Why does integration by parts yield $0=1$ here?

Question

I tried doing integration by parts on this integral: $$I = \int \frac{1}{x\log{x}}dx$$ With these substitutions: $$u = \frac{1}{\log{x}} \rightarrow du = -\frac{1}{x\log^2{x}}$$ $$dv = \frac{1}{x}dx \rightarrow v=\log{x}$$

We get: $$\begin{align} I &= uv - \int vdu \\ I &= \log{x}\frac{1}{\log{x}} - \int \log{x}(-\frac{1}{x\log^2x})dx \\ I &= 1 + \int \frac{1}{x\log{x}}dx\\ I &= 1 + I\\ 0 &= 1 \end{align} $$ How does this happen?

This is one of the reasons that we tell people to put "$+C$"s everywhere. — Chappers, Sep 18 '17 at 14:57
Your substitution with $u= \frac 1{log(x)} $works fine you evalutae $I=-\int \frac {du} u $ no need here for integration by part — user577215664, Sep 18 '17 at 14:58

score 7 · Accepted Answer · answered Sep 18 '17 at 14:46

When you compare indefinite integrals you should expect that they can differ by a constant term -- the constant of integration.

In particular, when you subtract "the same" indefinite integral in the last line, there should be a $C_2-C_1$ left on one of the sides.

score 2 · Answer 2 · answered Sep 18 '17 at 14:53

2

Note: $$\int \frac{1}{x\ln{x}}dx=\int \frac{1}{\ln{x}}d(\ln{x})=\ln{\ln{x}}+C_1=\ln{\ln{x}}+C_2+1.$$

answered Sep 18 '17 at 14:53

farruhota

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Why does integration by parts yield $0=1$ here?

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