# Value of $n$ for which an improper integral is convergent.

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 $+\infty$).

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To make it more formal, you could consider ${(b+1)^n\over (2b^2+n)^{3/4} } = {b^n\over b^{3/2}} \cdot {\bigl(1+{1\over b}\bigr)^n\over \bigl(2 +{n\over b^2}\bigr)^{3/4}}$. –  David Mitra Jul 5 '12 at 23:29
@DavidMitra, thank you for this suggestion. I edited the original post to include it. –  anonymous Jul 6 '12 at 11:58

Your method is correct and it is easy to formalize your neglecting of minor terms.

We can see that $$\ln\left(\dfrac{(b+1)^n}{(2b^2+n)^{3/4}}\right) - \ln\left( \frac{b^n}{(2b^2)^{3/4}}\right)= \ln \left( (1+1/b)^n (1+n/(2b^2))^{-3/4} \right) \to \ln 1=0$$

as $b\to \infty.$ So the limit of the first term equals the limit of the second term if either limit exists.

However, this is probably not the best way to do this problem. The point of these types of exercises is to get you used to estimating the value of an integral, by doing something to get a feel for the growth of the integrand. Often we can't explicitly integrate the term.

Here, if you put the terms on a common denominator you get $$\frac{n^2 + 2nx^2 - 3x(x+1)}{(x+1)(2x^2+n)}.$$

The critical fact used often is: $\displaystyle \int^{\infty}_1 \frac{1}{x^a} dx$ converges only for $a>1.$

So the $\frac{n^2}{(x+1)(2x^2+n)}$ part clearly converges. So we only need to worry about $$\frac{2nx^2 - 3x(x+1) }{(x+1)(2x^2+n)}.$$ If $n=3/2$ then $2nx^2=3x^2$ is able to cancel with the $-3x^2$ term in the next term, so we have a linear term in the top, cubic denominator, so behaviour like $1/x^2$ which is convergent. But if not, then we always get some quadratic term in the numerator, so the behaviour is like some constant term times $1/x$, which diverges. It is instructive to formalize these ideas with some inequalities, which I will leave you to try.

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