The Sine Integral, $\operatorname{Si}(x)$, is defined as $$\operatorname{Si}(x)=\int_0^x \frac{\sin t}{t} \ \mathrm{d}t $$

It follows that $$\int_0^{\pi} \operatorname{Si}(x) \ \mathrm{d}x = \bigg[x \operatorname{Si}(x) \bigg]_0^{\pi}-\int_0^{\pi} \sin x \ \mathrm{d}x = \pi \operatorname{Si} (\pi)-2$$ and further $$\int_0^{\pi} \operatorname{Si}^2 (x) \ \mathrm{d}x = \bigg[ x \operatorname{Si}^2(x) \bigg]_0^{\pi} -2\int_0^{\pi} \operatorname{Si} (x) \sin x \ \mathrm{d}x \\ = \pi \operatorname{Si}^2(\pi) -2 \left( \big[ -\cos x \operatorname{Si} (x) \big]_0^{\pi}+\int_0^{\pi} \frac{\sin x \cos x}{x} \ \mathrm{d}x \right)\\ = \pi \operatorname{Si}^2(\pi)-2 \operatorname{Si} (\pi)-\int_{2x=0}^{2\pi}\frac{\sin 2x}{2x} \ \mathrm{d}(2x)\\ = \pi\operatorname{Si}^2(\pi)-2 \operatorname{Si} (\pi)-\operatorname{Si} (2\pi)$$

Following a similar procedure on $\int_0^{\pi} \operatorname{Si}^3(x) \ \mathrm{d}x$ doesn’t quite work, because it requires evaluating the integral $\int_0^{\pi} \operatorname{Si}^2(x) \sin x \ \mathrm{d}x$, or equivalently, $\int_0^{\pi} \frac{\operatorname{Si}(x) \sin 2x}{x} \ \mathrm{d}x$.

This leads me to ask, can (and how can) $\int_0^{\pi} \operatorname{Si}^3(x) \ \mathrm{d}x$ be expressed in terms of elementary or special functions? How about the general case, i.e. $$\color{purple}{\int_0^{\pi} \operatorname{Si}^n(x) \ \mathrm{d}x} \; ? $$

  • 3
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Pedro
    Jun 15, 2021 at 10:02

5 Answers 5


Here is an observation that seems to hint a negative answer to OP's question.

Considering that $\pi$ does not play a particular role in the special values for $\operatorname{Si}(\cdot)$, the question would be almost equivalent to investigating the antiderivatives of $\operatorname{Si}(x)^k$'s. Now let us write

$$F(x) = \int_{x}^{\infty} \frac{\sin t}{t} \, \mathrm{d}t = \frac{\pi}{2} - \operatorname{Si}(x)$$

for the complementary version of $\operatorname{Si}(\cdot)$. In light of the binomial theorem, it is clear that $\int\operatorname{Si}(x)^n\,\mathrm{d}x$ can be computed if all of $\int F(x)^k\,\mathrm{d}x$ for $k=0,\dots,n$ can be computed. So we will shift our focus to finding the antiderivatives of $F(x)^k$'s.

Now employing OP's technique, we find that

\begin{align*} \int F(x) \, \mathrm{d}x &= x F(x) + \int \sin x \, \mathrm{d}x \\ &= x F(x) - \cos x + C \end{align*}


\begin{align*} \int F(x)^2 \, \mathrm{d}x &= x F(x)^2 + 2\int F(x) \sin x \, \mathrm{d}x \\ &= x F(x)^2 - 2F(x)\cos x - \int \frac{2\sin x\cos x}{x} \, \mathrm{d}x \\ &= x F(x)^2 - 2F(x)\cos x + F(2x) + C. \end{align*}

These computations, together with the asymptotic formula for $F(x)$, reveal the closed-forms for the following integrals:

$$ I_1 = \int_{0}^{\infty} F(x) \, \mathrm{d}x = 1 \qquad\text{and}\qquad I_2 = \int_{0}^{\infty} F(x)^2 \, \mathrm{d}x = \frac{\pi}{2}. $$

