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.

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- Jul 18th 2008, 09:18 PMrman144Converges 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.

- Jul 18th 2008, 09:24 PMMathstud28
- Jul 18th 2008, 09:28 PMrman144Reply
No, I'm saying that "x" can be any value greater than one and less than positive infinity

- Jul 18th 2008, 09:31 PMMathstud28
Yeah, Did you notice that I changed the variables, it won't change the result.

You want to show that

$\displaystyle \int_1^{\infty}\bigg[\arctan\left(\frac{x}{n+\frac{1}{2}}\right)-\frac{x}{n}\bigg]~dn<\infty\quad\forall{x}>1$ - Jul 18th 2008, 09:33 PMrman144Reply
Sorry, I'm kind of new to this stuff. Yeah, you are correct. Still, how can I prove it diverges/converges?

- Jul 18th 2008, 09:58 PMMathstud28
Lets look at the integral

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

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

Now we can see that the first and last terms are

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

Now let us consider

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

If we use integration by parts once we get

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

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

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

and

$\displaystyle dx=dz$

So this would gives us

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

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

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

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

Now making the appropriate backsubs we get

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

Putting this all back together we get

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

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

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

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

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

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

(Whew)

Can you go from there? - Jul 18th 2008, 10:03 PMrman144Reply
Yeah, thanks a lot. This site saves lives

- Jul 18th 2008, 10:06 PMMathstud28
- Jul 18th 2008, 10:14 PMrman144Reply
Thanks, already done.

- Jul 19th 2008, 12:08 AMrman144Reply
I know that this is beating a dead horse, but I seem to still have a problem. The integral test keeps telling me the series converges, but the graph of the equation is basically telling me that it diverges. Am I just messing myself up on this one, or did I make a mistake with the integral test?

- Jul 19th 2008, 12:10 AMMathstud28