I recently try this $$\int\frac{\mathop{\rm Si}(x)}x\,dx$$

Mathematica failed to calculate it.

Integrate[SinIntegral[x]/x, x]

Mma 11.3 failed,but 12.0 can do it.

Sympy gives the answer

from sympy import *
x = symbols("x")
print(integrate(Si(x)/x, x))

x*hyper((1/2, 1/2), (3/2, 3/2, 3/2), -x**2/4)

So I wanted to use the result in Mma.

enter image description here

I have searched in sympy's document,but still can't find solution.


How can I directly get this function in mathematica?

I saw numpy can be used as follows:(add import numpy as np to the beginning,otherwise it will return wrong)

enter image description here

Which maybe helpful.


enter image description here

12.0 will return an object.But I temporaily don't know how to use it.(picture is from others)

  • $\begingroup$ Mathematica 12.0 can solve this integral... So why use SymPy? $\endgroup$
    – ktm
    Jan 7, 2020 at 14:51
  • 3
    $\begingroup$ @user6014 because it's useful to put several tools together. Besides,I temporarily use 11.3. $\endgroup$ Jan 7, 2020 at 15:24
  • 1
    $\begingroup$ @user6014 Some users may not have access to Mathematica 12.0 yet either, unfortunately. It can be quite expensive. $\endgroup$
    – Michael A
    Jan 7, 2020 at 15:31
  • 1
    $\begingroup$ Are you interested in sympy-based solutions that work only in Mathematica 12.0? Mathematica's Python interface became much better in 12.0. $\endgroup$
    – Szabolcs
    Jan 8, 2020 at 8:31
  • $\begingroup$ @Szabolcs ,So it's very interesting. 11.3 can't solve it, but 12.0 can. 11.3 can't interface with python well, but 12.0 can.Seems the solution is to upgrade to 12.0.😂 $\endgroup$ Jan 8, 2020 at 9:18

1 Answer 1


As an alternative, you may use Rubi:

Int[SinIntegral[x]/x, x]

1/2 x HypergeometricPFQ[{1, 1, 1}, {2, 2, 2}, -I x] + 1/2 x HypergeometricPFQ[{1, 1, 1}, {2, 2, 2}, I x]

  • 1
    $\begingroup$ so it seems like this: 11.3 use Rubi can solve it. 12.0 can solve it directly. 12.0 can also use sympy to solve it. $\endgroup$ Jan 8, 2020 at 9:33

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