ∅) > sqrt(∅) when ∅< ∏

but when (∅> ∏): sqrt( ∏ - ∅) > sqrt(∅)

how does this apply on the integral of the problem. *checked* oopps.

the interval is from 0 to pi, so sqrt( ∏ - ∅) > sqrt(∅) when ∅< ∏ applies.

6. Mar 15, 2013

### SammyS

Staff Emeritus
Oops is right, but you continue writing incorrect statements. (Also, please use θ, not the symbol for the empty set, ∅ .)

If θ = π/2 , then $\displaystyle \ \sqrt{\pi-\theta\,}=\sqrt{\theta\,}\ .$

If 0 < θ < π/2, then $\displaystyle \ \sqrt{\pi-\theta\,}>\sqrt{\theta\,}\ .$

If π/2 < θ < π, then $\displaystyle \ \sqrt{\pi-\theta\,}<\sqrt{\theta\,}\ .$

However, I don't see what any of this has to do with your problem.