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I was trying to compute the following integration, $$I = \int_0^{\infty} dt_2\int_0^{\infty} dt_1 \frac{1}{t-t_1-t_2-t_3}\exp{(i\Omega(t_1+t_1+t_1)-i\omega_0t_1)}$$

I have taken $\Omega = 5.12$ and $\omega_0=2.35$. Instead of doing double integration in one NIntegrate, if was trying the following,

int1[t_?NumericQ, t2_?NumericQ, t3_?NumericQ]:= NIntegrate[(t - t1 - t2 - t3)^-1*Exp[I*(\[CapitalOmega]*(t1 + t1 + t1) - \[Omega]0*t1)], {t1, 0, \[Infinity]}, Method -> "LocalAdaptive"] int2[t_, t3_] := NIntegrate[int1[t, t2, t3], {t2, 0,\[Infinity]}] 

But the evaluation of int2[1,1] gives the following error,

enter image description here

I have a difficult time understanding the error. Does anyone have a solution to this?

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    $\begingroup$ For one thing, t3 has no value -- for NIntegrate to work, all variables must be assigned values. $\endgroup$ Commented Aug 23, 2022 at 20:56
  • $\begingroup$ Should $t_1+t_1+t_1$ be $t_1+t_2+t_3$ in both the integration and Mathematica code? $\endgroup$ Commented Aug 23, 2022 at 20:58
  • $\begingroup$ I have edited the question. Actually, int2 have two parameters, so I want to evaluate at any value of t and t3. I tried for t=1 and t3=1 $\endgroup$ Commented Aug 23, 2022 at 20:59
  • $\begingroup$ @JimB No, I get it why you think that but that is just 3*t1, I just copy paste from my notebook. $\endgroup$ Commented Aug 23, 2022 at 21:00
  • $\begingroup$ Is it because the power -1 is not in parenthesis? $\endgroup$ Commented Aug 24, 2022 at 0:55

1 Answer 1

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Do first integration anylytically with Integrate to see, second integral does not converge.

f = (t - t1 - t2 - t3)^-1* Exp[I*(\[CapitalOmega]*(t1 + t1 + t1) - \[Omega]0* t1)] /. {\[CapitalOmega] -> 512/100, \[Omega]0 -> 235/100}; int1[t_, t2_, t3_] = Integrate[f, {t1, 0, \[Infinity]}, Assumptions -> Thread[{t, t2, t3} > 0]] (* ConditionalExpression[-I E^( 1301/100 I (t - t2 - t3)) (\[Pi] + I ExpIntegralEi[-(1301/100) I (t - t2 - t3)]), t < t2 + t3] *) 

NIntegrate Integrate faile both. Series shows, int1 to go to infinity according to 1/t2, means it does not converge. Behavior at zero is integrable.

NIntegrate[int1[1, t2, 1] // Normal, {t2, 0, Infinity}, MaxRecursion -> 500, AccuracyGoal -> 5] ser0 = Series[int1[1, t2, 1], {t2, 0, 0}, Assumptions -> t2 > 0] // Normal (* EulerGamma - (I \[Pi])/2 + Log[1301/100] + Log[t2] *) ser1 = Series[int1[1, t2, 1], {t2, Infinity, 1}] // Normal (* -((100 I)/(1301 t2)) *) 
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