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I did this: $$\int_{1}^t x^{-1}dx=\int_{1}^t\lim_{n\rightarrow -1}{x^n}dx =\lim_{n\rightarrow -1}\int_{1}^t{x^n}dx $$ just to have a way to approximate $\ln t$. $$\ln{t}=\lim_{h\rightarrow 0}\frac{x^{h}-1}{h}$$

The second expression may be correct, but I was told I cannot say $\int_a^b \lim_{n\rightarrow -1}x^{n}dx=\lim_{n\rightarrow -1}\int_a^b x^n dx$ without previously prooved some statements. If so, what are those statement and where can I find information about this?

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You can't do that when the exponent is $-1$, can you write $\int x^{-1}=\dfrac{x^{-1+1}}{-1+1}$? – Inceptio May 15 '13 at 3:58
The result $\ln x=\lim\limits_{h\to0}\frac{x^h-1}{h}$ is true though. Assuming you are allowing yourself the knowledge that $\frac{d}{dh}x^h=x^h\ln x$, this is true by an application of L'Hospital's Rule. – alex.jordan May 15 '13 at 4:18
I didn't do that @Inceptio . The idea was approching x^-1 to x^-0.9999 so I could integrate. In fact, you could get a nice approximation of $\ln x$ with $\frac{x^{0.00001}-1}{0.00001}$. – Alan May 15 '13 at 4:21
up vote 4 down vote accepted

In general, you cannot interchange two limits. In this case, recall that the integration is a actually a limit, for instance, the limit of a Riemann sum. Hence, you need to be careful, in general: $$\int_a^b \lim_{n \to \infty} f_n(x) dx \neq \lim_{n \to \infty} \int_a^b f_n(x) dx$$

You may want to read through the following posts to figure out under what conditions, you can swap two limits.

When can the order of limit and integral be exchanged?

Can a limit of an integral be moved inside the integral?

when can we exchange order of two limits

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I may read this now. Thanks for helping :) – Alan May 15 '13 at 4:24

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