# Proper use of comparison theorum?

1. Sep 18, 2011

### PCSL

$$\int_{0}^{\frac{pi}{2}} \frac{cos(x)}{x^{1/2}} dx$$

$$\int_{0}^{\frac{pi}{2}} \frac{cos(x)}{x^{1/2}} dx \le \int_{0}^{\frac{pi}{2}} \frac{1}{x^{1/2}} dx$$

This would mean that the original equation was convergent. Was my reasoning correct in making the numerator 1 since the max value of cos(x) is 1?

Also, are there any hints you guys can give me for using the comparison theorem? Thanks.

2. Sep 18, 2011

### LCKurtz

Yes. It looks like you have the idea. Generally if you think something converges you would overestimate the numerator and/or underestimate the denominator to get an integrand that is both

1. Easier
2. Convergent.

And everything is just the opposite if you are trying to show an integral diverges.

3. Sep 18, 2011

### PCSL

Thanks!

Would this one be correct?

$$\int_{\frac{1}{2}}^{1} \frac{1}{x(1-x)^{\frac{1}{3}}} dx \le \int_{\frac{1}{2}}^{1} \frac{1}{(1-x)^{\frac{1}{3}}} dx$$

Also, I'm a little confused on how I would break up

$$\int_{0}^{\frac{1}{3}} \frac{sec(\pi x)}{1-3x} dx$$

Could I have a hint?

4. Sep 18, 2011

### Dick

1/(x*(1-x)^(1/3)) is GREATER than 1/((1-x)^(1/3)) on [1/2,1). Do you see why? (1/x) is GREATER than 1. For the second one, sketch a graph. What are the max and min values of sec(pi*x) on [0,1/3]?

5. Sep 18, 2011

### PCSL

Haha, now the second one makes total sense but I'm lost on the first one (1/(x(1-x)^(1/3). I have to find an equation that is greater than the original correct?

6. Sep 18, 2011

### Dick

I assume your given function is (1/(x(1-x)^(1/3)). Split it into (1/x)*(1/(1-x)^(1/3)). What are the max and min values of (1/x) for x in the range [1/2,1]?

7. Sep 18, 2011

### PCSL

Could I do

$$\int_{\frac{1}{2}}^{1} \frac{1}{x(1-x)^{\frac{1}{3}}} dx \le \int_{\frac{1}{2}}^{1} \frac{1}{2(1-x)^{\frac{1}{3}}} dx$$

8. Sep 18, 2011

### Dick

Close. 1/x <= 2. So (1/x)*(1/(1-x)^(1/3))<=2*(1/(1-x)^(1/3)). I hope that's what you meant.

9. Sep 18, 2011

### PCSL

Yes! Thank you for your help!