Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is this a sufficient proof for this integral?:

$$\int \frac{dx}{x} = \ln |x| + \mathcal{C}$$

Let $$x = e^{u} : $$

$$\int \frac{dx}{x} = \int du = u + \mathcal{C} = \ln |x| + \mathcal{C}$$

I'm not sure :S I don't know if my logic's a bit wishy washy/ circular with this :S And when I've looked, most websites seem to say it's a definition rather than a result but...I guess I just want a proof for this. Anyone have any ideas/ can validate whether the above is correct?

Thanks very much

share|improve this question
1  
I take it you are trying to prove $\int(dx/x)=\log x+C$. But this follows from the derivative of $\log x$ being $1/x$ --- are you not allowed to use that? –  Gerry Myerson Jun 11 '13 at 13:06
1  
Yes, sorry, that's what I was trying to prove yes. DOH! So if I can prove the derivative of ln x is 1/x then it follows that the anti-derivative of 1/x is lnx + c? I think I remember something like that...the uniqueness theorem of anti-derivatives or something? –  user78416 Jun 11 '13 at 13:07
    
your proof above seems logical enough though...i mean at least i don't see why it wouldn't be..hmm –  mathguy Jun 11 '13 at 13:09
    
Your proof is right, there is a small mistake though. The substitution $x=e^u$ only works for positive $x$. You need to treat the case $x$ negative separately. –  N. S. Jun 11 '13 at 13:15
    
The Fundamental Theorem Of Calculus (alternatively, the definition of $\int f(x)\,dx$). –  Gerry Myerson Jun 12 '13 at 9:03

3 Answers 3

Your proof is valid but you can add more precision:

  • if $x\in(0,+\infty)$ we pose $x=e^u$ and we find $$\int \frac{dx}{x} = \ln x+ \mathcal{C}$$
  • if $x\in(-\infty,0)$ we pose $x=-e^u$ and we find $$\int \frac{dx}{x} = \ln (-x) + \mathcal{C}$$ then we conclude.
share|improve this answer
    
Ah, excellent. Thanks very much! –  user78416 Jun 11 '13 at 13:51
    
You're welcome. –  Sami Ben Romdhane Jun 11 '13 at 13:52

Looks fine to me. By Taylor's Theorem, you can also write

$$\frac{1}{x} = \sum_k (-1)^k (x-1)^k$$

and integrate in the radius of convergence term by term, recognizing the result on the right-hand side. Assuming $x > 0$,

$$\begin{split} \int \frac{dx}{x} &= \int \sum_{k=0}^\infty (-1)^k (x-1)^k dx \\ &= \sum_{k=0}^\infty (-1)^k \int (x-1)^k dx \\ &= \sum_{k=0}^\infty \frac{(-1)^k}{k+1} (x-1)^{k+1} + C \\ &= \ln x +C \end{split} $$

share|improve this answer
    
That's an interesting approach! Thanks –  user78416 Jun 11 '13 at 13:51
1  
@user78416 the problem is that to "recognize" the series for the $\ln x$ in the last step, you need to know that $(\ln x)' = 1/x$ which is the crux of your question :). But it does approach it from a slightly different side... –  gt6989b Jun 11 '13 at 13:54
    
@gt6989b, the absolute value is missing in $\ln(x)$ above. –  Amad27 Oct 9 at 13:29

You solved the equation correctly (with the small negligence for negative x, which has been corrected by Sami).

If you want to (or are allowed to) use

$$ \frac{\partial}{\partial x} ln(x) = \frac{1}{x}, $$

then you might be interested in this proof.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.