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Is it possible to interchange summation and integral on $$\int^{1}_0 \sum^{\infty}_{i=1} \frac{(-1)^{i+1}x^{i-1}}{i} dx ?$$

The Fubini's theorem says that for general $f_n$, if $\int \sum |f_n| < \infty$ or $\sum \int |f_n| < \infty$, then $\int \sum f_n = \sum \int f_n$. I tried with this theorem, but it doesn't work.

Could someone tell me the method to be able to interchange the summation and integral in this case? Or just a hint is sufficient.

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    $\begingroup$ If the series converges uniformly on $[0,1]$, you can interchange summation and integration. $\endgroup$
    – Daniel Fischer
    Commented Oct 4, 2015 at 21:57
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    $\begingroup$ Yes, there is one. However, You can also use Fubini, since $\sum \int \lvert f_i\rvert < \infty$ with $f_i(x) = \frac{(-1)^{i+1} x^{i-1}}{i}$. $\endgroup$
    – Daniel Fischer
    Commented Oct 4, 2015 at 21:59
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    $\begingroup$ @PaulSinclair It does converge uniformly on $[0,1]$. It is an alternating series, and the remainder of the series is thus bounded by $\frac{x^{n-1}}{n}$ at each $x$, which gives the uniform bound of $\frac{1}{n}$. $\endgroup$
    – Daniel Fischer
    Commented Oct 4, 2015 at 22:10
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    $\begingroup$ I don't know if it has a name. It's just that $$\lim_{n\to\infty}\int_a^b f_n(x)\,dx = \int_a^b \lim_{n\to\infty} f_n(x)\,dx$$ when the convergence is uniform. That follows immediately from $$\biggl\lvert \int_a^b f(x) - g(x)\,dx \biggr\rvert \leqslant \int_a^b \lvert f(x) - g(x)\rvert\,dx \leqslant \int_a^b \lVert f-g\rVert_\infty\,dx = (b-a)\cdot \lVert f-g\rVert_\infty$$ for $-\infty < a < b < +\infty$. $\endgroup$
    – Daniel Fischer
    Commented Oct 4, 2015 at 22:22
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    $\begingroup$ @PaulSinclair Abel's Theorem states that if $\sum_n^\infty a_n$ is a convergent series of constants, then the power series $\sum_n^\infty a_nx^n$ converges uniformly for $0\le x\le 1$. $\endgroup$
    – Mark Viola
    Commented Oct 4, 2015 at 22:33

1 Answer 1

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Abel's Theorem states that if $\sum_n^\infty a_n$ is a convergent series of constants, then the power series $\sum_n^\infty a_nx^n$ converges uniformly for $0\le x\le 1$.

In addition, we have that if a sequence of functions $f_n(x)$ converges uniformly to $f(x)$ on $x\in[a,b]$, then

$$\lim_{n\to \infty}\int_a^b f_n(x)\,dx=\int_a^b f(x)\,dx$$

Therefore, we see that $\sum_{i=1}^{\infty}\frac{(-1)^{i+1}x^{i-1}}{i}$ converges uniformly on $[0,1]$ and

$$\bbox[5px,border:2px solid #C0A000]{\int_0^1 \sum_{i=1}^{\infty}\frac{(-1)^{i+1}x^{i-1}}{i}\,dx=\sum_{i=1}^{\infty}\frac{(-1)^{i+1}}{i}\int_0^1 x^{i-1}\,dx=\sum_{i=1}^{\infty}\frac{(-1)^{i+1}}{i^2}}$$

It is straightforward to show that

$$\sum_{i=1}^{\infty}\frac{(-1)^{i+1}x^{i-1}}{i}=\frac{\log(1+x)}{x}$$

Putting all of this together, we note that

$$\begin{align} \int_0^1\frac{\log (1+x)}{x}\,dx&=\sum_{i=1}^{\infty}\frac{(-1)^{i+1}}{i^2}\\\\ &=-\text{Li}_2(-1)\\\\ &=\frac{\pi^2}{12} \end{align}$$

where $\text{Li}_2(x)$ is the dilogarithm function.

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