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How do you compute

$$\int\frac 1x \, dx$$

without knowing its anti-derivative to start with?

Is there a way to do it by parts or substitution?

Michael Hardy
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Giulio Crisanti
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    Are you asking for a derivation of the fact that $$\int \frac{1}{x} dx = \log x + C?$$ – Edward Evans Apr 16 '16 at 01:31
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    Expand as a geometric series $\frac{1}{x} = \frac{1}{1-(1-x)} = 1 + (1-x) + (1-x)^2 + \cdots$. Integrating gives $(x-1) - \frac{1}{2} (x-1)^2 + \frac{1}{3} (x-1)^3 - \cdots$. This is a different way to write the antiderivative. And if you know your Taylor series, you may recognize this as the expansion of $\ln x$ around $x=1$. – Barry Smith Apr 16 '16 at 01:38
  • By parts or by substitution wouldn't work... I suspect anything that would involve an antiderivative would fail, because, well, at some point you'll need to anti-differentiate and get $\log x$. I was thinking you could start with a Riemann sum instead. This answer seems to hint that it could work... I haven't fully grokked the work there. – zahbaz Apr 16 '16 at 01:39
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    @BarrySmith I was thinking that, but it still assumes that you know $\frac{d}{dx}\ln x = \frac1x$ in a sense... – zahbaz Apr 16 '16 at 01:40
  • Not if you're happy with just leaving it as a Taylor series. – Barry Smith Apr 16 '16 at 01:40
  • You maybe could show that $F(x):=\frac{1}{h}\left(\int_1^{x+h} \frac{1}{t}, dt- \int_1^x \frac{1}{t}, dt\right) = \frac{1}{h}\left(\int_x^{x+h} \frac{1}{t}, dt\right)$ satisfies $F(xy) = F(x) + F(y)$ and then $F(e)=1$ to conlclude $F(x) = \ln(x)$. – Quinn Culver Apr 16 '16 at 01:41
  • There are several routes to defining $e^x$ and $\ln x$. Sometimes $\ln x$ is defined as the integral of $\frac 1x$. Sometimes $e^x$ is the Taylor series, sometimes it is $\lim_{n \to \infty}(1+\frac xn)^n$. I think there are others. You choose one definition, then the rest become theorems. You need to state which definition you are using so we know what we have to work with. Once these are proven, we often lose track of which one was the definition. – Ross Millikan Apr 16 '16 at 02:08
  • in my opinion, you'll have at some point to prove that $f(x) = \int_1^x \frac{dt}{t}$ is a logarithm i.e. $f(xy) = f(x)+f(y)$ hence there exists $a \in ]0,\infty[$ such that $a^{f(x)} = x$ for every $x \in ]0,\infty[$. then that $a = e$ is another story, and the usual construction is to show that the derivative of $f^{-1}(y) = a^y$ is itself : $(f^{-1}(y))' = \frac{1}{f'(f^{-1}(y))} = f^{-1}(y)$ since $f'(x) = \left(\int_1^x \frac{dt}{t}\right)' = \frac{1}{x}$. finally, show that there exists a number $e$ such that $(e^x)' = e^x$ and you are done : $ \int_1^x \frac{dt}{t} = \ln(x)$ – reuns Apr 16 '16 at 03:59

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Substitution could work: Set $x=e^t$ so that $dx=e^tdt$ then the integral becomes $\int1dt$ which is $t+C$ Backsub from $x=e^t$ gives $t=\ln x$. Ofcourse you would need to implement the absolute value since $x=e^t$ does not produce negative values for $x$, but the given function $y=1/x$ is of course well defined for $x<0$. My last statement may not be some standard of "rigor" but I don't know how much closer one can get wrt the OP's suggested integration techniques...

imranfat
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