Determine $\int^{\infty}_1 \frac1{x^2+x} \mathrm d x$

Question

Determine $$\int^{\infty}_1 \dfrac{1}{x^2+x} \mathrm d x$$

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My work

We can use partial fraction decomposition to find that $\dfrac{1}{x(x+1)} = \dfrac{1}{x} - \dfrac{1}{x+1}$.

$\therefore \lim_{x \to \infty} \int_1^x \dfrac{1}{x^2+x} dx = \lim_{x \to \infty} \int_1^x \dfrac{1}{x} - \dfrac{1}{x+1} dx$

$= \lim_{x \to \infty} (\ln(|x|) - 0) - \lim_{x \to 0} (\ln(|x + 1|) + \ln(2))$

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My understanding is that, since $\lim_{x \to \infty} \ln(|x|) = \infty$ and $\lim_{x \to \infty} \ln(|x + 1|) = \infty$, the integral should be divergent. However, the solutions suggest that the integral converges to $\ln(2)$?

I would greatly appreciate it if people could please take the time to explain where I am making an error.

Answer 1

Actually, it is easy to see that the integral must be convergent, since $1/x^2>1/(x^2+x)$ and $\int_1^{\infty} dx/x^2 = 1$. But one must be careful to only work with limits that exist: "$\infty-\infty$" is meaningless.

This is rather like the integral version of a telescoping series. Let's look at $$ \int_1^{M} \frac{dx}{x^2+x} = \int_1^{M} \frac{dx}{x} - \int_1^{M} \frac{dx}{1+x}. $$ Changing variables in the second gives $$ \int_1^{M} \frac{dx}{1+x} = \int_2^{M+1} \frac{dx}{x}, $$ so in fact, $$ \int_1^{M} \frac{dx}{x^2+x} = \int_1^2 \frac{dx}{x} - \int_M^{M+1} \frac{dx}{x}. $$ The first of these is of course $\log{2}$. The second tends to zero, since the integral is bounded by the maximum of the function multiplied by the length of the interval of integration, so in particular it is bounded by $1/M \to 0$ as $M \to \infty$. Hence the integral converges to $\log{2}$ as $M \to \infty$.

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The same obviously works for $f(x)-f(x+a)$ for any $f$ that converges to $0$ as $x \to \infty$.

Answer 2

$$\int_1^n\dfrac1{x^2+x}dx=\ln\left|\dfrac x{x+1}\right|_1^n=\ln\dfrac n{n+1}-\ln(2^{-1})$$

Now $\lim_{n\to\infty}\dfrac n{n+1}=\lim_{n\to\infty}\frac1{1+\dfrac1n}=?$

Alternatively, $$\int_1^\infty\dfrac1{x^2+x}dx=\int_1^\infty\dfrac4{(2x+1)^2-1^2}dx$$

Set $2x+1=\sec u$

Answer 3

Don't put $x$ as a limit of integration if $x$ is also the variable of integration.

Do this instead:

\begin{align*} \int_1^{+\infty} \frac{dx}{x^2 + x} &= \lim_{A \to +\infty} \int_1^A \frac{dx}{x^2+x}\\[0.3cm] &= \lim_{A \to +\infty}\left(\int_1^A \frac{dx}x - \int_1^A\frac{dx}{x+1}\right)\\[0.3cm] &= \lim_{A \to +\infty} \left(\ln|x|\bigg|_1^A - \ln|x+1|\bigg|_1^A\right)\\[0.3cm] &= \lim_{A\to+\infty} \left[\left(\ln |A| - \ln 1\right) - \left(\ln |A+1| - \ln 2\right)\right]\\[0.3cm] &= \lim_{A \to +\infty} \left(\ln| A| - \ln |A+1|\right) + \ln2\\[0.3cm] &= \lim_{A \to +\infty} \ln\left|\frac A{A+1}\right| + \ln2\\[0.3cm] &= \ln\left|\lim_{A \to +\infty} \frac A{A+1}\right| + \ln2\\[0.3cm] &= \ln 1 + \ln 2\\[0.3cm] &= \ln 2 \end{align*}

Answer 4

The mistake you are making is separating the terms of the integral. While it is true that the integrals of $x\mapsto{1\over x}$ and $x\mapsto{1\over x+1}$ are divergent on $[1, +\infty[$, their difference is not. As you noticed :

$${1\over{x(x+1)}} = {1\over x} - {1\over x+1}$$ of which a primitive is $x\mapsto \ln{x} - \ln{(x+1)}$ on $\Bbb R_+^*$, and this function has a limit at $+\infty$ (which is 0) so your integral converges.