# Evaluation of $\int_{0}^{1} \frac{dx}{1+\sqrt[n]{x}}$ for $n\in\mathbb{N}$

So some quick checks seem to indicate that

$$\int_{0}^{1} \frac{dx}{1+\sqrt[n]{x}} = C_{n} + (-1)^{n+1}\log(2^{n})$$

Where $C_{n}\in\mathbb{Q}$. The denominators of $C_{2n}$ appear to correspond to oeis:A117664. Is it possible to obtain $C_{n}$ for arbitrary $n$ in closed form? (I'm actually trying to evaluate $\prod_{n=1}^{\infty} \int_{0}^{1} \frac{2 dx}{1+\sqrt[n]{x}}$ numerically, but it's easier to evaluate if I can evaluate the integrals symbolically.)

• The substitution $u=1+\sqrt[n]{x}$ gives you $\displaystyle{\int_1^2\frac{n(u-1)^{n-1}}{u}du}$, which you could expand and integrate term by term. This at least confirms that it is rational $+(-1)^{n+1}\log(2^n)$, but I don't know if it leads to a nice "closed form" for $C_n$. – Jonas Meyer May 30 '11 at 8:03

Substituting $x=u^n$ gives $$\int_0^1 \frac{nu^{n-1}}{1+u}\,du \;=\; \int_0^1\sum_{k=0}^\infty (-1)^k u^k nu^{n-1}\,du$$ Integrating term-by-term yields the following formula: $$C_n \;=\; n(-1)^n \sum_{k=1}^{n-1} \frac{(-1)^{k+1}}{k}.$$ The summation on the right is a partial sum of the alternating harmonic series. For an explicit formula, the best you can do is express these numbers in terms of the harmonic numbers or the digamma function (see MathWorld).
• For the record, I plugged in the digamma function and got:$\int_{0}^{1} \frac{dx}{1+\sqrt[n]{x}} = 1 + \frac{n}{2}\left[\psi_{0}\left(\frac{1}{2}+\frac{n}{2}\right)-\psi_{0}\left(1+\frac{n}{2}\right)\right]$, and the numerics check out satisfactorily. – graveolensa Jun 2 '11 at 17:44
Another method, similar to Jim's solution, is to evaluate this separating it into two cases with $n$ even and $n$ odd and work it out.
Let $I_n = \displaystyle \int_{0}^{1} \frac{u^{n-1}}{1+u} du$
If $n=2k+1$, then $$\int_0^1 \frac{u^{2k}}{1+u} du = \int_0^1 \frac{u^{2k}-1}{1+u} du + \int_0^1 \frac{1}{1+u} du$$ $$\int_0^1 \frac{u^{2k}-1}{1+u} du = - \int_0^1 \left(\sum_{m=0}^{2k-1} (-u)^m du \right) = \sum_{m=0}^{2k-1} (-1)^{m+1} \int_0^1 u^m du = \sum_{m=0}^{2k-1} \frac{(-1)^{m+1}}{m+1}$$ $$\int_0^1 \frac{1}{1+u} du = \log(2)$$ Hence, if $n$ is odd we get $$I_{n} = n \log(2) + n \times \sum_{m=1}^{n-1} \frac{(-1)^m}{m}$$ If $n=2k$, then $$\int_0^1 \frac{u^{2k-1}}{1+u} du = \int_0^1 \frac{u^{2k-1}+1}{1+u} du - \int_0^1 \frac{1}{1+u} du$$ $$\int_0^1 \frac{u^{2k-1}+1}{1+u} du = \int_0^1 \left(\sum_{m=0}^{2k-2} (-u)^m du \right) = \sum_{m=0}^{2k-2} (-1)^{m} \int_0^1 u^m du = \sum_{m=0}^{2k-2} \frac{(-1)^{m}}{m+1}$$ $$\int_0^1 \frac{1}{1+u} du = \log(2)$$ Hence, if $n$ is even we get $$I_{n} = -n \log(2) + n \times \sum_{m=1}^{n-1} \frac{(-1)^{m-1}}{m}$$ Hence $$C_n = n \times (-1)^n \times \sum_{m=1}^{n-1} \frac{(-1)^{m-1}}{m}$$ In Jim's answer, though it is correct and can be argued out, we still need to prove that we can swap the limit and the integrals. However in this method there is no infinite sum and hence makes the argument a bit easier.