# Converges or Diverges?

• July 18th 2008, 09:18 PM
rman144
Converges or Diverges?
I need some help determining whether or not the equation in the attachment diverges or converges. I'm trying to teach myself the integral test, and cannot seem to get this one to work. Any help would be much appreciated.
• July 18th 2008, 09:24 PM
Mathstud28
Quote:

Originally Posted by rman144
I need some help determining whether or not the equation in the attachment diverges or converges. I'm trying to teach myself the integral test, and cannot seem to get this one to work. Any help would be much appreciated.

So in other words you are asking for

$\int_1^{\infty}\bigg[\arctan\left(\frac{a}{x+\frac{1}{2}}\right)-\frac{a}{x}\bigg]$?
• July 18th 2008, 09:28 PM
rman144
No, I'm saying that "x" can be any value greater than one and less than positive infinity
• July 18th 2008, 09:31 PM
Mathstud28
Quote:

Originally Posted by Mathstud28
So in other words you are asking for

$\int_1^{\infty}\bigg[\arctan\left(\frac{a}{x+\frac{1}{2}}\right)-\frac{a}{x}\bigg]$?

Quote:

Originally Posted by rman144
No, I'm saying that "x" can be any value greater than one and less than positive infinity

Yeah, Did you notice that I changed the variables, it won't change the result.

You want to show that

$\int_1^{\infty}\bigg[\arctan\left(\frac{x}{n+\frac{1}{2}}\right)-\frac{x}{n}\bigg]~dn<\infty\quad\forall{x}>1$
• July 18th 2008, 09:33 PM
rman144
Sorry, I'm kind of new to this stuff. Yeah, you are correct. Still, how can I prove it diverges/converges?
• July 18th 2008, 09:58 PM
Mathstud28
Quote:

Originally Posted by rman144
Sorry, I'm kind of new to this stuff. Yeah, you are correct. Still, how can I prove it diverges/converges?

Lets look at the integral

$\int\bigg[\arctan\left(\frac{a}{x+\frac{1}{2}}\right)-\frac{a}{x}\bigg]~dx$

$=\int\bigg[\frac{\pi}{2}-\arctan\left(\frac{x+\frac{1}{2}}{a}\right)-\frac{a}{x}\bigg]$

Now we can see that the first and last terms are

$\frac{\pi{x}}{2}$ and $a\ln|x|$ respectively.

Now let us consider

$\int\arctan\left(\frac{x+b}{a}\right)~dx$

If we use integration by parts once we get

$\int\arctan\left(\frac{x+b}{a}\right)~dx$

$=x\arctan\left(\frac{x+b}{a}\right)-a\int\frac{x}{(x+b)^2+a^2}$

Now if we let $z=x+b\Rightarrow{x=z-b}$

and

$dx=dz$

So this would gives us

$a\int\frac{z-b}{z^2+a^2}~dz$

$=a\int\frac{z}{z^2+a^2}~dz-ab\int\frac{dz}{z^2+a^2}$

Now by letting $z^2+a^2=u$ and $z=a\tan(\theta)$ respectively we come up with

$\frac{a}{2}\ln(z^2+a^2)-b\arctan\left(\frac{z}{a}\right)$

Now making the appropriate backsubs we get

$\frac{a}{2}\ln\left((x+b)^2+a^2\right)-b\arctan\left(\frac{x+b}{a}\right)$

Putting this all back together we get

$\int\arctan\left(\frac{x+b}{a}\right)$

$=x\arctan\left(\frac{x+b}{a}\right)-\frac{a}{2}\ln\left((x+b)^2+a^2\right)+b\arctan\le ft(\frac{x+b}{a}\right)$

$=\left(x+b\right)\arctan\left(\frac{x+b}{a}\right)-\frac{a}{2}\ln\left((x+b)^2+a^2\right)+C$

So putting this completely back together and remembering that $b=\frac{1}{2}$ we get

$\int\bigg[\arctan\left(\frac{a}{x+\frac{1}{2}}\right)-\frac{a}{x}\bigg]~dx$

$=\frac{\pi{x}}{2}-\left(x+\frac{1}{2}\right)\arctan\left(\frac{x+\fr ac{1}{2}}{a}\right)+\frac{a}{2}\ln\left(\left(x+\f rac{1}{2}\right)^2+a^2\right)-a\ln|x|$

(Whew)

Can you go from there?
• July 18th 2008, 10:03 PM
rman144
Yeah, thanks a lot. This site saves lives
• July 18th 2008, 10:06 PM
Mathstud28
Quote:

Originally Posted by rman144
Yeah, thanks a lot. This site saves lives

It's not the site, its the people that volunteer their time here. (Nod)

EDIT: You also might want to consider combinging the last two natural logs so that it is more obvious as the integrals behavior as x tends toward infinity.
• July 18th 2008, 10:14 PM
rman144