Mathematica couldn't solve following using just integrate function. I checked on interpolating as well, that also turned to be a dead end. Can someone help me with evaluating this integral? If it helps one can differentiate with respect to t and get an integro-differential equation and solve that. I tried both.

a,b,τ,t are all real numbers. Initial condition is σ[0] =0.5;

enter image description here

σ[t] == Assuming[{Element[{τ, t, a, b}, Reals], τ >= 0, t >= 0}, Integrate[Integrate[a (-1 + Exp[b (-1/(t - τ)^2 - 1/(I + t - τ)^2 + π^2 Csch[π (t-τ)]^2)]), {τ, 0, t}] σ[t], {t, 0, t}]]

Integro-differential equation is of the form,

eqn = Derivative[1][σ][t] == a*Assuming[{Element[{τ,t,a,b},Reals],τ>= 0,t>=0},Integrate[(Exp[b (-1/(t - τ)^2 - 1/(I + t - τ)^2 + π^2 Csch[π (t-τ)]^2)]-1)*σ[t],{τ,0,t}];
  • $\begingroup$ Please, always share your code in copyable form. $\endgroup$ – Henrik Schumacher Nov 30 '19 at 23:13
  • $\begingroup$ @HenrikSchumacher Done. Thanks for pointing it out. $\endgroup$ – Kobe Dec 1 '19 at 0:07
  • 3
    $\begingroup$ Have you tried the free Rubi integration add on package for Mathematica at rulebasedintegration.org to see if it can make any progress if you attempt just your inner integral and only do this as an indefinite integration? If it can make no progress then that may mean symbolic integration is very unlikely to succeed. $\endgroup$ – Bill Dec 1 '19 at 2:38
  • $\begingroup$ @Bill Thanks for input. I checked that out and it couldn't solve it also. Thanks for introducing me to the package. :) $\endgroup$ – Kobe Dec 1 '19 at 9:02

Think, you can only solve it numericaly.

Define the inner integral and solve the differential equation for t,a,b values of interest. (takes a few seconds).

nint1[ts_?NumericQ, a_?NumericQ, b_?NumericQ] := 
      a (-1 + Exp[
  b (-1/(ts - \[Tau])^2 - 
     1/(I + ts - \[Tau])^2 + \[Pi]^2 Csch[\[Pi] (ts - \
  \[Tau])]^2)]), {\[Tau], 0, ts}]

ndsol := NDSolve[{Derivative[1, 0, 0][\[Sigma]][t, a, b] == 
nint1[t, a, b]*\[Sigma][t, a, b], \[Sigma][0, a, b] == 
1/2}, \[Sigma], {t, 0, 10}, {a, 1/100, 3}, {b, 1/100, 4}]

\[Sigma]sol[t_, a_, b_] = \[Sigma][t, a, b] /. First@ndsol

 Plot[{Re[\[Sigma]sol[t, a, b]], Im[\[Sigma]sol[t, a, b]]}, {t, 0, 
  10}, PlotRange -> All, PlotStyle -> {Blue, Red}], {{a, 1}, 1/100, 
 3}, {{b, 1}, 1/100, 4}]

enter image description here

(Sorry for the bad layout. Had no more time.)

| improve this answer | |
  • $\begingroup$ Thank you very much for this. You have been very helpful. I have been trying this for so long. $\endgroup$ – Kobe Dec 2 '19 at 8:47
  • $\begingroup$ Thank you. I'm glad I could help. $\endgroup$ – Akku14 Dec 3 '19 at 13:26
  • $\begingroup$ Additional tip: Since for all a,b values NIntegrate starts from zero for every wanted t-value from 0 to 10, you can accelerate calculation a lot, if you force intermediate integration results to be stored. Define nint2[ts_?NumericQ, a_?NumericQ, b_?NumericQ] := nint2[ts, a, b] = NIntegrate[ a (-1 + Exp[ b (-1/(ts - \[Tau])^2 - 1/(I + ts - \[Tau])^2 + \[Pi]^2 Csch[\[Pi] (ts - \ \[Tau])]^2)]), {\[Tau], 0, ts}] and use nint2 in NDSolve . $\endgroup$ – Akku14 Dec 3 '19 at 16:56

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