# Symbolic integration of Exponential

How can I solve this integration? I want to solve this integration with c, B, H, Y, b, w surviving as constants in the result:

ClearAll[a, b, t, R1, R2 c, t, H, Y, w]
syms = {c, B, H, Y, b, w}
a = Exp[((t)^(c) - 1)]

R1 = Integrate[a^(-3*w - 3), t]


Second question

a = Exp[(1/\[Beta]) *((t^\[Beta]) - 1)];
R1 = a*(2/(3*w))*
Exp[(2/(3*w))*Integrate[Exp[-m*t^k], t] /. {m -> 3*(w + 1)/\[Beta],
k -> \[Beta]}]

f = D[R1, t]

p1 = Plot[f /. {\[Beta] -> .58, w -> -1.1}, {t, 0, 10},
AxesOrigin -> {0, 0}]

• I don't think there is anything wrong with your code. Mathematica can't do the integral. However, it can do it for specific integer values of c. Does that help? Dec 2, 2018 at 20:26

It is necessary to use the well-known integral

Integrate[Exp[-m*t^k], t]

(*Out[]= -((t (m t^k)^(-1/k) Gamma[1/k, m t^k])/k)*)


Let's transform

a = Exp[((t)^(c) - 1)];(* a^(-3*w - 3)=Exp[3*w+3]*Exp[-3*(w+1)*t^c]*);


Consequently

R1=Exp[3 + 3*w]*Integrate[Exp[-m*t^k], t] /. {m -> 3*(w + 1), k -> c}
(*Out[]= -((3^(-1/c) E^(3 + 3 w) t (t^c (1 + w))^(-1/c)
Gamma[1/c, 3 t^c (1 + w)])/c)*)


Let's check that the derivative D[R1,t]coincides with the original function.

R1 = -((3^(-1/c) E^(3 + 3 w) t (t^c (1 + w))^(-1/c) Gamma[1/c,
3 t^c (1 + w)])/c);
f = D[R1, t]
(*Out[]=E^(3 + 3 w - 3 t^c (1 + w)) + (
3^(-1/c) E^(3 + 3 w) t^c (1 + w) (t^c (1 + w))^(-1 - 1/c)
Gamma[1/c, 3 t^c (1 + w)])/c - (
3^(-1/c) E^(3 + 3 w) (t^c (1 + w))^(-1/c) Gamma[1/c, 3 t^c (1 + w)])/
c*)
{p1 = Plot[f /. {c -> .58, w -> -1.1}, {t, 0, 10},
AxesOrigin -> {0, 0}],
p2 = Plot[a^(-3*w - 3) /. {c -> .58, w -> -1.1}, {t, 0, 10},
PlotStyle -> Orange, AxesOrigin -> {0, 0}], Show[p1, p2]}


• thank you, how can ı get the differentiation of R1 with m, c surviving as constants in the result Dec 3, 2018 at 8:46
• @merve Do you want to calculate the derivative D[R1,c]? Dec 3, 2018 at 12:10
• not actually D[R1,c], I want to calculate D[R1,t] and then plot R1 versus t (with the range of 0-10). with c= 0.58, w = -1.1 values but I couldn't. Dec 3, 2018 at 12:21
• I updated the code added pictures Dec 3, 2018 at 13:46
• This is due to the fact that the function f[t] is complex, so use p1 = Plot[Re[f /. {\[Beta] -> .58, w -> -1.1}], {t, 0, 10}, AxesOrigin -> {0, 0}]` Dec 3, 2018 at 18:21