Similarly, if $\int F(x)^3 \, \mathrm{d}x$ admits a closed-form expression involving $F(\cdot)$, then we can expect that the integral

$$ I_3 = \int_{0}^{\infty} F(x)^3 \, \mathrm{d}x $$

will admit an elementary closed-form (or at least assume a simpler formula than $\int_{0}^{\pi}\operatorname{Si}(x)^3\,\mathrm{d}x$). However, we can prove that

$$I_3 = \frac{\pi^2}{4} - \frac{3}{2}\log^2 2 - \frac{3}{4}\operatorname{Li}_2\left(\frac{1}{4}\right) = 1.545982100082988\dots $$

holds, where $\operatorname{Li}_2(\cdot)$ is the dilogarithm. Since this expression seems non-elementary, I am skeptical about the ideal that $\operatorname{Si}(x)^3$ has an elementary antiderivative.

Addendum: Proof of the formula for $I_3$. We find that

\begin{align*} \int F(x)^3 \, \mathrm{d}x &= x F(x)^3 + 3\int F(x)^2 \sin x \, \mathrm{d}x \\ &= x F(x)^3 - 3F(x)^2\cos x - 6\int \frac{F(x)\sin x\cos x}{x} \, \mathrm{d}x \\ &= x F(x)^3 - 3F(x)^2\cos x - 3\int \frac{F(x)\sin(2x)}{x} \, \mathrm{d}x. \end{align*}

From this, we get

$$ I_3 = \frac{3\pi^2}{4} - 3\int_{0}^{\infty} \frac{F(x)\sin(2x)}{x} \, \mathrm{d}x. $$

To compute this, we adopt the Feynman's trick. More precisely, we consider the parametrized integral

$$ I_3(s) = \frac{3\pi^2}{4} - 3\int_{0}^{\infty} \frac{F(x)\sin(sx)}{x} \, \mathrm{d}x. $$


\begin{align*} I_3'(s) &= -3\int_{0}^{\infty} F(x)\cos(sx) \, \mathrm{d}x \\ &= -\frac{3}{s}\int_{0}^{\infty} \frac{\sin x \sin(sx)}{x} \, \mathrm{d}x \\ &= -\frac{3}{2s}\int_{0}^{\infty} \frac{\cos((s-1)x) - \cos((1+s)x)}{x} \, \mathrm{d}x \end{align*}

The last integral can be computed using the Frullani's integral, yielding

\begin{align*} I_3'(s) &= -3\int_{0}^{\infty} F(x)\cos(sx) \, \mathrm{d}x \\ &= -\frac{3}{s}\int_{0}^{\infty} \frac{\sin x \sin(sx)}{x} \, \mathrm{d}x \\ &= \frac{3}{2s}(\log\left|s-1\right| - \log(s+1)). \end{align*}

So it follows that

\begin{align*} I_3 = I_3(2) &= I_3(0) + \int_{0}^{2} I_3'(s) \, \mathrm{d}s \\ &= \frac{3\pi^2}{4}+ \int_{0}^{2} \frac{3}{2s}(\log\left|s-1\right| - \log(s+1)) \, \mathrm{d}s. \end{align*}

By computing the last integral using the dilogarithm, we can complete the proof of the formula.


This note is divided into

  • Part I: an elaboration of my comments (which have now been moved to chat), and

  • Part II: in agreement with Sangchul's claim that an elementary closed form does not exist.

