I'm new to Wolfram Mathematica and I was trying to evaluate this integral in particular:

$$\int_{0}^{x}\frac{e^t \operatorname{erf}(\sqrt{t})\sqrt{\pi t}+1}{\sqrt{\pi t x -\pi t^2}}\mathrm dt$$

The code I tried first was:

Integrate[(E^t*Erf[Sqrt[t]]*Sqrt[Pi*t] + 1)/Sqrt[Pi*t*x - Pi*t^2], {t, 0, x}]

But the output just returns the input; what am I doing wrong? I tried using Assumptions -> x > 0 but it doesn't work either. Can someone help me?

  • 2
    $\begingroup$ "the output just returns the input" - that means Mathematica doesn't know what to do with it. Did you have any reason to expect a closed form? $\endgroup$ Jul 5, 2022 at 21:15
  • $\begingroup$ You need to use E rather than e; however, it still does not evaluate. $\endgroup$
    – Bob Hanlon
    Jul 5, 2022 at 21:53
  • $\begingroup$ @J.M. I solved the half derivative of $e^x$ and I wanted to solve the half derivative of the answer as well to see if it checked out. So basically what I'm trying to do is to find the half derivative of $\frac{e^x erf(\sqrt{x})\sqrt{\pi x}+1}{\sqrt{\pi x}}$ $\endgroup$ Jul 5, 2022 at 22:26
  • $\begingroup$ @BobHanlon Thank you, fixed it... It's sad that it doesn't work. $\endgroup$ Jul 5, 2022 at 22:29
  • 1
    $\begingroup$ @DavidG.Stork It's the fractional derivative with order = 1/2. $\endgroup$ Jul 6, 2022 at 0:10

2 Answers 2


Re-expressing in terms of the regularized incomplete gamma function makes things work:

Assuming[x > 0, Integrate[Exp[t] (1 - GammaRegularized[-1/2, t])/Sqrt[x - t], {t, 0, x}]]
   E^x Sqrt[π]

Nevertheless, Mathematica should still have been smart enough to do something about this; you should perhaps report this to Support.


Suppose we want to find an anti derivative of e^x. We might try this.
enter image description here
Which Gives
e^t - 1
That's not wrong, but not what we really want, so instead we do this:
enter image description here
which gets us the e^t without the constant.

Now, if we take the fractional integral of e^t from 0 to x using Cauchy's formula
enter image description here
we get enter image description here the derivative of which you have already found to be enter image description here

However, like above, if we change limits of integration in Cauchy's formula to go from negative infinity to x:
enter image description here
we instead get just e^x as the half antiderivative of e^x and therefore e^x is also the half derivative of e^x.



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