# 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? – mikado Dec 2 '18 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 – merve Dec 3 '18 at 8:46
• @merve Do you want to calculate the derivative D[R1,c]? – Alex Trounev Dec 3 '18 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. – merve Dec 3 '18 at 12:21
• I updated the code added pictures – Alex Trounev Dec 3 '18 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}]` – Alex Trounev Dec 3 '18 at 18:21