Bug introduced in 10.2.0 or earlier and fixed by 11.0.

Integrate[Exp[c x]/x, {x, -a, a}]


ConditionalExpression[-ExpIntegralEi[-a c] + ExpIntegralEi[a c], Re[a] > 0 && Im[a] == 0]

but the integral does not actually converge for any values of c or a (because near x = 0, Exp[c x] ~ 1, so the integrand has a 1/x singularity). If I replace either c or a with an actual number, then Mathematica correctly says that the integral does not converge - even if I replace c with a positive real number, satisfying the returned condition.


3 Answers 3


Expanded in response to comment by OP

The result in the question also is obtained with Mathematica "10.3.0 for Microsoft Windows (64-bit) (October 9, 2015)".

Integrate[Exp[c x]/x, {x, -a, a}]
(* ConditionalExpression[-ExpIntegralEi[-a c] + ExpIntegralEi[a c], 
   Re[a] > 0 && Im[a] == 0] *)

which can be simplified for a > 0 && c > 0.

FullSimplify[s1, a > 0 && c > 0]
(* 2 SinhIntegral[a c] *)

(The same result is obtained for a > 0 && c < 0 but not for a > 0 && c == 0.) This is the Cauchy Principal Value result.

Integrate[Exp[c x]/x, {x, -a, a}, Assumptions -> a > 0, PrincipalValue -> True]
(* 2 SinhIntegral[a c] *)

which seems strange, because PrincipalValue -> False is the default. Thus,

Integrate[Exp[c x]/x, {x, -a, a}, Assumptions -> a > 0 && c > 0]

returns unevaluated with the expected error message

Integrate::idiv: Integral of E^(c x)/x does not converge on {-a,a}. >>

Giving a and c specific real values likewise produces this error message, as noted by the OP's comment. Perhaps, Integrate becomes confused, when no constraints are placed on the parameters a and c.

  • $\begingroup$ The Cauchy principal value is a very specific method for dealing with divergent integrands, and I do not think it is standard to assume it is being used when not specified. And more to the point, why does Mathematica assume it should only use the CPV when c and a are left as variables, but not for specific numerical values? $\endgroup$
    – tparker
    Dec 5, 2015 at 10:02
  • $\begingroup$ I assumed the setting for PrincipalValue was Automatic, but you're right, it's False. I guess a user-setting for PrincipalValue makes Integrate choose an approach that does not try to avoid the singularity. $\endgroup$
    – Michael E2
    Dec 5, 2015 at 15:24
  • 1
    $\begingroup$ @MichaelE2 Setting PrincipalValue -> False does not change the result cited in the question. Perhaps, this might be described as a bug. By the way, +1 for your answer. $\endgroup$
    – bbgodfrey
    Dec 5, 2015 at 15:32
  • $\begingroup$ Of course. Somehow I confused myself. I'll update my answer. $\endgroup$
    – Michael E2
    Dec 5, 2015 at 15:38

It appears that this is a bug, and Mathematica is incorrectly evaluating the Cauchy Principal Value of the integral despite the default PrincipalValue -> False (and continues to do so even if you manually specify this).


Since PrincipalValueis by default set it to False (thanks, @bbgodfrey), it helps to indicate that a is a positive real number:

Integrate[Exp[c x]/x, {x, -a, a}, Assumptions -> a > 0]

Mathematica graphics

The way to look at Integrate is that it tries to find a generic integral over a complex line. Sometimes you have to give it extra information to get it to think right. Sometimes even when you do, it still has trouble analyzing the complex-number situation.

Compare with this complex line integral that goes around zero:

Integrate[Exp[c x]/x, {x, -a, I, a}, Assumptions -> a > 0]
 -ExpIntegralEi[-a c] + ExpIntegralEi[a c],
 (c == 0 || Im[c] <= 0 || Re[c] <= 0 || a Im[c] + Re[c] <= 0) &&
  (Re[c] >= 0 || c == 0 || Im[c] <= 0 || a Im[c] <= Re[c])]

Note: This integral is off by Pi I, so this falls into the doesn't-quite-get-it-right category.

A more direct way to get divergence, assuming you know where the singularity is:

Integrate[Exp[c x]/x, {x, -a, 0, a}]

Mathematica graphics


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