> >>> J.diff(x).doit() > f(x) > > Thanks, it indeed does the job. I just didn't know this method...
Now, I have a second problem wich a slight modification : >>> J1 = f(x-t).integrate((t,0,x)) >>> print J1.diff(x) f(0) + Integral(Subs(Derivative(f(_xi_1), _xi_1), (_xi_1,), (-t + x,)), (t, 0, x)) This result, even with the nice MathJax rendering in IPython Notebook is pretty difficult to understand and the "doit" trick doesn't help. I did a separate pen and paper derivation and came to the conclusion that J1.diff(x) is actually f(x) However, I do not trust my integration skills which is the reason why I asked sympy for this case. Is it a kind of symbolic integration that is not supported by current sympy ? Or is just me doing bad integration ;-) ... Best, Pierre -- You received this message because you are subscribed to the Google Groups "sympy" group. To view this discussion on the web visit https://groups.google.com/d/msg/sympy/-/zDVV1g-b-LQJ. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/sympy?hl=en.
