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

1. PLEASE HELP!!! Finding a value for C, such that the integral Converges.

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

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

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 $C$ such that $\int\limits_0^\infty\left(\frac{x}{x^2+1}-\frac{C}{3x+1}\right)dx$ converges. Let's start to evaluate this improper integral as following:

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

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

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

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

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

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

Roy