Closed Form of $a_n = \int_0^1 \ln(1+x^n) dx$

Question

I want to know the closed form of : $$a_n = \int_0^1 \ln(1+x^n)dx, \quad \forall n \in \mathbb{N}$$

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I found :

$$0<a_n<\frac{1}{n+1}, \quad \lim_{n\to\infty} a_n =0$$

I started from \begin{align} &\ln (1+x)=\sum_{k=0}^\infty \frac{(-1)^k x^{k+1}}{k+1} \\ & \Rightarrow \ln(1+x^n)=\sum_{k=0}^\infty \frac{(-1)^k x^{nk+n}}{k+1} \\ & a_n = \int_0^1 \ln(1+x^n)dx = \sum_{k=0}^\infty \frac{(-1)^k}{k+1}\int_0^1 x^{n+nk}dx \\ &=\sum_{k=0}^\infty \frac{(-1)^k}{k+1} \times \frac{1}{n+nk+1} \\ &=\sum_{k=0}^\infty \frac{(-1)^k}{k+1} \times \frac{1}{(n+1)(k+1)-k} \end{align}

But I stucked here. Is there any closed (or approximated) from exist?

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These are some results for litte $n$ : \begin{align} & a_1 = 2\ln 2 - 1 \\ & a_2 = \ln2 - 2 + \frac{\pi}{2} \\ & a_3 = 2\ln 2 - 3 + \frac{\pi \sqrt{3}}{3} \\ \end{align}

Answer 1

I think this way will be better. As your solution, my solution also uses power series expansion:

By integration by parts, we see that $$a_n=\ln2-\int_{0}^1\frac{nx^n}{1+x^n}=\ln2-n(1-\int_0^1\frac{dx}{1+x^n}).$$

Now, note that $$n(1-\int_0^1\frac{dx}{1+x^n})=n(1-\int_0^1\sum_{k=0}^{\infty}(-1)^kx^{nk})\\ =n(1-\sum_{k=0}^{\infty}\frac{(-1)^k}{nk+1})\\ =\sum_{k=1}^{\infty}\frac{(-1)^kn}{nk+1}$$ Now: $$\lim_{n\to\infty}a_n=\ln2-\lim_{n\to\infty}\sum_{k=1}^{\infty}\frac{(-1)^kn}{nk+1}\\ =\ln2-\sum_{k=1}^\infty\frac{(-1)^k}{k}=0$$

Answer 2

Just for your curiosity.

There is a closed form $$a_n=\sum_{k=0}^\infty \frac{(-1)^k}{k+1} \times \frac{1}{(n+1)(k+1)-k}=\log (2)-\Phi \left(-1,1,1+\frac{1}{n}\right) $$ where appears the Hurwitz-Lerch transcendent function (which still hides an infinite summation).

However, we can use asymptotics and get $$a_n=\frac{\pi ^2}{12 n}-\frac{3 \zeta (3)}{4 n^2}+\frac{7 \pi ^4}{720 n^3}-\frac{15 \zeta (5)}{16 n^4}+\frac{31 \pi ^6}{30240 n^5}-\frac{63 \zeta (7)}{64 n^6}+O\left(\frac{1}{n^7}\right)$$

For illustration purposes $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 2 & 0.25875865857 & 0.26394350735 \\ 3 & 0.19975226361 & 0.20009372535 \\ 4 & 0.16099050589 & 0.16103912992 \\ 5 & 0.13470442094 & 0.13471504382 \\ 6 & 0.11577477354 & 0.11577782305 \\ 7 & 0.10150281021 & 0.10150386842 \\ 8 & 0.09036040460 & 0.09036082676 \\ 9 & 0.08142081049 & 0.08142099791 \\ 10 & 0.07408996034 & 0.07409005089 \\ 11 & 0.06796964628 & 0.06796969313 \\ 12 & 0.06278300534 & 0.06278303100 \\ 13 & 0.05833159222 & 0.05833160696 \\ 14 & 0.05446946187 & 0.05446947069 \\ 15 & 0.05108688988 & 0.05108689534 \\ 16 & 0.04809979505 & 0.04809979855 \\ 17 & 0.04544266327 & 0.04544266556 \\ 18 & 0.04306369143 & 0.04306369297 \\ 19 & 0.04092137953 & 0.04092138059 \\ 20 & 0.03898209081 & 0.03898209155 \end{array} \right)$$

