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I was asked to find the mclaurin series of $\int_0^x\frac{\arctan (t)}{t}dt$

using the known mclaurin for arctan: $\arctan(t)=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}t^{2n-1}}{2n-1}$

Ok, so what I did is use the known formula: $$\int_0^x\frac{\arctan (t)}{t}=\int_0^x\frac{\sum_{n=1}^{\infty} \frac{(-1)^{n+1}t^{2n-1}}{2n-1}}{t}dt=\int_0^x\sum_{n=1}^{\infty}\frac{(-1)^{n+1}t^{2n-2}}{2n-1}dt$$

This is nothing more than $\int_0^x1-\frac{t^2}{3}+\frac{t^4}{5}-\cdots dt$.

If this was a finite sum, then yes, for sure I can divide it into separate integrals. But this is an infinite sum. The integrand is a polynomial, an integrable and even continuous function so I don't see any reason why we can't separate that integral of the sum into the sum of the integrals, but it's not apparent to me why it's obvious that we can do that either.

But if we can, then the mclaurin series we were looking for is $\sum_{n=1}^{\infty}\int_0^x\frac{(-1)^{n+1}t^{2n-2}}{2n-1}dt$

is this correct? if so - can we always separate the integral of the sum into sum of integrals? even infinite sums?

Michael Hardy
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Oria Gruber
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    It is a nice question, but there is a very nice answer here already :-) http://math.stackexchange.com/questions/83721/when-can-a-sum-and-integral-be-interchanged – Ant Feb 15 '15 at 20:19

1 Answers1

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If $\displaystyle\int_a^b \sum_{n=1}^\infty |a_n(t)|\,dt<\infty$ then $\displaystyle\int_a^b \sum_{n=1}^\infty a_n(t)\,dt = \sum_{n=1}^\infty \int_a^b a_n(t)\,dt$.

This is actually a special case of Fubini's theorem.

Michael Hardy
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  • I don't see the relation to fubini's theorem to be honest. fubini's theorem tells us when it's possible to calculate a double integral using iterated single integrals. we dont have a double integral in this question – Oria Gruber Feb 15 '15 at 20:39
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    @OriaGruber an infinite sum can be seen as an integral calculated with respect to the counting measure http://en.wikipedia.org/wiki/Counting_measure – Ant Feb 15 '15 at 20:47
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    Or a special case of the dominated convergence theorem. – PhoemueX Feb 15 '15 at 21:48
  • @OriaGruber : Fubini's theorem doesn't care about any distinction between integrals and sums. In the Lebesgue theory, they're all integrals. ${}\qquad{}$ – Michael Hardy Feb 16 '15 at 06:05
  • I just fixed a pretty bad typo in this answer: absolute value should be in the "if" part but not in the "then" part. – Michael Hardy Apr 20 '20 at 18:18
  • The expansion of $\arctan(t)$ is true for $|t|\leq 1$ therefore $|x|\leq 1$ – FDP Apr 20 '20 at 23:45