# Problem of the Week #60 - May 20th, 2013

Status
Not open for further replies.

#### Chris L T521

##### Well-known member
Staff member
Thanks again to those who participated in last week's POTW! Here's this week's problem!

-----

Problem: Find the value of the constant $C$ for which the integral
$\int_0^{\infty}\left(\frac{1}{\sqrt{x^2+4}}-\frac{C}{x+2}\right)\,dx$
converges. Then evaluate the integral for this value of $C$.

-----

#### Chris L T521

##### Well-known member
Staff member
This week's question was correctly answered by MarkFL (Sudharaka gets honorable mention since he pretty much had it but the conclusion on the value of C was incorrect).

You can find MarkFL's solution below:

I would choose to write the improper integral as:

$\displaystyle L=\lim_{t\to\infty}\left(\int_0^t \frac{1}{\sqrt{x^2+4}}- \frac{C}{x+2}\,dx \right)$

Applying the anti-derivative form of the FTOC, we may write

$\displaystyle L=\lim_{t\to\infty}\left(\left(\ln(t+\sqrt{t^2+4})-C\ln(t+2) \right)-\left(\ln(2)-C\ln(2) \right) \right)$

Applying logarithmic properties, there results:

$\displaystyle L=\lim_{t\to\infty}\left(\ln\left(\frac{2^C(t+ \sqrt{t^2+4})}{2(t+2)^C} \right) \right)$

In order for $L$ to exist, we see that we require the denominator to be of degree 1, hence $C=1$:

$\displaystyle L=\ln\left(\lim_{t\to\infty}\left(\frac{1+ \sqrt{1+\frac{4}{t^2}}}{1+\frac{2}{t}} \right) \right)$

$\displaystyle L=\ln(2)$

Status
Not open for further replies.