Integrals evaluating to zero

Questions
1

Hi all,

I am having trouble computing some integrals of type Fourier transforms. I was always able to compute them but since version 13 the very same code evaluates these integrals to zero and I have no idea why.

One example is below:

Integrate[(E^(-k \[Pi] + (k \[Gamma])/2) (-1 + E^(k \[Gamma])))/(-1 + E^(-k \[Pi])) Exp[I k x], 
{k, \[Infinity], \[Infinity]}, Assumptions -> 0 < \[Gamma] < \[Pi]/5]

Any ideas?

Thanks!

2

I have no idea if this might help you or not, but

FullSimplify[
  Integrate[(E^(-k Pi+(k γ)/2)(-1+E^(k γ)))/(-1+E^(-k Pi))Exp[I k x],{k,-Infinity,0},Assumptions->0<γ<Pi/5]+
  Integrate[(E^(-k Pi+(k γ)/2)(-1+E^(k γ)))/(-1+E^(-k Pi))Exp[I k x],{k,0,Infinity},Assumptions->0<γ<Pi/5],
 0<γ<Pi/5]

returns

ConditionalExpression[Csch[x-I/2*γ]*Csch[x-3*I/2*γ]*Sin[γ],γ>2*Im[x]&&2*(Pi+Im[x])>3*γ]

Is there any chance that might be something that you could use

3

Try:

 FourierTransform[(E^(-k \[Pi] + (k \[Gamma])/2) (-1 + 
       E^(k \[Gamma])))/(-1 + E^(-k \[Pi])), k, x, 
  Assumptions -> 0 < \[Gamma] < \[Pi]/5, FourierParameters -> {1, -1}]

 (*(2 Sin[\[Gamma]])/(-Cos[\[Gamma]] + Cosh[2 x + 2 I \[Gamma]])*)
4

Thanks! This worked

5

Thanks! This works too

6

I filed this as a bug report.

7

Okay, after much scratching of heads I found the issue. Your lower bound was intended to be -Infinity but was in fact (positive) Infinity.

With that sorted out, here is one way to improve matters: add an assumption that x is real-valued. Below is that I get in version 13.2.1.

In[1]:= InputForm[Timing[ft = Integrate[                                        
E^(-k*Pi + k*gamma/2)*(-1 + E^(k*gamma))/(-1 + E^(-k*Pi))*Exp[I*k*x],           
{k, -Infinity, Infinity},                                                       
Assumptions -> {0 < gamma < Pi/5, Element[x,Reals]}]]]                          

(* Out[1]//InputForm= {46.270603, -Cot[gamma/2 + I*x] + Cot[(3*gamma)/2 + I*x]} *)