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I need to evaluate the following expression,

$\frac{1}{\Gamma(1-\alpha)} \frac{d}{d x} \int_{0}^{x}(x-t)^{-\alpha}(\exp(\lambda t)) d t, 0<\alpha<1, \lambda\in \mathbb{C}$

using Mathematica to obtain $\sum_{k=0}^{\infty} \frac{\lambda^{k} x^{k-\alpha}}{\Gamma(k+1-\alpha)}$.

So I used the following code

Assuming[
 Re[x] > 0 && Im[x] == 0 && Re[- α] > -1, 
 1/Gamma[1 - α] D[Integrate[(x - t)^(-α) Exp[λ t], {t, 0, x}], x]
]

And obtain the following output

(λ^α (x λ)^-α + E^(x λ) λ^α (Gamma[1 - α] - Gamma[1 - α, x λ]))/Gamma[1 - α]

Can anyone please help me to obtain the equation

$\sum_{k=0}^{\infty} \frac{\lambda^{k} x^{k-\alpha}}{\Gamma(k+1-\alpha)}$

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  • 4
    $\begingroup$ A definite integral of a function $f(t)$ with respect to $t$ in the limits $0\le t\le x$ cannot depend on $t$ $\endgroup$ – yarchik Oct 11 '19 at 8:16
  • $\begingroup$ if you do this to your result Map[FullSimplify, Series[%1, {x, 0, 5}, Assumptions -> x > 0] // Normal // Expand] you can check that it works $\endgroup$ – chris Oct 13 '19 at 15:52
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I guess, you can take the n-th term of the expansion of the exponent (which is as you know, (l*t)^n/n! where I put l instead of lambda. Then integrate it:

Assuming[
 n > 0 && n \[Element] Integers && Re[x] > 0 && Im[x] == 0 && 
  Re[-\[Alpha]] > -1, 
 1/Gamma[1 - \[Alpha]] D[
   Integrate[(x - t)^(-\[Alpha])*(l*t)^n/n!, {t, 0, x}], x]]

(*  (l^n x^(n - \[Alpha]) (1 + n - \[Alpha]))/Gamma[
 2 + n - \[Alpha]]   *)

and here you get it.

Have fun!

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