1. Showing convergence of Integral

Ok, I just want to make sure I got it right:

I need to show if the Integral $\int_0^{\Pi/2} \frac {1} {sinx}$ is convergent.
So I solved the Integral: = ln[tan $\frac {x} {2}$]
then I put in the values $\Pi/2$ and $\alpha$ lim ->0

ln[tan $\frac {\Pi} {4}$] - ln[tan $\frac {\alpha} {2}$] =1

so the Integral is convergent and the limit is 1. Am I right? Is that the right way to do it?

2. That integral is divergent.

Just see that $\sin(x)\sim{x}$ as $x\rightarrow{0}$ and the function $\frac{1}{\sin(x)}$ is contiunous in $(0, \frac{\pi}{2}]$

Thus, since $\int_0^{\delta}{\frac{dx}{x}}$ is divergent ( $\delta>0$), then so is: $\int_0^{\pi/2}{\frac{dx}{\sin(x)}}$

And apart from that: $\int\frac{dx}{\sin(x)}=\ln\left(\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}\right)+k$ but I'd use the method above

3. Originally Posted by JanW
Ok, I just want to make sure I got it right:

I need to show if the Integral $\int_0^{\Pi/2} \frac {1} {sinx}$ is convergent.
So I solved the Integral: = ln[tan $\frac {x} {2}$]
then I put in the values $\Pi/2$ and $\alpha$ lim ->0

ln[tan $\frac {\Pi} {4}$] - ln[tan $\frac {\alpha} {2}$] =1

so the Integral is convergent and the limit is 1. Am I right? Is that the right way to do it?
PaulRS's Response was very good, but if I was doing this I would have broken it up first

$\int_0^{\frac{\pi}{2}}\frac{dx}{\sin(x)}=\int_0^{1 }\frac{dx}{\sin(x)}+\int_1^{\frac{\pi}{2}}\frac{dx }{\sin(x)}$

Now as Paul stated as $x\to{0}$ we have that $\sin(x)\sim{x}$

and since

$\int_0^{1}\frac{dx}{x}$ is divergent by the special p-value test

and $\int_1^{\frac{\pi}{2}}\frac{dx}{\sin(x)}$ is convergent

The sum of a convergent and divergent integral is a divergent one

4. Great, thanks for the detailed explenation. Just one question, what do you mean with the p-value test? Sorry about my ignorance..

5. Originally Posted by JanW
Great, thanks for the detailed explenation. Just one question, what do you mean with the p-value test? Sorry about my ignorance..
It is the test that says an integral of the form

$\int_0^{1}\frac{dx}{x^{\beta}}$

Converges iff $\beta<1$

6. Wait do you mean $\beta$ >1
And what about if I have the function : $\int_0^{1}\frac{dx}{1+x^{\beta}}$ with $\beta$ = 1 ?
Is it now convergent or divergent?

7. Originally Posted by JanW
Wait do you mean $\beta$ >1
And what about if I have the function : $\int_0^{1}\frac{dx}{1+x^{\beta}}$ with $\beta$ = 1 ?
Is it now convergent or divergent?
There is no longer a discontinuity there, look where on $[0,1]$ is there a problem for $\frac{1}{1+x^{\beta}}$

and no, $\beta>1$ is for integrals of the form

$\int_1^{\infty}\frac{dx}{x^{\beta}}$

8. Ok, thanks a lot for your help!