Part I: Writing the expression as a double integral, we have \begin{align}\int_0^\pi\frac{\operatorname{Si}(x)\sin2x}x\,dx&=\int_0^\pi\int_0^1\frac{\sin yx\sin2x}{yx}\,dy\,dx\\&=\frac12\int_0^1\int_0^\pi\frac{\cos((y-2)x)-\cos((y+2)x)}{yx}\,dx\,dy\\&=\frac12\int_0^1\frac{\operatorname{Ci}((y-2)\pi)-\log(y-2)-(\operatorname{Ci}((y+2)\pi)-\log(y+2))}y\,dy.\end{align} This follows from the series expansion $$\operatorname{Ci}(x)=\gamma+\log x+\sum_{n\ge1}\frac{(-1)^nx^{2n}}{2n(2n)!}\tag1$$ so that \begin{align}\lim_{x\to0}[\operatorname{Ci}((y+2)x)-\operatorname{Ci}((y-2)x)]&=\lim_{x\to0}\left[\log((y+2)x)-\log((y-2)x)+\sum_{n\ge1}\frac{f(n,y)x^{2n}}{2n(2n)!}\right]\\&=\log(y+2)-\log(y-2)\end{align} where \begin{align}f(n,y)&=(-1)^n((y+2)^{2n}-(y-2)^{2n})\\&=(-1)^n\sum_{0\le k\le2n}\binom{2n}ky^{2n-k}2^k(1-(-1)^k)\\&=(-1)^n\sum_{1\le k\le n}\binom{2n}{2k-1}y^{2n-2k+1}2^{2k}.\end{align} Using $(1)$ again, we obtain \begin{align}\int_0^\pi\frac{\operatorname{Si}(x)\sin2x}x\,dx&=-\frac12\int_0^1\sum_{n\ge1}\frac{(-1)^n\pi^{2n}f(n,y)}{2n(2n)!y}\,dy\\&=-\frac14\sum_{n\ge1}\sum_{1\le k\le n}\int_0^1\frac{(-1)^n\pi^{2n}4^k}{n(2k-1)!(2n-2k+1)!}y^{2n-2k}\,dy\\&=-\frac14\sum_{n\ge1}\sum_{1\le k\le n}\frac{(-1)^n\pi^{2n}4^k}{n(2n-2k+1)(2k-1)!(2n-2k+1)!}\end{align} where the sum-integral interchange is justified by Fubini. This can be rewritten in one summation as $$\int_0^\pi\frac{\operatorname{Si}(x)\sin2x}x\,dx=-2\sum_{n\ge1}\frac{(-1)^n\pi^{2n}}{(2n)!(2n-1)}{}_3F_2\left(\frac12-n,\frac12-n,1-n;\frac32,\frac32-n;4\right)$$ by expressing the denominator in terms of Pochhammer symbols. Here is verification.

Part 2: It is important to note that the upper limit of $\pi$ does not simplify the integral, as mentioned in Sangchul's answer. From Part I, we have \begin{align}\int_0^\pi\frac{\operatorname{Si}(x)\sin2x}x\,dx=\lim_{\epsilon\to0^+}\int_\epsilon^1\frac{\operatorname{Ci}((y-2)\pi)-\log(y-2)-(\operatorname{Ci}((y+2)\pi)-\log(y+2))}{2y}\,dy.\end{align} The antiderivatives of $\log(y\pm2)/y$ can be readily expressed in terms of dilogarithms, so the only issue stems from the evaluation of $$\int_\epsilon^1\frac{\operatorname{Ci}((y\pm2)\pi)}y\,dy=\int_{(\epsilon\pm2)\pi}^{(1\pm2)\pi}\frac{\operatorname{Ci}(w)}{w\mp2\pi}\,dw.$$ This is why I strongly doubt there is an elementary closed form for $\int_0^\pi\operatorname{Si}^3(x)\,dx$, as $\operatorname{Ci}$ is non-periodic and the limits do not allow for any simplifications to be made.


As it was mentioned in the OP, \begin{align} &I=\int\limits_0^\pi \operatorname{Si}^3(x)\,\text dx \;\overset{\text{IBP}}{=\!=}\;\operatorname{Si}^3(x) x\bigg|_0^\pi -3\int\limits_0^\pi x\operatorname{Si}^2(x)\operatorname{sinc}(x)\,\text dx =\pi\operatorname{Si}^3(\pi) +3\int\limits_0^\pi \operatorname{Si}^2(x)\,\text d\cos x\\[4pt] &\overset{\text{IBP}}{=\!=}\;\pi\operatorname{Si}^3(\pi)+3\operatorname{Si}^2(x)\cos x \bigg|_0^\pi - 6\int\limits_0^\pi \cos x\operatorname{Si}(x)\operatorname{sinc}(x)\,\text dx =\pi\operatorname{Si}^3(\pi)-3\operatorname{Si}^2(\pi)-6I_1, \end{align} where \begin{align} &I_1=\int\limits_0^\pi \operatorname{Si}(x)\operatorname{sinc}(2x)\,\text dx =\int\limits_0^\pi \int\limits_0^x\operatorname{sinc}(y)\operatorname{sinc}(2x)\,\text dy\,\text dx =\int\limits_0^\pi \int\limits_y^\pi\operatorname{sinc}(y)\operatorname{sinc}(2x)\,\text dx\,\text dy\\[4pt] &=\dfrac12\int\limits_0^\pi \int\limits_y^\pi\operatorname{sinc}(y)\operatorname{sinc}(2x)\,\text d(2x)\,\text dy =\dfrac12\int\limits_0^\pi \operatorname{sinc}(y)\operatorname{Si}(2x)\bigg|_y^\pi\,\text dy\\[4pt] &=\dfrac12\int\limits_0^\pi \operatorname{sinc}(y)(\operatorname{Si}(2\pi)-\operatorname{Si}(2y))\,\text dy, \end{align} $$I_1=\dfrac12\operatorname{Si}(\pi)\operatorname{Si}(2\pi)-\dfrac12\int\limits_0^\pi \operatorname{sinc}(x)\operatorname{Si}(2x)\,\text dx.\tag1$$ Applying the known series in the form of $$\operatorname{Si}(2x)=\pi\sum\limits_{k=0}^\infty J^2_{\large k+^1/_2}(x) \tag2,$$ where $$J_{\large^1/_2}(x)=\sqrt{\dfrac2{\pi x}}\sin x,\quad J_{\large^3/_2}(x) =\sqrt{\dfrac2{\pi x}}(\operatorname{sinc} x-\cos x),\tag3$$ $$J_{\large m+^3/_2}(x) = \dfrac{2m+1}x J_{\large m+^1/_2}(x)-J_{\large m-^1/_2}(x),\quad m=1, 2\dots,\tag4$$ allows to get high accuracy closed-form approximations.

For example, from the equality $$ \begin{align} &S_{10}=\int\limits_0^\pi\left(\sum\limits_{k=0}^9 J^2_{\large k+^1/_2}(x)\right)\operatorname{sinc} x \,\text dx = \dfrac1{28605549772800 \pi^{17}}\\[4pt] &\times\bigg(198641616776068257792000000 - 51388022909702134656000000 \pi^2\\[4pt] &+ 4492572854595378448896000 \pi^4 - 155376628399064274048000 \pi^6\\[4pt] &+ 2259641487353598835200 \pi^8 - 12461916450169368000 \pi^{10}\\[4pt] &+ 24885770867920800 \pi^{12} + 186094688685480 \pi^{14}\\[4pt] &+ \pi^{16} (-64011445084 + 112455910172835(\operatorname{Ci}(\pi) - \operatorname{Ci}(3\pi)+\ln3))\bigg)\\[4pt] &\approx 0.76480\,65196\,02706\,51930\,94484 \end{align} $$ should $$I\approx \pi \operatorname{Si}^3(\pi)-3\operatorname{Si^2}(\pi) -3\operatorname{Si}(\pi)\operatorname{Si}(2\pi)+3\pi S_{10},$$ $$I\approx 8.99407\,09143\,86010\,29167.$$ Numeric integration gives $$I\approx 8.99407\,09143\,90144\,50150.$$

However, the closed form of I was not obtained.

  • 1
    $\begingroup$ The Mathjax isn’t rendering at the end. Can you fix it? $\endgroup$
    – Vishu
    Jun 18, 2021 at 12:00
  • $\begingroup$ @Tavish I don't see this, but updated the answer. $\endgroup$ Jun 18, 2021 at 12:17

I am not sure my result is entirely correct or not but if there is any mistakes then please tell me in the comments.

So the integral is, $\textstyle\displaystyle{\int_{0}^{\pi}\operatorname{Si}^n(x)dx}$

I think we can start to simplify this using integration by parts, more specifically the DI method (If you don't know what DI method is then for introduction you can watch this and this video.

So let's start-

$\begin{array} \ & \text{D} & \text{I}\\ + & \operatorname{Si}^n(x) & 1\\ - & n\operatorname{Si}^{n-1}(x)\frac{\sin(x)}{x} & x \end{array}$




Let's do this process again to find some pattern.

$\begin{array} \ & \text{D} & \text{I}\\ + & \sin(x)\operatorname{Si}^{n-1}(x) & 1\\ - & \begin{align}\cos(x)\operatorname{Si}^{n-1}(x) & +(n-1)\operatorname{Si}^{n-2}(x)\frac{\sin^2(x)}{x}\end{align} & x \end{array}$



Here, $\textstyle\displaystyle{nx\operatorname{Si}^{n-1}(x)\sin(x)\bigg]_{0}^{\pi}=0}$

And let's do it for the last time

$\begin{array} \ & \text{D} &\text{I}\\ + & \sin^2(x)\operatorname{Si}^{n-2}(x) & 1\\ - & \begin{align}2\cos(x)\operatorname{Si}^{n-2}(x) & +(n-2)\operatorname{Si}^{n-3}(x)\frac{\sin^3(x)}{x}\end{align} & x \end{array}$


$\textstyle\displaystyle{=\pi\operatorname{Si}^n(\pi) +n\int_{0}^{\pi}x\cos(x)\operatorname{Si}^{n-1}(x)dx+n(n-1)x\sin^2(x)\operatorname{Si}^{n-2}(x)\bigg]_{0}^{\pi}-2n(n-1)\int_{0}^{\pi}x\cos(x)\operatorname{Si}^{n-2}(x)dx+n(n-1)(n-2)\int_{0}^{\pi}\sin^3(x)\operatorname{Si}^{n-3}(x)dx}$

Here, $\textstyle\displaystyle{n(n-1)x\sin^2(x)\operatorname{Si}^{n-2}(x)\bigg]_{0}^{\pi}=0}$

So now let's find patterns, at each step:-

$\pi\operatorname{Si}^n(\pi)$ stays, the term which is in the form $\textstyle\displaystyle{\int_{0}^{\pi}\sin^k(x)\operatorname{Si}^{n-k}(x)dx}$ is broken into


$\textstyle\displaystyle{\frac{kn!}{(n-k)!}\int_{0}^{\pi}x\cos(x)\operatorname{Si}^{n-k}(x)dx}$ and

$\textstyle\displaystyle{\frac{n!}{[n-(k+1)]!}\int_{0}^{\pi}\sin^{k+1}(x)\operatorname{Si}^{n-(k+1)}(x)dx}$ where

$\textstyle\displaystyle{\frac{n!}{(n-k)!}x\sin^k(x)\operatorname{Si}^{n-k}(x)\bigg]_{0}^{\pi}}$ vanishes to $0$,

$\textstyle\displaystyle{\frac{n!k}{(n-k)!}\int_{0}^{\pi}x\cos(x)\operatorname{Si}^{n-k}(x)dx}$ stays and

$\textstyle\displaystyle{\frac{n!}{[n-(k+1)]!}\int_{0}^{\pi}\sin^{k+1}(x)\operatorname{Si}^{n-(k+1)}(x)dx}$ is broken down further.

So finally we get,


$\textstyle\displaystyle{=\pi\operatorname{Si}^n(\pi)$ $+\sum_{k=1}^{n}\frac{n!k}{(n-k)!}\int_{0}^{\pi}x\cos(x)\operatorname{Si}^{n-k}(x)dx}$

This is as far as I have come. If I can work out more then I would edit my answer to add that.

