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Re: Mathematica not considering an assumption

matematica65 at wrote:
> Assuming[a < 0,Integrate[x^a, {x, 5, 6}]}]
> This Integral, for a = -1, is equal to Log(6/5)
> Why doesn't Mathematica report it?

[I'm using version 5.2, but that probably doesn't make any difference

First, if you replace a with -1 in the definite integral, it is hardly
surprising that Mathematica gets the answer we all expect:

In[6]:= Integrate[x^(-1), {x, 5, 6}]

Out[6]= Log[6/5]

It might also be worthwhile to note that

In[7]:= Assuming[a == -1, Integrate[x^a, {x, 5, 6}]]

Out[7]= Log[6/5]

Now, in your title, you seem concerned with "Mathematica not considering an
assumption". I can understand, perhaps, why you say that because both

Integrate[x^a, {x, 5, 6}] and Assuming[a < 0, Integrate[x^a, {x, 5, 6}]]

produce the same result, namely,

Out[9]= (-5^(1 + a) + 6^(1 + a))/(1 + a)

Thus, since your assumption, a < 0, did not affect the result, you seem to
have concluded that Mathematica was (incorrectly?) ignoring your
assumption. And I suppose you realize that merely replacing a with -1 in
Out[9] yields Indeterminate, rather than your desired Log[6/5].

But you need to ask yourself:
How could the result of Integrate[x^a, {x, 5, 6}] or
Assuming[a < 0, Integrate[x^a, {x, 5, 6}]] be improved?

One possibility which might occur to you is that the result should be given
using Piecewise so that, when a == -1, we get Log[6/5], and otherwise, we
get Out[9]. But does that constitute an improvement? You may think so,
while the people at Wolfram Research (and other CASs, for that matter)
might disagree.

Another possibility, due to the fact that the singularity in Out[9] is
removable, is that the result could be given as

Limit[(-5^(1 + u) + 6^(1 + u))/(1 + u), u -> a]

But again we must ask if that would constitute an improvement or not.

David W. Cantrell

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