MHB What is the value of the constant C for the converging integral?

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The discussion centers on determining the value of the constant C that ensures the convergence of the integral \[\int_0^{\infty}\left(\frac{1}{\sqrt{x^2+4}}-\frac{C}{x+2}\right)\,dx\]. MarkFL provided the correct solution, while Sudharaka received an honorable mention for nearly solving the problem but miscalculating the value of C. The integral's convergence is crucial for proper evaluation, and the community engaged in analyzing the conditions required for this. Ultimately, the focus is on finding the appropriate C and evaluating the integral based on that value.
Chris L T521
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Thanks again to those who participated in last week's POTW! Here's this week's problem!

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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 $C$.

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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 $C=1$:

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

$\displaystyle L=\ln(2)$
 

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