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I think I'm having a bad day. I was just trying to use integration by parts on the following integral: $$\int \frac{dx}{x\log(x)}$$ Which yields $$\int \frac{dx}{x\log(x)} = 1 + \int \frac{dx}{x\log(x)}$$ Now, if I were to subtract $$\int \frac{dx}{x\log(x)}$$ from both sides, it would seem that $0 = 1$. What is going on here?

Aside: I do know the integral evaluates to $$\log(\log(x))$$ plus a constant.

Michael Hardy
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providence
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    You're missing a constant n the expression immediately after the "I know the integral valuates to...". This solves your conundrum – DonAntonio Mar 20 '13 at 10:16
  • Explanation is in Yahoo Answers about 3 years ago. Search "integral paradox" – GEdgar Mar 20 '13 at 12:20

3 Answers3

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You do not show details, but an indefinite integral always involves an arbitrary constant (${}+C$). So the calculation quite likely is correct, and a contradiction has not been reached. A problem of a similar character is common when integrating trigonometric functions. One does the integral in two different ways, and gets two different answers. However, in such situations, the answers differ by a constant, so can be both correct.

However, unfortunately the result indicates that one has made no progress towards finding the answer. This sort of thing happens moderately often in attempted integrations by parts. After some calculations, one reaches the conclusion that the integral is equal to $\dots$ the integral.

Remark: I do not think integration by parts can play a useful role here. Let $u=\log x$. Then $du=\frac{dx}{x}$, so in one step we get to $\int\frac{du}{u}$. Thus our integral is $\log(|\log x|)+C$.

Michael Hardy
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André Nicolas
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  • I am not actually trying to evaluate the integral. I want to know why this perplexing result occurs when I proceed as above. – providence Mar 20 '13 at 10:16
  • @André Nicolas: do you mean it's actually okay to say $\int f(x) dx = \int f(x)dx +c$? It makes sense, I'm just clarifying. – harry Apr 11 '21 at 15:52
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In a general case we have $$\int\frac{f'(x)}{f(x)}dx=\log|f(x)|+C,$$ and in our case, take $f(x)=\log(x)$ so $f'(x)=\frac{1}{x}$ to find the desired result.

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Hint: $dx/x = d(\log{x})$ So substitute $u=\log{x}$ rather than integrate by parts.

Ron Gordon
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  • Yes, my trouble is not with evaluating the integral. I just want to know why, if I attempt to do it by parts, I seem to have $1 = x$ where $x \ne 1$. – providence Mar 20 '13 at 10:15
  • You should get something like $$1 + \int dx \log{x} (d/dx) (1/\log{x})$$ – Ron Gordon Mar 20 '13 at 10:20