# Thread: Finding a value for C, such that the integral Converges.

$\displaystyle int(x/(x^2+1)-C/(3*x+1), x = 0 .. infinity)$

Please help me here. I thought that I would split the integral up into the two separate fractions and found that the first fraction diverged. I am unsure as to how I would go about finding the value of C.

2. Originally Posted by nyhc_is_br00tal
$\displaystyle int(x/(x^2+1)-C/(3*x+1), x = 0 .. infinity)$

Please help me here. I thought that I would split the integral up into the two separate fractions and found that the first fraction diverged. I am unsure as to how I would go about finding the value of C.
So the question is find value $\displaystyle C$ such that $\displaystyle \int\limits_0^\infty\left(\frac{x}{x^2+1}-\frac{C}{3x+1}\right)dx$ converges. Let's start to evaluate this improper integral as following:

$\displaystyle \int\limits_0^\infty\left(\frac{x}{x^2+1}-\frac{C}{3x+1}\right)dx$

$\displaystyle =\lim\limits_{b\rightarrow \infty} \int\limits_0^b\left(\frac{x}{x^2+1}-\frac{C}{3x+1}\right)dx$

$\displaystyle =\lim\limits_{b\rightarrow \infty}\left[\int\limits_0^b\left(\frac{x}{x^2+1}\right)dx-\int\limits_0^b\left(\frac{C}{3x+1}\right)dx\right]$

$\displaystyle =\lim\limits_{b\rightarrow \infty}\left[\frac{1}{2}\ln(b^2+1)-\frac{C}{3}\ln(3b+1)\right]$

$\displaystyle =\lim\limits_{b\rightarrow \infty}\left[\ln\frac{\sqrt{b^2+1}}{(3b+1)^\frac{C}{3}}\right]$

Now consider the expression $\displaystyle \frac{\sqrt{b^2+1}}{(3b+1)^\frac{C}{3}}$, since the highest power of $\displaystyle b$ in the numerator is 1 (note that $\displaystyle \sqrt{b^2}=b$), we need the highest power of $\displaystyle b$ in the denominator also to be 1 in order for the limit to converge. Otherwise, as $\displaystyle b\rightarrow\infty$, the expression $\displaystyle \frac{\sqrt{b^2+1}}{(3b+1)^\frac{C}{3}}$ will either goto $\displaystyle 0$ or $\displaystyle \infty$, in either cases, the limit diverges. Hence when $\displaystyle C=3$, the given integral converges.

Roy