Examples of non-elementary integrals, but whose definite integral IS solvable with power series.

Question

In a high-school level calculus course you learn about Taylor series and some basic integration techniques. In my experience, most definite-integration exercises boiled down to finding an antiderivative and then evaluating at the endpoints using the F.T.C.

Sometime after this, I learned that non-elementary integrals existed. This meant that the finding-antiverivative technique wasn't going to work on these types of integrals. However, I was surprised to find out some of these non-elementary integrals had definite integrals with concise closed forms!

Most of these integrals are usually solved using more advanced techniques like multivariable calculus and complex analysis, but not all of them. This got me thinking...

How many examples of non-elementary integrals, but whose definite integral is solvable with power series, can I find?

I believe these types of integrals would be great examples in a high-school course, since they break away from the monotony of the "Find antiderivative and plug-in values" recipe used in most exercises, with the added bonus of sometimes giving very insightful solutions in the process.

One example of this type of integral is $\int_0^1 \frac{\ln(x)}{x-1} \mathrm{d}x = \frac{\pi^2}{6}$. Since the antiderivative of $\frac{\ln(x)}{x-1}$ is given in terms of a polylogarithm function, it isn't elementary. However, using the power series for $\frac{1}{x-1}$ and interchanging the sum and integral we can achieve the result given in the answer above.

Besides the example above, I haven't managed to find many other examples like this. Does anyone know of some other definite integrals like the above? Any and all suggestions are welcome. Thank you very much!

Answer 1

After looking at Inside interesting integrals, as suggested by @Sal in the comments, I scoured this site looking for similar integrals as the ones presented in Nahin's notes. These are some that I've found so far:

1. From this answer by projectilemotion:

$$ \int_0^{1} \frac{\ln^2(x)}{1+x^2}\mathrm{d}x = \sum_{n\ge 0}(-1)^n\int_0^1 \ln^2(x)x^{2n}\mathrm{d}x = 2\sum_{n\ge 0}\frac{(-1)^n}{(2n+1)^3} = \frac{\pi^3}{16} $$

2. From this answer by Jack D'Aurizio, evaluating a special value of the inverse tangent integral:

$$\int_0^1 \frac{\arctan(x)}{x} \mathrm{d}x = \sum_{n \ge 0}\frac{(-1)^n}{(2n+1)}\int_{0}^{1} x^{2n}\mathrm{d}x =\sum_{n \ge 0}\frac{(-1)^n}{(2n+1)^2} = G $$

3. From this answer by Tunk-Fey:

$$\int_0^1 \ln(x)\ln(1-x)\mathrm{d}x = -\sum_{n\ge 1}\frac{1}n\int_0^1 x^n\ln (x)\mathrm{d}x = \sum_{n\ge 1}\frac{1}{n(n+1)^2} =2-\frac{\pi^2}{6}$$

Answer 2

Here's a couple of integrals that can be done using series:

$$1+\frac{(-1)^n}{n!}\int_0^1\log^n(x)\log(1-x)\,dx = \sum_{k=1}^n(\zeta(k+1)-1)$$

(The particular case of $n=1$ is in Robert Lee answer).

$$\int_0^1\frac{\log^n(x)}{(1-x)^m}\,dx = \frac{(-1)^n n!}{(m-1)!}\sum_{j=1}^m\begin{bmatrix}m-1\\m-j\end{bmatrix}\zeta(n-m+j+1)$$

where $\begin{bmatrix}n\\k\end{bmatrix}$ denotes the unsigned Stirling numbers of the first kind.

Even if its solutions doesn't involve series, I like this completely elementary example of a definite integral with a non-elementary antiderivative.

\begin{align} \int_0^\pi \log\sin x\,dx & = 2\int_0^{\pi/2} \log\sin 2u\,du = 2\int_0^{\pi/2} \log(2\sin u\cos u)\,du\\ &=\pi\log 2+2\left(\int_0^{\pi/2}\log\sin u\,du+\int_0^{\pi/2}\log\cos u\,du\right)\\ &= \pi\log2+2\left(\int_0^{\pi/2}\log\sin u\,du+\int_{0}^{\pi/2}\log\sin(u+\pi/2)\,du\right)\\ &= \pi\log2+2\left(\int_0^{\pi/2}\log\sin u\,du+\int_{\pi/2}^{\pi}\log\sin(v)\,dv\right)\\ &=\pi\log 2 + 2\int_0^\pi\log\sin x\,dx \end{align}

So $$\int_0^\pi\log\sin x\, dx = -\pi \log 2$$

Answer 3

Here are a couple of more \begin{align} \int_{0}^{\infty}\frac{\sin x}{e^x-1}\,dx =& \sum_{k=1}^\infty\int_{0}^{\infty}\sin x\>e^{-kx}\,dx = \sum_{k=1}^\infty\frac{1}{k^2+1} =\frac\pi2\coth\pi -\frac12\\ \\ \int_0^1 \frac{x^{-a}-x^a}{1-x}dx=&\sum_{k=1}^\infty\int_0^1(x^{k-a-1}-x^{k+a-1})dx =\sum_{k=1}^\infty\frac{2a}{k^2-a^2}= \frac{1}{a}-\pi \cot a\pi\\ \\ \int_0^\infty\frac{\cosh ax}{\cosh x}{d}x =&\sum_{k\ge1}\int_0^\infty (-1)^{k+1} (e^{(a-2k+1)x}+ e^{(-a-2k+1)x}){d}x\\ =& \frac12\sum_{k\in Z}\frac{(-1)^k}{\frac{a+1}2+k} = \frac\pi2 \sec\frac{\pi a}2 \end{align}

Answer 4

My favorite is the following: $$\int_0^\infty \ln\left(1-e^{-x}\right) dx.$$

To solve this, note that on the domain $(0, \infty)$, defining $u := e^{-x}$, we have $|u| < 1$ so we can use the Taylor series $\ln\left(1-u\right) = -\left[\frac{u^1}{1} + \frac{u^2}{2} + \dots\right] = -\sum_{i = 1} ^ \infty \frac{u^i}{i} $. This series is absolutely convergent on this domain. Therefore, we can write

$$\int_0^\infty \ln\left(1-e^{-x}\right) dx = -\int_0^\infty \sum_{i = 1} ^ \infty \frac{\left(e^{-x}\right)^i}{i} dx \\ = -\sum_{i = 1} ^ \infty \frac{1}{i} \int_0^\infty e^{-ix} dx \\= -\sum_{i = 1} ^ \infty \frac{1}{i} \frac{1}{i} \\ = -\sum_{i = 1} ^ \infty \frac{1}{i^2} = -\frac{\pi^2}{6}.$$

Cool, right?