How to show $\int_0^1 \left( \frac{1}{\ln(1+x)} - \frac{1}{x} \right) \,dx$ converges?

Question

I need to show that $$\int_0^1 \left( \frac{1}{\ln(1+x)} - \frac{1}{x} \right) \,dx$$ converges, given that $$\lim_{x\rightarrow0^+} \left( \frac{1}{\ln(1+x)} - \frac{1}{x} \right) = \frac{1}{2}$$

I'm not sure how to do this because my text gave a very brief treatment on applying the limit comparison test for the improper integral of the second kind. Here is what I've tried:

I define the test function to be $g(t)=1$. Therefore,

$$\frac{\left( \frac{1}{\ln(1+x)} - \frac{1}{x} \right)}{g(t)} = \left( \frac{1}{\ln(1+x)} - \frac{1}{x} \right) \tag{1}$$

So, $(1)$ tends to $\frac{1}{2}$ as $t$ tends to $0^+$ and $$\int_0^1 g(t) \,dt$$ is convergent. Therefore, by the limit comparison test, the given integral is convergent.

Did I do the test correctly?

EDIT

My text's solution (which I find to be vague) is:

What does 'repair' the integrand mean?

Answer 1

I'll give a more general answer, which should explain what “repairing the function” means.

Suppose $f$ is a continuous function defined on the interval $(a,b]$ (where $a<b$) and that $$ \lim_{x\to a^+}f(x)=r $$ is finite.

Define $g(x)=f(x)$ for $a<x\le b$ and $g(a)=r$. Then $g$ is a continuous function on $[a,b]$. Set \begin{align} F(x)&=\int_{x}^b f(t)\,dt &&\text{for $x\in(a,b]$}\\ G(x)&=\int_{x}^b g(t)\,dt &&\text{for $x\in[a,b]$} \end{align} The two functions $F$ and $G$ are continuous (even differentiable, by the fundamental theorem of calculus) and coincide on $(a,b]$, so $$ \lim_{x\to a^+}F(x)=\lim_{x\to a^+}G(x)=G(a) $$

Answer 2

For first, $\frac{1}{\log(1+x)}-\frac{1}{x}$ is continuous on $\mathbb{R}^+$. Now:

$$ \lim_{x\to 0^+}\left(\frac{1}{\log(1+x)}-\frac{1}{x}\right) = \lim_{x\to 0^+}\frac{x-\log(1+x)}{x\log(1+x)} = \lim_{x\to 0^+}\frac{\frac{x^2}{2}+O(x^3)}{x^2+O(x^3)}=\frac{1}{2}$$ hence $f(x)$ is continuous and bounded over $(0,1)$, and that implies that $f(x)$ is integrable.

Answer 3

I do not know if this is the kind of answer you expect; so, forgive me if I am off-topic.

Consider Taylor series, built at $x=0$, of $$\frac{1}{\log(1+x)}-\frac{1}{x} =\frac{x-\log(1+x)}{x\log(1+x)}$$ For the numerator, you have $$\frac{x^2}{2}-\frac{x^3}{3}+\frac{x^4}{4}-\frac{x^5}{5}+O\left(x^6\right)$$ and for the denominator $$x^2-\frac{x^3}{2}+\frac{x^4}{3}-\frac{x^5}{4}+O\left(x^6\right)$$ Performing the long division, you then have $$\frac{1}{\log(1+x)}-\frac{1}{x} =\frac{1}{2}-\frac{x}{12}+\frac{x^2}{24}-\frac{19 x^3}{720}+\frac{3 x^4}{160}-\frac{863 x^5}{60480}+O\left(x^6\right)$$ Integrating from $0$ to $1$, the approximation is $\frac{847319}{1814400}\approx 0.466997$ while the exact solution would be $\text{li}(2)-\gamma\approx 0.467948$.

Answer 4

We have $$\int_{0}^{1}\left(\frac{1}{\log\left(1+x\right)}-\frac{1}{x}\right)dx=\lim_{\epsilon\rightarrow0^{+}}\int_{\epsilon}^{1}\left(\frac{1}{\log\left(1+x\right)}-\frac{1}{x}\right)dx$$ now $\frac{1}{\log\left(1+x\right)}-\frac{1}{x}$ is an decreasing function over $\left(\epsilon,1\right)$, so $$\lim_{\epsilon\rightarrow0^{+}}\int_{\epsilon}^{1}\left(\frac{1}{\log\left(1+x\right)}-\frac{1}{x}\right)dx\leq\lim_{\epsilon\rightarrow0^{+}}\max_{x\in\left[\epsilon,1\right]}\left(\frac{1}{\log\left(1+x\right)}-\frac{1}{x}\right)\left(1-\epsilon\right)\leq\lim_{\epsilon\rightarrow0^{+}}\left(\frac{1}{\log\left(1+\epsilon\right)}-\frac{1}{\epsilon}\right)=\frac{1}{2}.$$

Answer 5

I think "repair" means this. Define function $$ f(x) = \begin{cases} \displaystyle\frac{1}{\ln(1+x)} - \frac{1}{x},\qquad x>0 \\ \displaystyle\frac{1}{2},\qquad\qquad x=0 \end{cases} $$ Then $f$ is continuous on $[0,1]$, so it is integrable.