prove $\int_0^1 \frac{x^\sqrt{2}}{1+x} dx= \frac{1}{\sqrt{2} + 1} - \frac{1}{\sqrt{2}+2} + \frac{1}{\sqrt{2}+3}-\frac{1}{\sqrt{2}+4}+\ldots$

Question

Prove $$\int_0^1 \frac{x^\sqrt{2}}{1+x} dx= \frac{1}{\sqrt{2} + 1} - \frac{1}{\sqrt{2}+2} + \frac{1}{\sqrt{2}+3}-\frac{1}{\sqrt{2}+4}+\ldots\,.$$

I know that $\sum_{k=0}^\infty (-1)^k x^{\sqrt{2}+k} = \frac{x^\sqrt{2}}{1+x}$ is a uniform convergence for $|x| < 1$. But the integral range includes $x = 1$. How can I approach it?

Answer 1

For all $k\geq 1$ we have $$ \frac 1{1+x}-\sum_{n=0}^k(-x)^n=\frac{(-x)^{k+1}}{1+x}. $$ By multiplying this relation with $x^{\sqrt 2}$ and integrating it between $x=0$ and $x=1$ we can estimate the error between the integral and the truncated series as $$ \left|\int_{0}^1\frac{x^{\sqrt 2}}{1+x}\mathrm d x - \sum_{n=0}^{k}\frac{(-1)^{n}}{\sqrt 2+1+n}\right|=\int_{0}^1\frac{x^{k+1+\sqrt 2}}{1+x}d x\leq\int_{0}^1x^{k+1+\sqrt 2} d x=\frac{1}{2+\sqrt 2+k}, $$ which tends to $0$ as $k\to+\infty$.

Answer 2

The shortest proof involves taking summands in pairs.

Let $r=\sqrt{2}$. Let $$f_n(x)=\sum_{k=0}^{2n-1} (-1)^k x^{r+k}= \sum_{\ell=0}^{n-1} x^{r+2\ell}(1-x) \,.$$ Since for $x \in [0,1]$, the continuous functions $f_n(x)$ are nondecreasing in $n$ and converge to the continuous function $f(x)=\frac{x^r}{1+x}$ pointwise on $[0,1]$, Dini's theorem [1] ensures that the convergence is uniform, so integration term by term is valid without invoking Lebesgue theory.

[1] https://en.wikipedia.org/wiki/Dini%27s_theorem