Answer 3

From where you stopped $$\sum_{k=0}^\infty \frac{(-1)^k}{k+1} \frac{1}{(n+1)(k+1)-k},$$ if you multiply out the denominator, you see that it can be easily factored, so that we have $$\frac 1n\sum_{k=0}^\infty \frac{(-1)^k}{(k+1)(k+1+\frac 1n)},$$ which after performing the shift $k\mapsto k-1$ becomes $$\frac 1n\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k(k+\frac 1n)},$$ which should have a closed form, since the series is very similar to $$\sum_{k=1}^\infty \frac{\frac 1n}{k(k+\frac 1n)},$$ which has a closed form.

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See, for example, what Wolfram Alpha gives, but I think there is an elementary representation since the last series above also has an elementary closed format. I shall attempt to look for it; if I find it (before you or someone else, of course) I'll edit this to let you know.

Answer 4

We have \begin{align} \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)(k+z)} &= \sum_{k=0}^\infty \frac{1}{(k+1)(k+z)} - \sum_{k=0}^\infty \frac{1-(-1)^k}{(k+1)(k+z)} = \\ &= \sum_{k=0}^\infty \frac{1}{(k+1)(k+z)} - \sum_{m=0}^\infty \frac{2}{(2m+2)(2m+1+z)} = \\ &= \sum_{k=0}^\infty \frac{1}{(k+1)(k+z)} - \frac12 \sum_{m=0}^\infty \frac{1}{(m+1)(m+\frac{1+z}{2})} = \\ &= \frac{\psi(z)+\gamma}{z-1} - \frac12 \frac{\psi(\frac{1+z}{2})+\gamma}{\frac{1+z}{2}-1} = \\ &= \frac{\psi(z)-\psi(\frac{1+z}{2})}{z-1}\end{align} where $\psi(z)$ is the digamma function and $\gamma$ is Euler-Mascheroni constant.

We have then \begin{align} \int_0^1 \ln(1+x^n)dx &=\sum_{k=0}^\infty \frac{(-1)^k}{(k+1)(n+nk+1)} = \\ &= \frac{1}{n} \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)(k+1+\frac{1}{n})} = \\&= \frac{1}{n} \frac{\psi(1+\frac{1}{n})-\psi(1+\frac{1}{2n})}{(1+\frac{1}{n})-1} = \\ &= \psi\big(1+\frac{1}{n}\big)-\psi\big(1+\frac{1}{2n}\big)\end{align}

It turns out that for any $n\in\mathbb N$ it can be expressed in terms of elementary functions. This is because \begin{align} \psi\big(1+\frac{1}{n}\big) &= -\gamma + \sum_{k=0}^\infty\frac{\frac1n}{(k+1)(k+1+\frac1n)} = \\ &= -\gamma + \sum_{k=0}^\infty\Big(\frac{1}{k+1}-\frac{1}{k+1+\frac1n}\Big) = \\ &= -\gamma + \sum_{k=0}^\infty\int_0^1 (x^k-x^{k+\frac1n})dx = \\ &= -\gamma + \int_0^1 \frac{1-x^{\frac1n}}{1-x}dx = \\ &= -\gamma + \int_0^1 \frac{(1-y)ny^{n-1}}{1-y^n}dy = \\ &= -\gamma + \int_0^1 \frac{ny^{n-1}}{\prod_{k=1}^{n-1}(y-e^{i\frac{2\pi k }{n}})}dy \end{align} and the last integral can be calculated using the partial fraction decomposition.

Answer 5

Here is a closed form involving a finite sum of trigonometric functions.

Continuing Quruitay's answer, you can also use the partial fraction decomposition of $$\frac{1}{x^n+1} = -\frac{1}{n}\sum\limits_{k=0}^{n-1} \frac{\zeta_{2n}^{2k+1}}{x-\zeta_{2n}^{2k+1}}$$ to find the integral $$\int_0^1\frac{1}{x^n+1} \mathrm{d}x = -\frac{1}{n}\sum\limits_{k=0}^{n-1}\zeta_{2n}^{2k+1}\ln\left(\zeta_{2n}^{-2k-1}-1\right)$$

Now $\zeta_{2n}^{-2k-1} - 1 = 2\left|\sin\left(\frac{2k+1}{2n}\pi\right)\right|e^{-\frac{n+2k+1}{2n}\pi i}$ so we get the sum as: $$-\frac{1}{n}\sum\limits_{k=0}^{n-1}\zeta_{2n}^{2k+1}\left[\ln 2 + \ln\left|\sin\left(\frac{2k+1}{2n}\pi\right)\right|-\frac{n+2k+1}{2n}\pi i\right]$$

We assume $n \gt 1$. From Vieta's formulas applied to $x^n + 1$ we know that $\sum\limits_{k=0}^{n-1}\zeta_{2n}^{2k+1} = 0$, so we can ignore the constant $\ln 2$ term, as well as the constant $-\frac{n+1}{2n}\pi i$ term.

