Prove that $\sum_{n=0}^\infty \frac{(-1)^n}{n+x}=\int_0^1 \frac{t^{x-1}}{t+1}\,dt$

Question

I'm trying to prove that:$$\sum_{n=0}^\infty \frac{(-1)^n}{n+x}=\int_0^1 \frac{t^{x-1}}{t+1} \,dt$$ If we use term by term differentiation, we obtain: $$\int_0^1 \frac{t^{x-1}}{t+1}dt=\int_0^1 t^{x-1} \left( \sum_{n=0}^\infty (-1)^n t^n \right )dt=\sum_{n=0}^{\infty}(-1)^n \int_0^1t^{n+x-1} \, dt=\sum_{n=0}^\infty \frac{(-1)^n}{n+x}$$ However, I the function $t \mapsto (-1)^n t^{n+x-1}$ is not uniformly convergent on the compact set $[0,1]$, so I can't justify the above. Is this correct?

Answer 1

No, you cannot use this argument. The best you can do is restrict to the interval $[0,1-\delta]$ ($\delta>0$) so that the geometric series converges uniformly. This leaves you with $$\int_0^1 \frac{t^{x-1}}{t+1}dt = \lim_{\delta\to 0^+} \int_0^{1-\delta} \frac{t^{x-1}}{t+1}dt = \lim_{\delta\to 0^+} \sum_{n=0}^\infty \frac{(-1)^n(1-\delta)^{n+x}}{n+x}.$$ You again need some uniformity to interchange the limit and the infinite sum.

Answer 2

Note that, for $N>\max\{1,-x\}$, \begin{eqnarray} &&\bigg|\int_0^1 \frac{t^{x-1}}{t+1}dt-\sum_{n=0}^N \frac{(-1)^n}{n+x}\bigg|\\ &=&\bigg|\int_0^1 \frac{t^{x-1}}{t+1}dt-\int_0^1 t^{x-1} \left( \sum_{n=0}^N (-1)^n t^n \right )dt\bigg|\\ &=&\bigg|\int_0^1 \frac{t^{x-1}}{t+1}dt-\int_0^1 t^{x-1}\frac{1-(-t)^{N+1}}{1+t}dt\bigg|\\ &=&\bigg|\int_0^1 (-1)^{N+1}\frac{t^{x+N}}{t+1}dt\bigg|\\ &\le&\int_0^1 t^{x+N}dt\\ &=&\frac{1}{x+N+1}. \end{eqnarray} Now letting $N\to\infty$ gives $$\int_0^1 \frac{t^{x-1}}{t+1}dt=\sum_{n=0}^\infty \frac{(-1)^n}{n+x}.$$

Answer 3

We assume $x>0$ in what follows.

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Consider the following geometric series $$t^{x-1}-t^x+t^{x+1}-\dots=\frac{t^{x-1}}{1+t}$$ The convergence is uniform in any interval of the form $[0, y] $ where $0<y<1$. It follows that the above series can be integrated term by term in $[0,y]$ and we have thus $$\sum_{n=0}^{\infty}(-1)^n\frac{y^{n+x}}{n+x}=\int_{0}^{y}\frac{t^{x-1}}{1+t}\,dt$$ for $0<y<1$. This can be rewritten as $$\sum_{n=0}^{\infty} (-1)^n\frac{y^n}{n+x}=y^{-x}\int_{0}^{y}\frac{t^{x-1}}{1+t}\,dt$$ The series on left side is convergent for $y=1$ and the limit of the expression on right as $y\to 1^{-}$ exists and hence by Abel's theorem we have the desired equality.