[?] Difference between DawsonF[5] and DawsonF[5.0] integration?

Questions
1

What’s the difference between DawsonF[5] and DawsonF[5.0]?
The former doestn’t calculate.
Thanks.

In[815]:= DawsonF[5]

Out[815]= DawsonF[5]

In[816]:= DawsonF[5.0]

Out[816]= 0.102134

In[820]:= DawsonF[1 + I]

Out[820]= DawsonF[1 + I]

In[819]:= DawsonF[1.0 + I]

Out[819]= 0.990373 - 0.638873 I

9033fig1.png
9033fig1.png385×238 2.42 KB

2

Also E^2 does not calculate, just as DawsonF[5] and Dawson[1+I]. They are exact symbolic numbers for which no automatic calculations are done, if you evaluate them in isolation. However, you may try to find different symbolic expressions, for example with

DawsonF[5] // FunctionExpand
DawsonF[1 + I] // FunctionExpand

The floating point in Dawson[5.0] triggers numeric algorithms. You can do it also with N[DawsonF[5]].

3

I get it. Thanks a lot.