Message Boards Message Boards

0
|
8210 Views
|
17 Replies
|
5 Total Likes
View groups...
Share
Share this post:

Inconsistency in evaluating a definite integral?

Consider the integral $ \int_1^{\infty } \log \left(1-\frac{\alpha }{x}\right) \, dx$ where $0<\alpha <1$

This integral is divergent as can be verified by evaluating the indefinite integral: $x \log \left(1-\frac{\alpha }{x}\right)-\alpha \log (x-\alpha )$ between the two limits.

However, the Mathematica input: $\text{Assuming}\left[0<\alpha <1,\int_1^{\infty } \log \left(1-\frac{\alpha }{x}\right) \, dx\right]$

produces the output: $(\alpha -1) \log (1-\alpha )-\alpha$
POSTED BY: Joel Storch
17 Replies
Sort By:
Bernd Günther
Bernd Günther
Posted 5 years ago

Thanks!
POSTED BY: Bernd Günther
Bernd Günther
Bernd Günther
Posted 5 years ago

One more item (this time a limit instead of an integral):

Limit[ParabolicCylinderD[\[Nu], -t]/(
 Sqrt[2 Pi]/Gamma[-\[Nu]] t^(-\[Nu] - 1) Exp[t^2/4]) , 
 t -> \[Infinity]]

Mathematica evaluates to

-\[Infinity] Gamma[-\[Nu]] Gamma[1 + \[Nu]] Sin[\[Pi] \[Nu]]

The correct value would be -1, according to Gradshteyn-Ryzhik formula 9.246 part 2.
POSTED BY: Bernd Günther

Numerical checks make me suspect that the limit should be 1 rather than -1. But clearly the current behavior is wrong. I'll file a bug report for that.
POSTED BY: Daniel Lichtblau
Bernd Günther
Bernd Günther
Posted 5 years ago

Anyway, I have more gems of this sort:

Integrate[Csc[t] Log[1 + x Sin[t]], {t, -(\[Pi]/2), \[Pi]/2}, 
 Assumptions -> 0 < x < 1]

Mathematica evaluates this to 0. The correct result would be

\[Pi] ArcSin[x]

as can be seen from series expansions.
POSTED BY: Bernd Günther

This will also be filed as a bug.
POSTED BY: Daniel Lichtblau
Bernd Günther
Bernd Günther
Posted 5 years ago

Thanks!
POSTED BY: Bernd Günther
Bernd Günther
Bernd Günther
Posted 5 years ago

Compelling reason? No, but not forbidden either. One can enter formulae like this either with escape syntax or using the class room editor.
POSTED BY: Bernd Günther
Bernd Günther
Bernd Günther
Posted 5 years ago

If that helps: 

Simplify[\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(r1\), \(r2\)]\(
\*FractionBox[\(1\), \(
\*SuperscriptBox[\(r\), \(2\)]\ 
\*SqrtBox[\(\((r - r1)\) \((r2 - r)\)\)]\)] \[DifferentialD]r\)\), 
 Assumptions -> (r1 > 0) && (r2 > 0) && (r2 > r1)]

2nd formula:

Simplify[\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(a\), \(b\)]\(
\*FractionBox[\(1\), \(
\*SuperscriptBox[\(r\), \(2\)]\ 
\*SqrtBox[\(\((r - a)\) \((b - r)\)\)]\)] \[DifferentialD]r\)\), 
 Assumptions -> (a > 0) && (b > 0) && (b > a)]

3rd one:

Integrate[1/(r^2 Sqrt[(r - r1) (r2 - r)]), {r, r1, r2}, 
 Assumptions -> (r1 > 0) && (r2 > 0) && (r2 > r1)]
POSTED BY: Bernd Günther

Is there a compelling reason to enter the first two using the box formatting? (This is a rhetorical question. The answer is "no". The problems with that formatting are that it is not readable and also does not copy/paste into a raw kernel on my Linux machine.) General rule: the simpler the code provided, the more likely one is to get a response.

As for the issue at hand, it appears that the integral, when not provided the assumptions, is off by a sign.
POSTED BY: Daniel Lichtblau
Bernd Günther
Bernd Günther
Posted 5 years ago

There seem to be more inconsistencies, for example (sorry, I must include this as picture) the following. The funny thing is that the result depends minutely on how the formula is entered:

enter image description here
POSTED BY: Bernd Günther

I find it impossible to believe that the code could not have been included in this message. I am unable to copy/paste an image.
POSTED BY: Daniel Lichtblau

Reported as a bug.
POSTED BY: Daniel Lichtblau

Mathematica gets the parametric integral wrong:

Assuming[0 < a < 1, Integrate[Log[1 - a/x], {x, 1, \[Infinity]}]]

but it gets it right when the parameter a is a number:

With[{a = 1/2}, Integrate[Log[1 - a/x], {x, 1, \[Infinity]}]]

If you force Evaluate on the integral, the wrong result will win the evaluation race, before the parameter a gets a numeric value. Remove Evaluate and the parameter a will become numeric before the integral is evaluated:

Manipulate[ Integrate[Log[1 - a/x], {x, 1, [Infinity]}], {a, 0.001, 0.999}]
POSTED BY: Gianluca Gorni

I have used

Manipulate[Evaluate[Assuming[0 < a < 1, Integrate[Log[1 - a/x], {x, 1, \[Infinity]}]]], {a, 0.001, 0.999}]

and for all the values of a $\in$ [0.001; 0.999] the integral values are finite. It's very confusing...
POSTED BY: Valeriu Ungureanu

I am unclear as to your statement, "It seems to be a correct output". The integral is either convergent or divergent, it can't be both. The evaluation of the indefinite integral is correct as can be verified by differentiation. Clearly, this approach proves that the integral is divergent. As you verified, the input: $\text{Assuming}\left[0<\alpha <1,\int_1^{\infty } \log \left(1-\frac{\alpha }{x}\right) \, dx\right]$ yields a finite result; a contradiction ! Note that if we evaluate the integral with $\alpha$=1/2 i.e $\int_1^{\infty } \log \left(1-\frac{1}{2 x}\right) \, dx$, Mathematica says that the integral diverges.
POSTED BY: Joel Storch

It seems to be a correct output as both the limits (x->1,x-> $\infty$) tend to $-\infty$. I attached the nb file with the code I used for an experiment.

Experiment Results...

Attachments:
POSTED BY: Valeriu Ungureanu

Inconsistency in Evaluating Definite Integral
POSTED BY: Valeriu Ungureanu
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Tag limit exceeded
Note: Only the first five people you tag will receive an email notification; the other tagged names will appear as links to their profiles.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract