I'm interested in evaluating

\begin{equation} \int_0^x \frac{Si(t)}{t}\;dt \end{equation}


\begin{equation} Si(x) = \int_0^x \frac{\sin t}{t}\;dt \end{equation}

I've found a nice method for evaluating $Si(x)$ using Pade and Chebyshev-Pade approximations in https://arxiv.org/pdf/1407.7676.pdf but I'm unable to find an existing equivalent method of the integral above.

I can probably derive the small $x$ Pade approximation without too much difficulty, as the integral has a relatively simple Taylor series expansion, but I'm unsure how to approach the large $x$ part.

Does anyone have a reference for calculatin such an integral? Or even a good reference for deriving the large $x$ Chebyshev-Pade approximation for a similar integral?

  • 2
    $\begingroup$ Have you seen this question and answer? I think your integral falls into the same category, especially the second answer. $\endgroup$
    – Bort
    Commented Aug 12, 2019 at 11:33
  • 1
    $\begingroup$ @Bort That's a neat answer, but I think its geared towards infinite integrals, rather than definite integrals. If you can see a way to adapt it to this case, I'd love a pointer. $\endgroup$ Commented Aug 13, 2019 at 2:52

1 Answer 1


For large $x$, you can use the same approach as the one in the paper you cite. They define these auxiliary functions that link $\mathrm{Si},\mathrm{Ci}$ with $\sin,\cos$, and the point of why this works is that the relationship between these two pairs of functions is in general a $2\times 2$ matrix, but this matrix only has two independent functions in its components.

Using Mathematica (code gist), I computed the same auxiliary functions for your case. The integral you give is $$ \mathrm{Si}(x)\log x - \int_0^x t^{-1}\sin t\log t\,\mathrm{d}t. $$ Consider the four integrals $$ \int_x^\infty t^{-1}\big( \sin t, \cos t, \sin t\log t, \cos t\log t \big)\,\mathrm{d}t. $$ These are equal to $$ \begin{gathered} f_1 \cos x + f_2\sin x,\\ f_2\cos x - f_1\sin x,\\ -f_3\cos x+f_1\cos x\log x+f_4\sin x+f_2\sin x\log x,\\ -f_4\cos x+f_2\cos x\log x-f_3\sin x-f_1\sin x\log x. \end{gathered} $$ To compute this, I applied integration by parts to each integral up to order $O(x^{-10})$, and then read off the coefficients of $(\cos x,\sin x,\cos x\log x,\sin x\log x)$, which are the auxiliary functions. The power series for them are: $$ \begin{aligned} f_1 &= \frac{1}{x} - \frac{2}{x^3} + \frac{24}{x^5} - \frac{720}{x^7} + \frac{40320}{x^9}+\cdots,\\ f_2 &= \frac{1}{x^2} - \frac{6}{x^4} + \frac{120}{x^6} - \frac{5040}{x^8} + \frac{362880}{x^{10}} - \cdots,\\ f_3 &= \frac{3}{x^3} - \frac{50}{x^5} + \frac{1764}{x^7} - \frac{109584}{x^9}+\cdots,\\ f_4 &= \frac{1}{x^2} - \frac{11}{x^4} + \frac{274}{x^6} - \frac{13068}{x^8} + \frac{1026576}{x^{10}}-\cdots. \end{aligned} $$ Here $f_1,f_2$ should be equal to $f,g$ in the paper. To find closed forms for $f_{3,4}$, they are the $(\Im,-\Re)$ parts of the integral $$ \int_x^\infty t^{-1}\log(t/x)e^{\mathrm{i}(t-x)}\,\mathrm{d}t, $$ change the contour of integration to $(x,x+\mathrm{i}\infty)$ and use $t=x(1+\mathrm{i}v)$. Then $$ f_3 = \int_0^\infty \frac{\frac12 \log(1+v^2) + v\arctan v}{1+v^2}e^{-v x}\,\mathrm{d}v,\\ f_4 = \int_0^\infty \frac{-\frac12 v \log(1+v^2) + \arctan v}{1+v^2}e^{-v x}\,\mathrm{d}v, $$ which are easy to evaluate numerically.

Because there is a one-to-one mapping between the $f$'s and the four integrals (a system of four linear equations in $f$'s for each value of $x$), this is enough information to numerically compute the Pade approximations of the auxiliary functions $f_{1,2,3,4}$ just like they do in the paper.

  • $\begingroup$ Since you are using Mathematica, don't you think it's a little weird that, in Mathematica, Integrate[SinIntegral[z]/z, {z, 0, a}] shows an exact result in terms of HypergeometricPFQ? $\endgroup$
    – David
    Commented Aug 15, 2019 at 4:23
  • 1
    $\begingroup$ Worth noting the coefficients of f1, f2 are just pulled alternatively from n!, and those for f3,f4 look like they come from oeis.org/A000254, which can be generated by : $a_{n+1} =(n+1)*a_n+n!$. $\endgroup$ Commented Aug 15, 2019 at 5:04
  • 6
    $\begingroup$ @David From a numerical perspective there's a problem with using hypergeometric functions. It is a very general class of functions, encompassing pretty much all the standard ones (dlmf.nist.gov/15), and as a result it is very hard to compute accurately, requiring a combination of different approaches chosen depending on the argument and the parameters (dlmf.nist.gov/15.19; mpmath.org/doc/current/functions/hypergeometric.html). $\mathrm{Si}(x)$ can also be expressed terms of hypergeometrics (dlmf.nist.gov/6.11), but it's computed as in dlmf.nist.gov/6.18. $\endgroup$
    – Kirill
    Commented Aug 15, 2019 at 12:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.