I've found here the following integral.

$$I = \int_{0}^{1}\sin{(\pi (1-x))}x^x(1-x)^{1-x}\,dx=\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx=\frac{\pi e}{24}$$

I've never seen it before and I also didn't find the evaluation on math.se. How could we verify it?

If it is a well-known integral, then could you give a reference?

  • $\begingroup$ I just want to say good luck to whoever tries to figure this one out. $\endgroup$ – DanZimm Oct 5 '14 at 0:37
  • $\begingroup$ Damn, this is a good one. $\endgroup$ – Bennett Gardiner Oct 5 '14 at 2:47
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    $\begingroup$ I think I can show the first part. I'll take a crack at the second. $\endgroup$ – Mike Oct 5 '14 at 2:49
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    $\begingroup$ Aren't the two integrals trivially identical? Since $\sin(\pi-z) = \sin(z)$? $\endgroup$ – Bennett Gardiner Oct 5 '14 at 2:51
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    $\begingroup$ Possible duplicate of Two curious "identities" on $x^x$,$e$,and $\pi$ $\endgroup$ – Brevan Ellefsen Mar 4 '17 at 21:07

This one can be done with "residue at infinity" calculation. This method is shown in the Example VI of http://en.wikipedia.org/wiki/Methods_of_contour_integration .

First, we use $z^z = \exp ( z \log z )$ where $\log z$ is defined for $-\pi\leq \arg z < \pi$.

For $(1-z)^{1-z} = \exp ( (1-z)\log (1-z) )$, we use $\log (1-z)$ defined for $0\leq \arg(1-z) <2\pi$.

Then, let $f(z)= \exp( i\pi z + z \log z + (1-z) \log (1-z) )$.

As shown in the Ex VI in the wikipedia link, we can prove that $f$ is continuous on $(-\infty, 0)$ and $(1,\infty)$, so that the cut of $f(z)$ is $[0,1]$.

We use the contour: (consisted of upper segment: slightly above $[0,1]$, lower segment: slightly below $[0,1]$, circle of small radius enclosing $0$, and circle of small radius enclosing $1$, that looks like a dumbbell having knobs at $0$ and $1$, can someone edit this and include a picture of it please? In fact, this is also the same contour as in Ex VI, with different endpoints.)

On the upper segment, the function $f$ gives, for $0\leq r \leq 1$, $$\exp(i\pi r) r^r (1-r)^{1-r} \exp( (1-r) 2\pi i ).$$

On the lower segment, the function $f$ gives, for $0\leq r \leq 1$, $$\exp(i\pi r) r^r(1-r)^{1-r}. $$

Since the functions are bounded, the integrals over circles vanishes when the radius tend to zero.

Thus, the integral of $f(z)$ over the contour, is the integral over the upper and lower segments, which contribute to

$$\int_0^1 \exp(i\pi r) r^r (1-r)^{1-r} dr - \int_0^1 \exp(-i\pi r) r^r(1-r)^{1-r} dr$$

which is $$ 2i \int_0^1 \sin(\pi r) r^r (1-r)^{1-r} dr.$$

By the Cauchy residue theorem, the integral over the contour is $$ -2\pi i \textrm{Res}_{z=\infty} f(z) = 2\pi i \textrm{Res}_{z=0} \frac{1}{z^2} f(\frac 1 z).$$

From a long and tedious calculation of residue, it turns out that the value on the right is $$ 2i \frac{\pi e}{24}.$$ Then we have the result: $$ \int_0^1 \sin(\pi r) r^r (1-r)^{1-r} dr = \frac{\pi e}{24}.$$

  • $\begingroup$ Could we get closed forms of the integral with other integrand limits by similar long and tedious calculation? For example between $1$ and $3/2$ or between $3/2$ and $2$, or between $2$ and $5/2$, or any other not that nice arbitrary $a$ and $b$? $\endgroup$ – user153012 Oct 5 '14 at 20:12
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    $\begingroup$ It is of interest to note that \begin{align} \int_{0}^{1} e^{i \pi x} \, x^{x} (1-x)^{1-x} \, dx = i \, \frac{\pi e}{4!} \end{align} $\endgroup$ – Leucippus Oct 5 '14 at 20:55
  • $\begingroup$ @user153012 I do not know for sure, but the calculation would be much more difficult for other limits. $\endgroup$ – i707107 Oct 6 '14 at 1:37
  • $\begingroup$ @Leucippus That one is easy to see when you notice that the integral with $\cos \pi x$ vanishes. My solution came out like this when started computing that one instead. $\endgroup$ – i707107 Oct 6 '14 at 16:30
  • $\begingroup$ (For anyone scratching their heads, this is an old post don't downvote!) $\endgroup$ – Zach466920 Jun 16 '15 at 14:58