Moreover, we can show $\sum\limits_{k=0}^{n-1}k\zeta_{2n}^{2k+1} = \frac{n}{2i\sin\left(\frac{\pi}{n}\right)}$ using a general formula for $\sum\limits_{k=1}^{n-1} kX^k$. These results allow us to simplify the expression as:

$$\frac{\pi}{2n\sin\left(\frac{\pi}{n}\right)}-\frac{1}{n}\sum\limits_{k=0}^{n-1}\zeta_{2n}^{2k+1}\ln\left|\sin\left(\frac{2k+1}{2n}\pi\right)\right|$$

Using $\zeta_{2n}^{2k+1} = \cos\left(\frac{2k+1}{n}\pi\right)+i\sin\left(\frac{2k+1}{n}\pi\right)$ and multiplying out, we further obtain:

$$\frac{\pi}{2n\sin\left(\frac{\pi}{n}\right)}-\frac{1}{n}\sum\limits_{k=0}^{n-1}\left[\cos\left(\frac{2k+1}{n}\pi\right)+i\sin\left(\frac{2k+1}{n}\pi\right)\right]\ln\left|\sin\left(\frac{2k+1}{2n}\pi\right)\right|$$

The imaginary part should cancel since we expect the result to be real. In fact we can exploit the symmetry of the sum regarding $k \to n - 1 - k$ (each term in the imaginary part flips its sign, while each term in the real part remains the same), and therefore we can keep the real part only, and keep the sum up to $\big\lfloor\frac{n-1}{2}\big\rfloor$ while doubling each term. Note that if $n$ is odd, we have one unpaired term corresponding to $k = \frac{n-1}{2}$, but it turns out this term is equal to $0$ anyway, so we get:

$$\int_0^1\frac{1}{x^n+1} \mathrm{d}x = \frac{\pi}{2n\sin\left(\frac{\pi}{n}\right)}-\frac{1}{n}\sum\limits_{k=0}^{n-1}\cos\left(\frac{2k+1}{n}\pi\right)\ln\left|\sin\left(\frac{2k+1}{2n}\pi\right)\right|$$

$$\int_0^1\frac{1}{x^n+1} \mathrm{d}x = \frac{\pi}{2n\sin\left(\frac{\pi}{n}\right)}-\frac{2}{n}\sum\limits_{k=0}^{\big\lfloor\frac{n-1}{2}\big\rfloor}\cos\left(\frac{2k+1}{n}\pi\right)\ln\left(\sin\left(\frac{2k+1}{2n}\pi\right)\right)$$

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It's interesting to note that the sum is actually a Riemann sum for $\frac{1}{\pi}\int\limits_0^\pi \cos x\ln\left(\sin \frac{x}{2}\right) \mathrm{d}x = -\frac{1}{2}$, so we expect the limit of the sum to converge to $-\frac{1}{2}$. But $\lim\limits_{n\to\infty} \frac{\pi}{2n\sin\left(\frac{\pi}{n}\right)} = \frac{1}{2}$ also, so the subtraction gives $\lim\limits_{n\to\infty}\int\limits_0^1 \frac{1}{x^n+1} \mathrm{d}x = \frac{1}{2} - \left(-\frac{1}{2}\right) = 1$, which is not surprising.

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When $n$ is even, we can further exploit the symmetry $k \to \frac{n}{2} - 1 - k$ which gives $\cos\left(\frac{2\left(\frac{n}{2} - 1 - k\right)+1}{n}\pi\right)\ln\left(\sin\left(\frac{2\left(\frac{n}{2} - 1 - k\right)+1}{2n}\pi\right)\right) = -\cos\left(\frac{2k+1}{n}\pi\right)\ln\left(\cos\left(\frac{2k+1}{2n}\pi\right)\right)$.

We can pair up all terms except possibly the one corresponding to $k = \frac{n-2}{4}$, but it turns out this term would be $0$ anyways. So we get:

$$\int_0^1\frac{1}{x^n+1} \mathrm{d}x = \frac{\pi}{2n\sin\left(\frac{\pi}{n}\right)}-\frac{2}{n}\sum\limits_{k=0}^{\big\lfloor\frac{n-2}{4}\big\rfloor}\cos\left(\frac{2k+1}{n}\pi\right)\ln\left(\tan\left(\frac{2k+1}{2n}\pi\right)\right) \text{ when }n\text{ is even.}$$

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Using Quruitay's result we find that $$\int_0^1\ln\left(1+x^n\right)\mathrm{d}x = \ln 2 - n + \frac{\pi}{2\sin\left(\frac{\pi}{n}\right)} - \sum\limits_{k=0}^{n-1}\cos\left(\frac{2k+1}{n}\pi\right)\ln\left|\sin\left(\frac{2k+1}{2n}\pi\right)\right|$$

$$\int_0^1\ln\left(1+x^n\right)\mathrm{d}x = \ln 2 - n + \frac{\pi}{2\sin\left(\frac{\pi}{n}\right)} - 2\sum\limits_{k=0}^{\big\lfloor\frac{n-1}{2}\big\rfloor}\cos\left(\frac{2k+1}{n}\pi\right)\ln\left(\sin\left(\frac{2k+1}{2n}\pi\right)\right)$$

And if $n$ is even:

$$\int_0^1\ln\left(1+x^n\right)\mathrm{d}x = \ln 2 - n + \frac{\pi}{2\sin\left(\frac{\pi}{n}\right)} - 2\sum\limits_{k=0}^{\big\lfloor\frac{n-2}{4}\big\rfloor}\cos\left(\frac{2k+1}{n}\pi\right)\ln\left(\tan\left(\frac{2k+1}{2n}\pi\right)\right)$$

The results are valid for $n \gt 1$.

I checked the values numerically from $n=2$ to $n=20$ in Wolfram Alpha and they match the exact numerical values provided in Claude Leibovici's answer, so I am fairly confident in the result.

Answer 6

Integrate by parts $$a_n=\int_0^1 \ln({x^n+1})dx= -n+\ln2+ \int_0^1 \frac n{x^n+1}dx$$ With the roots $x_k= e^{i\frac{(2k-1)\pi}n} $ for $x^n+1=0$, integrate \begin{align} \int_0^1 \frac{n}{x^n+1}dx & = -\sum_{k=1}^n\int_0^1 \frac{x_k}{x-x_k} dx = -\sum_{k=1}^nx_k\ln(1-x_k^{-1})\\ &= -\sum_{k=1}^n x_k \left[i \frac{n-2k+1}{2n}+ \ln \left(2 \sin\frac{(2k-1)\pi}{2n}\right) \right] \end{align} Per $\sum_{k=1}^nx^k=0$ and the symmetry of $x^k$ to arrive at the close-form $$\int_0^1 \frac n{x^n+1}dx =\sum_{k=1}^{[\frac n2]} ( \pi-\theta_k )\sin\theta_k -\cos\theta_k \ln\sin^2\frac{\theta_k}2,\>\>\> \theta_k= \frac{2k-1}{n} \pi$$ Listed below are the results for the first few $n’s$

\begin{align} & a_2 = - 2 + \ln 2 + \frac\pi2\\ & a_3 = -3 +2\ln2 + \frac\pi{\sqrt3}\\ & a_4 = -4 +\ln 2+ \frac\pi{\sqrt2}+\sqrt2\ln(1+\sqrt2)\\ & a_5 = -5 +2\ln2 +\frac\pi{\sqrt{10}}\sqrt{5+\sqrt5}+\sqrt5\ln\frac{1+\sqrt5}2\\ & a_6 =-6+\ln2 + \pi +\sqrt3\ln(2+\sqrt3) \\ & a_7 =-7+\ln2 + \frac\pi2\csc\frac\pi7 +\frac{\ln\csc^2\frac\pi{14}}{\sec\frac\pi7} +\frac{\ln\csc^2\frac{3\pi}{14}}{\sec\frac{3\pi}7} +\frac{\ln\csc^2\frac{5\pi}{14}}{\sec\frac{5\pi}7} \end{align}