A question from the Calculus book that I'm self-studying is asking me to determine the value of $n$ for which the improper integral below is convergent:
$\int_1^{+\infty}\left( \frac{n}{x+1} - \frac{3x}{2x^2 + n} \right ) dx$
My attempt is below:
Using the definition of improper integral:
$\lim_{b\to+\infty} \int_1^{b}\left( \frac{n}{x+1} - \frac{3x}{2x^2 + n} \right ) dx$
I will first solve the indefinite integral:
$\int\left( \frac{n}{x+1} - \frac{3x}{2x^2+n} \right )dx = \int \frac{n}{x+1}dx-\int \frac{3x}{2x^2+n}dx$
For the first integral, $\int \frac{n}{x+1}dx=n\ln (x+1) + \text{constant}$. For the second integral, substituting $u = 2x^2+n$ (and therefore $dx=\frac{du}{4x}$), $\int \frac{3x}{2x^2+n}dx=\frac{3}{4}\int \frac{1}{u}du=\frac{3}{4}\ln(2x^2+n) + \text{constant}$. So, the result for the whole integral is:
$\int\left( \frac{n}{x+1} - \frac{3x}{2x^2+n} \right )dx = n\ln(x+1) - \frac{3}{4}\ln(2x^2+n) + \text{constant} = \ln\left(\frac{(x+1)^n}{(2x^2+n)^{3/4}}\right) + \text{constant}$
Thus, the value of the definite integral from $x = 1$ to $x = b$ is:
$\ln\left(\frac{(b+1)^n}{(2b^2+n)^{3/4}}\right)-\ln\left(\frac{(2)^n}{(2+n)^{3/4}}\right)$
Thus, the value of the improper integral is:
$\lim_{b\to+\infty} \left[\ln\left(\frac{(b+1)^n}{(2b^2+n)^{3/4}}\right)-\ln\left(\frac{2^n}{(2+n)^{3/4}}\right)\right]$
For this limit to exist, it only depends on the limit of the term $\ln\left(\dfrac{(b+1)^n}{(2b^2+n)^{3/4}}\right)$, because the term $\ln\left(\frac{2^n}{(2+n)^{3/4}}\right)$ is a constant. To calculate the limit of $\ln\left(\dfrac{(b+1)^n}{(2b^2+n)^{3/4}}\right)$ as $b\to +\infty$, I suppose we can neglect the terms $1$ in $(b+1)^n$ and $n$ in $(2b^2+n)^{3/4}$, because they become negligible as $b$ gets larger. Thus we get, for the limit of this term:
$\lim_{b\to+\infty} \ln\left(\frac{(b)^n}{(2b^2)^{3/4}}\right)$
For the above limit to exist, it seems that $n = 3/2$ is the only possible value, because $\lim_{b\to+\infty} \ln\left(\dfrac{(b)^{3/2}}{(2b^2)^{3/4}}\right) = \lim_{b\to+\infty} \ln\left(\dfrac{b^{3/2}}{2^{3/4}b^{3/2}}\right) = \lim_{b\to+\infty} \ln\left(\dfrac{1}{2^{3/4}}\right)$. If any different value for $n$ were chosen, the logarithm would approach $-\infty$ or $+\infty$, and therefore the limit would not exist.
For $n = 3/2$, the value of the limit is:
$\lim_{b\to+\infty} \left[\ln\left(\frac{(b)^{3/2}}{(2b^2)^{3/4}}\right)-\ln\left(\frac{2^{3/2}}{(2+3/2)^{3/4}}\right)\right]=\lim_{b\to+\infty} \left[\ln\left(\frac{1}{2^{3/4}}\right)-\ln\left(\frac{2^{3/2}}{(7/2)^{3/4}}\right)\right]=\frac{3}{4}\ln\frac{7}{16}$
This is equal to the answer given by the book. So, this informal argument of neglecting the terms $1$ and $n$ worked.
But I would like to ask if the reasoning above is correct, and if this informal way of finding the value of $n$ is good. Is there a more formal way?
Edit: A more formal way of showing that $n$ should be $3/2$ is suggested by David Mitra in the comments to this question below:
$\lim_{b\to+\infty} \ln \left( {(b+1)^n\over (2b^2+n)^{3/4} } \right) = \lim_{b\to+\infty} \ln \left( {b^n\over b^{3/2}} \cdot {\bigl(1+{1\over b}\bigr)^n\over \bigl(2 +{n\over b^2}\bigr)^{3/4}} \right)$
It is clear from above that $n$ should equal $3/2$. If it were a different value, the limit would not exist, because the expression inside the logarithm would approach either $0$ (causing the logarithm to approach $-\infty$) or $+\infty$ (causing the logarithm to tend to $+\infty$).