¶ x
−1
dx = ln
(
x
)
which is also correctly returned by the CAS. This is not a concern if you are doing problems by hand,
but if you are doing this type of symbolic integration in a program, it is worth checking for this condition
and trapping it as an error, or using the correct solution. Both Mathematica and MathCad return the
general result, too.
(Credit to Bhuvanesh Bhatt)
[2.7] Use when() in integration in AMS 2.03
With AMS version 2.03 and later, you can use the when() function in integration, for example,
á(when(x<1,0,1)x,0,2)
returns 1.
[2.8] Use expand() with respect to variable for faster results
The expand() function may work faster and more completely if you expand with respect to a variable in
the expression. For example, this takes over 11 minutes on my 92 w/Plus module AMS2.03:
expand(((3*a+2*b)/(2*a^2-5*b)*z-(7*a-2*b)/(3*a-b^2))^2)
and also warns that Memory full, simplification might be incomplete, but this takes less than 4 seconds:
expand(((3*a+2*b)/(2*a^2-5*b)*z-(7*a-2*b)/(3*a-b^2))^2,z)
and returns the same result, without the Memory full warning message. Note that the only
difference is that the z variable argument has been added in the second example. The result is
(
3a+2b
)
2
z
2
(
2a
2
−5b
)
2
−
2
(
3a+2b
)(
7a−2b
)
z
(
3a−b
2
)(
2a
2
−5b
)
+
(
7a−2b
)
2
(
3a−b
2
)
2
which is not completely expanded. If this result is again expanded:
expand((3*a+2*b)^2*z^2/(2*a^2-5*b)^2-2*(3*a+2*b)*(7*a-2*b)*z/((3*a-b^2)*(2*a^2-5
*b))+(7*a-2*b)^2/(3*a-b^2)^2)
the result (not shown here) is returned after about 5 seconds, completely expanded.
(Credit to Bhuvanesh Bhatt)
[2.9] Find more symbolic integrals, faster, with Real mode and constraints
Finding symbolic integrals can be more successful when you use constraints and set the Complex
Format to Real instead of Rectangular or Polar. This integral is a good example:
¶
0
∞
t
2
e
−st
dt
which is entered as
2 - 4
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