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Someone asked on another site about ways to evaluate $\int \ln x \ dx $ without using integration by parts.

My response was the following:

$$ \begin{align} \int\ln x \, dx & =\int\lim_{t\to 0}\frac{x^{t}-1}{t} \, dx \\ &= \lim_{t\to 0}\frac{1}{t}\int (x^{t}-1) \, dx \\ &=\lim_{t\to 0}\frac{1}{t}\left(\frac{x^{t+1}}{t+1}-x\right)+C \\ &=\lim_{t\to 0}\frac{x^{t+1}-x(t+1)}{t(t+1)}+C \\ &=\lim_{t\to 0}\frac{x^{t+1}\ln x-x}{2t+1}+C \\ &= x\ln x-x+C \end{align}$$

But I don't know how to justify moving the limit outside the indefinite integral.

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  • $\begingroup$ Because the limit variable is not the variable that you are integrating with respect to. $\endgroup$ – Ataraxia Jun 17 '13 at 16:39
  • $\begingroup$ @ZettaSuro That argument is not very good.. if that was the case then life would be really easy. $\endgroup$ – AD. Jun 17 '13 at 16:42
  • $\begingroup$ @ZettaSuro Well, because there are plenty of counterexamples showing that the argument is wrong. Suppose for simplicity that we worked on a bounded interval, then we would need some argument telling us more about the convergence - like uniform, bounded, monotonicity etc. $\endgroup$ – AD. Jun 17 '13 at 16:49
  • $\begingroup$ @AD. So the principle of linearity does not necessarily apply? $\endgroup$ – Ataraxia Jun 17 '13 at 16:50
  • $\begingroup$ @ZettaSuro No.. $\endgroup$ – AD. Jun 17 '13 at 16:56
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First of all, let's fix the interval of integration. We want to show that for any $0<a<b<\infty$, $$\int_a^b \ln x\,dx = \lim_{t\to 0}\int_a^b \frac{x^t-1}{t}\,dx \tag1$$ As nbubis said, it suffices to prove that we have uniform convergence. To this end, write $x=e^u$ and use the Taylor approximation $$|e^{tu}-1-tu|\le C(tu)^2 \tag2$$ for sufficiently small $tu$. The important point is that $|tu|\le |t|\max(|\log a|, |\log b|)$, which allows us to rewrite (2) as $$|e^{tu}-1-tu|\le \widetilde Ct^2 \tag3$$ for sufficiently small $t$, with $\widetilde C$ independent of $u$. Hence, $$ \sup_{x\in [a,b] } \left|\frac{x^{t}-1}{t} -\ln x\right|= \sup_{u\in [\log a,\log b]} \left|\frac{e^{tu}-1}{t} -u\right| \le \widetilde Ct $$ as desired.

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I've seen it said that if the series of functions $f_n$ converges uniformly (not only pointwise!) to the function $f$, then you can write:

$$\lim_{n\to \infty} \int f_n dx = \int \lim_{n\to \infty}f_n dx = \int fdx$$ I'll try and find the source, and add it here later.

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