Torsten Carleman $[2]$ proved in 1922 that

$$ \sum_{n=1}^\infty\left(a_1a_2\cdots a_n\right)^{1/n} < e\sum_{n=1}^\infty a_n, $$

where $a_n \geq 0$, $n=1,2,\dots$, and $0 < \sum_{n=1}^\infty a_n < \infty$. Thenceforth, this result is known as Carleman's inequality. There exists a number of refined versions of Carleman's original work $[3, 6]$. It has turned out that the following generalization is – from our point of view – important, which is proved by Yang $[7]$: $$ \sum_{n=1}^\infty\left(a_1a_2\cdots a_n\right)^{1/n} < e\sum_{n=1}^\infty \left(1-\sum_{k=1}^6 \frac{b_k}{(n+1)^k}\right)a_n, $$

with $b_1 = 1/2, b_2 = 1/24, b_3 = 1/48, b_4 = 73/5670, b_5 = 11/1280, b_6 = 1945/580608$. On the last page of his paper, Yang $[7]$ conjectured that if $$ \left(1+\frac{1}{x}\right)^x = e\left(1-\sum_{n=1}^\infty \frac{b_n}{(x+1)^n}\right), \quad x>0, $$

then $b_n > 0$, $n=1,2,\dots.$ In fact, the constants $b_4$ and $b_6$ are not corrent in Yang's work, the correct values are $b_4 = 73/5760$ and $b_6 = 3625/580608$. Later, this conjecture was proved and discussed by Yang $[8]$, Gylletberg and Ping $[4]$, and Yue $[9]$. They are using the recurrence $$ b_1 = \frac12, \quad b_n = \frac{1}{n}\left(\frac{1}{n+1} - \sum_{k=0}^{n-2} \frac{b_{n-k-1}}{k+2} \right), \quad n = 2,3,\dots. $$ The recurrence is given in a somewhat more compact form in Finch's manuscript $[3]$, as the following:

$$ b_0 = -1, \quad b_n = -\frac{1}{n}\sum_{k=1}^{n} \frac{b_{n-k}}{k+1}, \quad n = 1,2,\dots. $$

The first ten values of the sequence are listed in the next table. \begin{array} {|r|r|} \hline b_0 & -1 \\ \hline b_1 & 1/2 \\ \hline b_2 & 1/24 \\ \hline b_3 & 1/48 \\ \hline b_4 & 73/5760 \\ \hline b_5 & 11/1280 \\ \hline b_6 & 3625/580608 \\ \hline b_7 & 5525/1161216 \\ \hline b_8 & 5233001/1393459200 \\ \hline b_9 & 1212281/398131200 \\ \hline b_{10} & 927777937/367873228800 \\ \hline \end{array} The numerators are recorded as A249276, and the denominators as A249277 in the OEIS. I've calculated the $b_n$ sequence in the range $n=0,\dots,32$, the elements are listed here.

The following theorem is proved in general in the paper by Hu and Mortici $[5]$, and for the special cases $n=0$ and $n=1$ in the paper by Alzer and Berg $[1]$.

For all integer $n \geq 0$, we have

$$ \int_0^1 x^n\sin\left(\pi x\right)x^x\left(1-x\right)^{1-x}\,dx = b_{n+2}\pi e. $$

The special case $n=0$ answers my question.


  1. H. Alzer, C. Berg, Some classes of completely monotonic functions, Annales Academiæ Scientiarum Fennicæ Mathematica, 27, 2002, 445–460. (pdf)
  2. T. Carleman, Sur les fonctions quasi-analytiques, Comptes rendus du Ve Congres des Mathematiciens Scandinaves, Helsinki (Helsingfors), 1922, 181–196.
  3. S. Finch, Carleman's Inequality, manuscript, 2013.
  4. M. Gyllenberg, Y. Ping, On a conjecture by Yang, Journal of Mathematical Analysis and Applications, 264(2), 2001, 687–690.
  5. Y. Hu, C. Mortici, On the coefficients of an expansion of $(1+1/x)^x$ related to Carleman's inequality, manuscript, arXiv:1401.2236, 2014.
  6. M. Johansson, L.-E. Persson, A. Wedestig, Carleman's inequality - History, proofs and some new generalizations, Journal of Inequalities in Pure and Applied Mathematics, 4(3), 2003.
  7. X. Yang, On Carleman’s inequality, Journal of Mathematical Analysis and Applications, 253(2), 2001, 691–694.
  8. X. Yang, Approximations for constant $e$ and Their Applications, Journal of Mathematical Analysis and Applications, 262(2), 2001, 651–659.
  9. H. Yue, A Strengthened Carleman’s Inequality, Communications in Mathematical Analysis, 1(2), 2006, 115–119. (pdf)


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