  • $\begingroup$ You're the first person I've seen on MSE to use the DI method explicitly, nice :) $\endgroup$ Jul 4, 2021 at 8:59
  • $\begingroup$ @A-Level-Student. Thanks. I hate the u-dv method. $\endgroup$ Jul 4, 2021 at 12:08

Here is an alternative series expression for the integral

$$i = \int_{0}^{\pi} \frac{\sin(2x)}{x} \text{Si}(x )\,dx\tag{1}$$

which I obtained from replacing the $\sin$ by its power series and integrating termwise. A similar series was provided by TheSimpliFire.


$$c(n) = \int_0^{\pi } x^{2 n} \text{Si}(x) \, dx \\ = \frac{\pi ^{2 n+1} \left(2 (n+1) \text{Si}(\pi )-\pi \, _1F_2\left(n+1;\frac{3}{2},n+2;-\frac{\pi ^2}{4}\right)\right)}{2 (n+1) (2 n+1)}\tag{2}$$

which I have derived by looking for a recursive solution for $c(n)$ (or alternatively simply emloying Mathematica), I get

$$i = s := s_1 + s_2$$


$$s_1 =\sum _{n=0}^{\infty } \frac{(-1)^n (2 \pi )^{2 n+1} \text{Si}(\pi )}{(2 n+1) (2 n+1)!} =\text{Si}(\pi ) \text{Si}(2 \pi ) \tag{3}$$


$$s_2 =-\pi\sum _{n=0}^{\infty } \frac{(-1)^n (2 \pi )^{2 n+1} \, _1F_2\left(n+1;\frac{3}{2},n+2;-\frac{\pi ^2}{4}\right)}{2 (n+1) (2 n+1) (2 n+1)!}\tag{4}$$

The hypergeometric function $s_2$ is related to Bessel functions as can be seen from the first four summands

$$\begin{align}\left\{-4,\frac{1}{9} (-2) \left(4+\sqrt{2} \pi ^3 J_{-\frac{1}{2}}(\pi )+2 \sqrt{2} \pi ^2 J_{-\frac{3}{2}}(\pi )\right),\frac{1}{75} (-2) \left(48-\sqrt{2} \pi ^4 J_{-\frac{3}{2}}(\pi )+\sqrt{2} \pi ^5 J_{-\frac{5}{2}}(\pi )-8 \sqrt{2} \pi ^3 J_{-\frac{5}{2}}(\pi )\right),\frac{4 \left(-1440+\sqrt{2} \pi ^5 \left(\pi ^2-9\right) J_{-\frac{5}{2}}(\pi )+3 \sqrt{2} \pi ^4 \left(\pi ^2-16\right) J_{-\frac{7}{2}}(\pi )\right)}{2205}\right\}\end{align}\tag{4a}$$

They simplify, however, to these curious polynomials in $\pi$:

$$\begin{align}\left\{-4,\frac{4 \pi ^2}{9}-\frac{16}{9},-\frac{64}{25}+\frac{16 \pi ^2}{25}-\frac{4 \pi ^4}{75},\\ -\frac{256}{49}+\frac{64 \pi ^2}{49}-\frac{16 \pi ^4}{147}+\frac{8 \pi ^6}{2205}\right\}\end{align}\tag{4b}$$

(As expected) I was not able to calculate $s_2$ analytically.

But notice that the hypergeometric function appearing here is of smaller degree than that in the sum in 1 so that maybe some progress is possible.


The sum can also be written with a hypergeometric function with smaller parameters as

$$\begin{align}s=\left(\text{Si}(\pi )-\frac{1}{\pi }\right) \text{Si}(2 \pi )-\sum _{n=0}^{\infty } \frac{(-1)^n \left(2 (2 \pi )^{2 n}\right) \, _1F_2\left(n;\frac{1}{2},n+1;-\frac{\pi ^2}{4}\right)}{(2 n+1) (2 n+1)!}\end{align}\tag{5}$$

$\, _1F_2$ admits the representations

$$\begin{align}& \, _1F_2\left(n;\frac{1}{2},n+1;-\frac{\pi ^2}{4}\right)\\ & = n \sqrt{\pi } \sum _{k=0}^{\infty } \frac{1}{ (k+n) \Gamma \left(k+\frac{1}{2}\right)}\frac{z^k}{k!}|_{z\to -(\frac{\pi}{2})^2}\\ & =n \int_{0}^{1} x^{n-1}\cos(\pi \sqrt{x})\,dx=2 n \int_0^1 y^{2 n-1} \cos (\pi y) \, dy\end{align}\tag{6}$$


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