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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.

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    $\begingroup$ To the best of my knowledge, avoiding integration by parts all together is not possible. Of course, one can convert your integral into some other integral through a u-sub, but then that integral is to be evaluated without any integration by parts. Doubtful, and I am really looking forward to any answer to make that work. $\endgroup$ – imranfat Sep 29 '16 at 2:16
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    $\begingroup$ What on earth is happening in the edit history of this question? $\endgroup$ – JiK Sep 29 '16 at 13:24
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    $\begingroup$ @JiK looks like a conflict between "don't contradict the OP" editors and "OP obviously added irrelevant crap / is trying to ask a second question, try to rescue it" editors. $\endgroup$ – hobbs Sep 29 '16 at 17:00
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$$\int_1^t \ln(x)\,dx = \int_1^t\int_1^x\frac 1u\,du\,dx = \int_1^t\int_u^t\frac 1u\,dx\,du = \int_1^t\frac{t-u}{u}\,du = {\large[}t\ln(u)-u{\large]}_{1}^t$$

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    $\begingroup$ I realize adding an extra dimension and using Fubini's theorem isn't exactly "only using sum/difference rule and substitution/chain rule", but I feel it's still in the spirit of the question. $\ddot \smile$ $\endgroup$ – user137731 Sep 29 '16 at 2:38
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    $\begingroup$ +1 this is beautifully simple. $\endgroup$ – Simply Beautiful Art Sep 29 '16 at 13:24
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    $\begingroup$ Silly question - why do the indices of the integration change in that particular way when the integration order is changed? $\endgroup$ – templatetypedef Sep 29 '16 at 16:37
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    $\begingroup$ @templatetypedef - graph the region of integration in (x,u)-space. You will see that both regions are the same triangle with corners $(1,1), (t,1), (t,t)$. $\endgroup$ – Paul Sinclair Sep 29 '16 at 17:55
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    $\begingroup$ @templatetypedef: Paul Sinclair gives a good geometric answer; stated more algebraically, it’s because the conditions $1 \leq x \leq t$, $1\leq u \leq x$ are equivalent to the conditions $1 \leq u \leq t$, $u \leq x \leq t$. $\endgroup$ – Peter LeFanu Lumsdaine Sep 29 '16 at 17:59
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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 $$ x\log x-a\log a=\int_a^x\log tdt+\int_{\log a}^{\log x}e^tdt\\ \Rightarrow x\log x-a\log a-\int_{\log a}^{\log x}e^tdt=\int_a^x\log tdt\\ \Rightarrow x\log x-x+a-a\log a=\int_a^x\log tdt\\ \Rightarrow x\log x-x+c=\int_a^x\log tdt $$

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  • $\begingroup$ Almost right, except I believe the LHS of your first equation should be $bf(b) - af(a)$. After all, we have $\int_1^x \log t\,dt = x \log x - x + 1$. $\endgroup$ – arkeet Sep 29 '16 at 3:28
  • $\begingroup$ @arkeet there, that should also be more general $\endgroup$ – qbert Sep 29 '16 at 3:39
  • $\begingroup$ Sure, although I think it would be a satisfactory answer to just have an answer for one $a$. $\endgroup$ – arkeet Sep 29 '16 at 3:46
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    $\begingroup$ I need to remember this... $\endgroup$ – Simply Beautiful Art Sep 29 '16 at 13:28
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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 $a = 0$.

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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:

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

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    $\begingroup$ "Fermat's Method of Geometric Riemann Summation" - a curious name, given that Fermat died in 1665 while Riemann was born in 1826. $\endgroup$ – Paul Sinclair Sep 29 '16 at 18:00
  • $\begingroup$ Nice profile picture (+1) $\endgroup$ – clathratus Sep 12 at 23:33
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This may not necessarily be what you're looking for, but I got a kick out of it so I figured I'd share. We may evaluate the antiderivative by using integration by parts indirectly.

We seek the integral $$I_f(x)=\int f^{-1}(x)dx.$$ We set $x=f(u)$ to get $$I_f(x)=\int uf'(u)du.$$ Integration by parts gives $$I_f(x)=uf(u)-\int f(u)du$$ which is $$I_f(x)=xf^{-1}(x)-F\circ f^{-1}(x)+C.$$ Choosing $f^{-1}(x)=\ln x,$ $$I_{\ln}(x)=x\ln x-x+C.$$

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