Antiderivative of $\log(x)$ without Parts

Question

I understand how the antiderivative of $\log(x)$ can be obtained by Integration by Parts (i.e. product rule), but I was wondering how-if at all- it could be obtained only using sum/difference rule and substitution/chain rule.

Answer 1

$$\int_1^t \ln(x)\,dx = \int_1^t\int_1^x\frac 1u\,du\,dx = \int_1^t\int_u^t\frac 1u\,dx\,du = \int_1^t\frac{t-u}{u}\,du = {\large[}t\ln(u)-u{\large]}_{1}^t$$

Answer 2

There's a fun formula relating the integral of an invertible function and the integral of its inverse, namely $$ bf(b)-af(a)=\int_a^bf(t)dt+\int_{f(a)}^{f(b)}f^{-1}(t)dt $$ Which is easily seen from a picture. Then setting $b=x$, $0<x<a$ and $f(t)=\log t$ gives $$ x\log x-a\log a=\int_a^x\log tdt+\int_{\log a}^{\log x}e^tdt\\ \Rightarrow x\log x-a\log a-\int_{\log a}^{\log x}e^tdt=\int_a^x\log tdt\\ \Rightarrow x\log x-x+a-a\log a=\int_a^x\log tdt\\ \Rightarrow x\log x-x+c=\int_a^x\log tdt $$

Answer 3

We have:

$$\int_0^{t} x^a dx = \frac{t^{a+1}}{a+1}$$

Differentiating w.r.t. $a$ gives:

$$\int_0^{t} x^a \log(x)dx = \frac{t^{a+1}\left[(a+1)\log(t)-1\right]}{(a+1)^2}$$

The result then follows if we put $a = 0$.

Answer 4

Fermat's Method of Exhaustion can be used to obliterate this problem from first principles.

However, we will require the evaluation of a limit, which will be done so in advance.

Preliminary:

$\displaystyle \lim_{r \to 1} \frac{r \log{r}}{1-r} = -1$

Proof:

Consider the elementary inequality which holds true for all $r>0$:

$\displaystyle r-1 \le \log{r} \le 1-\frac{1}{r}$

The result follows by the squeeze theorem.

Consider a real number $0<r<1$, which will be used to construct the infinite geometric sequence of intervals.

Construct the lines $x = b, x = br, x = br^2, x = br^3, \cdots$ all the way to infinity.

The region bounded by $\log x, x = br^n, x = br^{n+1}$ is bounded strictly from below and above by the following inequality:

$\displaystyle b(1-r)r^{n+1} (\log{b}+(n+1)\log{r}) < \int_{br^{n+1}}^{br^n} \log{x}\, \text{d}x < b(1-r)r^n (\log{b}+n\log{r})$

Summing up these inequalities from $n=0$ to $\infty$ will give us strict upper and lower bounds on the integral.

After some elementary Arithmetico-Geometric Summation, we have the following:

$\displaystyle r \, b \, \log{b} + b\,\frac{r \log{r}}{1-r} < \int_0^b \log{x} \, \text{d}x < b\log{b} + b\,\frac{r \log{r}}{1-r}$

Take the limit as $r$ approaches $1$, and by the squeeze theorem, it can then be concluded:

$\displaystyle \int_0^b \log{x} \, \text{d}x = b\log{b} - b$