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I'm interested in evaluating

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

Where

\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?

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    $\begingroup$ Have you seen this question and answer? I think your integral falls into the same category, especially the second answer. $\endgroup$ – Bort Aug 12 at 11:33
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    $\begingroup$ @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. $\endgroup$ – Michael Anderson Aug 13 at 2:52
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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.

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  • $\begingroup$ 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 Aug 15 at 4:23
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    $\begingroup$ 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 : $a_{n+1} =(n+1)*a_n+n!$. $\endgroup$ – Michael Anderson Aug 15 at 5:04
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    $\begingroup$ @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 Aug 15 at 12:20

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