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

Solution 1:

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).

Solution 2:

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.