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!

  • $\begingroup$ Beyond exponential or logarithms, most Taylor Series get complicated rather quickly... even the series for sine and cosine can be rather complicated. A number of years ago I worked out a trick to solve the generalized Basel problem for $\zeta(4)$ by working with the power series $\sum_{k=0}^\infty \frac{x^{k}}{(4k)!}(-1)^k$, though this requires some creativity to figure out what function has this power series (namely $\cos\left(\frac{x^{1/4}}{\sqrt{2}}\right)\cosh\left(\frac{x^{1/4}}{\sqrt{2}}\right)$; one can derive this by multiplying roots of unity, giving a generalization to $\zeta(2n)$) $\endgroup$ Feb 27, 2022 at 3:05
  • 1
    $\begingroup$ Even after getting comfortable with integrals, I still find myself rarely resorting to power series of anything beyond the expected ones (rational functions, logs/exponentials, Laurent expansions of $\Gamma$ and $\zeta$, occasional hypergeometric series, etc) since its easy to integrate power series but often hard to recognize the resulting power series in terms of standard functions $\endgroup$ Feb 27, 2022 at 3:08
  • 1
    $\begingroup$ Chapter 5 of Inside interesting integrals by Nahin has quite a few examples of this $\endgroup$
    – Sal
    Feb 27, 2022 at 11:11
  • $\begingroup$ To do your example, Robert, you have to do $\sum n^{-2}=\pi^2/6$, right? So, you'll be expecting the high school students to take that on faith? $\endgroup$ Mar 2, 2022 at 4:46
  • $\begingroup$ @GerryMyerson, good point. At the time of writing my question, I hadn't found other integrals that used the Taylor series method to solve them. I agree that the Basel problem is more advanced than high-school maths, but my hope with the bounty was to find some better examples that fit the question's parameters $\endgroup$
    – Robert Lee
    Mar 2, 2022 at 5:03

4 Answers 4


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:

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

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

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

  1. From this answer by Tunk-Fey:

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

  • 4
    $\begingroup$ Are there elementary ways to evaluate those infinite sums? Isn't it "cheating" to say the answer is $G$, when you have no way to express $G$, other than as an infinite sum? $\endgroup$ Mar 2, 2022 at 4:44

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


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}

  • 2
    $\begingroup$ And is there an elementary way to evaluate those sums to get those $\coth$ and $\cot$ values? $\endgroup$ Mar 2, 2022 at 4:43
  • 1
    $\begingroup$ @GerryMyerson - I don’t suppose so. $\endgroup$
    – Quanto
    Mar 8, 2022 at 19:21

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?

  • 1
    $\begingroup$ Oops, thank you for pointing this out! My memory served me wrong - thank you! I have corrected my answer. $\endgroup$ Mar 7, 2022 at 20:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .