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Let's consider an indefinite integral

$$\int \frac{dx}{x\ln x}$$

It can be easily calculated to be $\ln(\ln x)+C$, e.g. via substitution $\ln x=t$ or directly from $\int\frac{f'(x)}{f(x)}dx=\ln|f(x)|+C$. So far so good.

But when integration by parts is employed: $u'=\frac{1}{x}$, $v=\frac{1}{\ln x}$, one gets

$$\int \frac{dx}{x\ln x}=1+\int \frac{dx}{x\ln x}$$

from which $0=1$. Even if we plug an arbitrary constant of integration in the r.h.s. of the last equality, we'll just get that $C$ should be $-1$ for the equality to be an identity (but in general the constant of integration can be put in at the very last step of integration; like here and here), and we'll still know nothing about the integral.

So, my question is: why does integration by parts fail in this case? Are there some assumptions not fulfilled that I have overseen here? An explanation that "it doesn't work so one has to use different methods" is no explanation.

Community
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corey979
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1 Answers1

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The problem is that you are not specifying specific boundaries of integration when integrating by parts. The constant $C$ on the left and right will not turn out to be the same when you do specify boundaries of integration.

Indefinite integrals as written do not specify a single function $f(x)$, rather they specify a whole family of functions of the form $\{f(x)+C: C \in \mathbb{R}\}$. Given any such function, $f(x)+C$, $f(x)+ C +1 = f(x) + (C+1) = f(x) + C'$ is also in the set, and thus is also equal to the "indefinite integral". That's why they are called "indefinite", because they don't specify any specific function.

Chill2Macht
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  • Of course $f(x)+C$ is an answer as good as $f(x)+C'$, but why does the integration by parts fail to obtain $f(x)=\ln(\ln x)$ in the above case? E.g., integrating $\sin^2x$ by parts allows to get a useful answer even though in the mid-step also the same integral as the starting one appears on the r.h.s. – corey979 Sep 11 '16 at 10:58
  • As you've written it, the expression $\int \frac{dx}{x \ln x}$ only obtains $f(x) = \ln(\ln x) + C$ with $C$ undetermined because no initial conditions are specified. If you specify the boundaries of integration, i.e. use a definite integral, it is easier to see the equality, because for any $f_1, f_2$ in the family $f_1(b) -f_1(a) = f_2(b) - f_2(a)$ because $C-C=C'-C'=0$. Your question is essentially like asking why is $g(x)=x+1$ a bijection from $\mathbb{R}$ to $\mathbb{R}$, except that instead of x a point in $\mathbb{R}$, now $x\in { \ln(\ln x) + C : C \in \mathbb{R} }$. – Chill2Macht Sep 11 '16 at 17:44
  • @corey979 The point being: indefinite integration is NOT a map from functions to functions -- it is a map from a function to a FAMILY of functions. Only DEFINITE integration, or indefinite integration with boundary conditions specifying the value of $C$, is a map from functions to functions. Your questions are based on false premises, so I can't give you an answer you will find satisfying until you realize how these premises are false. In the equation you wrote, any realizable boundary conditions for the LHS and the RHS would obviously have to be different. – Chill2Macht Sep 11 '16 at 17:48