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I'm looking to the result of the following integral: $$\int_0^{\infty} f(g)\exp(-tg) dg $$ where: $f(g) = \frac{ckg^{(c-1)}}{(1+g^c)^{k+1}}$ is a PDF not defined in 0 but with $\int_0^\infty f_G(g) = 1$, and $c$, $k$, $t$ are variables.

Knowing that for both the functions: $f(g)$ and $\exp(-tg)$ the integral exist, I don't understand with I am not able to compute integral of the product of the two functions using Mathematica with the following expression:

c = 0.6448;
k = 0.787;    
Integrate[c*k*(g^(c - 1)/(1 + g^c)^(k + 1))*Exp[-t*g], {g, 0, Infinity}].

I do not want to compute the integral in a numeric way, I'm just looking for an expression that simplify the integral, considering that this is just an inner integral. Thus I can proceed numerically with the computation of the others external integrals.

Any suggestion on how I can obtain a simplified expression?

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Your integral can be computed with MA for any rational $c$. First, notice that we are dealing here with a Laplace transform of a derivative. For the Laplace transform $$ \int_0^\infty f(t) e^{-st} t= F(s) $$ we have $$ \int_0^\infty f'(t) e^{-st} t= sF(s)-f(0^-) $$

Thus we can focus on a simpler integral: $\int_0^\infty\frac{1}{\left(g^c+1\right)^{k}} e^{-g t} dg$.

With enough patience, it can be computed for any rational $c$, for instance for $c=3/5$, and $t>0$ the result is:

$${\tiny \Gamma \left(1-\frac{3 k}{5}\right) \, _4F_6\left(\frac{k}{5}+\frac{1}{5},\frac{k}{5}+\frac{2}{5},\frac{k}{5}+\frac{3}{5},\frac{k}{5}+\frac{4}{5};\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\frac{k}{5}+\frac{1}{3},\frac{k}{5}+\frac{2}{3};\frac{t^3}{27}\right) t^{\frac{3 k}{5}-1}+\frac{20 \, _5F_7\left(1,\frac{6}{5},\frac{7}{5},\frac{8}{5},\frac{9}{5};\frac{4}{3},\frac{5}{3},\frac{6}{5}-\frac{k}{5},\frac{7}{5}-\frac{k}{5},\frac{8}{5}-\frac{k}{5},\frac{9}{5}-\frac{k}{5},2-\frac{k}{5};\frac{t^3}{27}\right) t^2}{(k-5) (k-4) (k-3) (k-2) (k-1)}-\frac{5^k \Gamma \left(-\frac{4}{15}\right) \Gamma \left(\frac{1}{3}\right) \Gamma \left(\frac{14}{15}\right) \Gamma \left(\frac{17}{15}\right) \Gamma \left(\frac{23}{15}\right) \Gamma \left(\frac{k}{5}-\frac{1}{3}\right) \Gamma \left(\frac{k}{5}-\frac{2}{15}\right) \Gamma \left(\frac{k}{5}+\frac{1}{15}\right) \Gamma \left(\frac{k}{5}+\frac{4}{15}\right) \Gamma \left(\frac{k}{5}+\frac{7}{15}\right) \, _4F_6\left(\frac{8}{15},\frac{11}{15},\frac{14}{15},\frac{17}{15};\frac{2}{3},\frac{8}{15}-\frac{k}{5},\frac{11}{15}-\frac{k}{5},\frac{14}{15}-\frac{k}{5},\frac{17}{15}-\frac{k}{5},\frac{4}{3}-\frac{k}{5};\frac{t^3}{27}\right)}{96 \pi ^4 \Gamma (k)}-\frac{5^k t \Gamma \left(\frac{1}{15}\right) \Gamma \left(\frac{2}{3}\right) \Gamma \left(\frac{19}{15}\right) \Gamma \left(\frac{22}{15}\right) \Gamma \left(\frac{28}{15}\right) \Gamma \left(\frac{k}{5}-\frac{2}{3}\right) \Gamma \left(\frac{k}{5}-\frac{7}{15}\right) \Gamma \left(\frac{k}{5}-\frac{4}{15}\right) \Gamma \left(\frac{k}{5}-\frac{1}{15}\right) \Gamma \left(\frac{k}{5}+\frac{2}{15}\right) \, _4F_6\left(\frac{13}{15},\frac{16}{15},\frac{19}{15},\frac{22}{15};\frac{4}{3},\frac{13}{15}-\frac{k}{5},\frac{16}{15}-\frac{k}{5},\frac{19}{15}-\frac{k}{5},\frac{22}{15}-\frac{k}{5},\frac{5}{3}-\frac{k}{5};\frac{t^3}{27}\right)}{624 \pi ^4 \Gamma (k)}+\frac{3^{-\frac{3 k}{5}-\frac{29}{10}} k (k+1) (k+2) (k+3) t^{\frac{3 k}{5}+\frac{7}{5}} \csc \left(\frac{\pi }{5}-\frac{k \pi }{5}\right) \Gamma \left(-\frac{k}{5}-\frac{7}{15}\right) \Gamma \left(-\frac{k}{5}-\frac{2}{15}\right) \, _4F_6\left(\frac{k}{5}+1,\frac{k}{5}+\frac{6}{5},\frac{k}{5}+\frac{7}{5},\frac{k}{5}+\frac{8}{5};\frac{6}{5},\frac{7}{5},\frac{8}{5},\frac{9}{5},\frac{k}{5}+\frac{17}{15},\frac{k}{5}+\frac{22}{15};\frac{t^3}{27}\right)}{16 \Gamma \left(\frac{k}{5}+\frac{4}{5}\right)}-\frac{3^{-\frac{3 k}{5}-\frac{1}{10}} k t^{\frac{3 k}{5}-\frac{2}{5}} \Gamma \left(\frac{2}{15}-\frac{k}{5}\right) \Gamma \left(\frac{7}{15}-\frac{k}{5}\right) \Gamma \left(\frac{4}{5}-\frac{k}{5}\right) \, _4F_6\left(\frac{k}{5}+\frac{2}{5},\frac{k}{5}+\frac{3}{5},\frac{k}{5}+\frac{4}{5},\frac{k}{5}+1;\frac{2}{5},\frac{3}{5},\frac{4}{5},\frac{6}{5},\frac{k}{5}+\frac{8}{15},\frac{k}{5}+\frac{13}{15};\frac{t^3}{27}\right)}{2 \pi }+\frac{3^{-\frac{3 k}{5}-\frac{7}{10}} k (k+1) t^{\frac{3 k}{5}+\frac{1}{5}} \Gamma \left(-\frac{k}{5}-\frac{1}{15}\right) \Gamma \left(\frac{4}{15}-\frac{k}{5}\right) \Gamma \left(\frac{3}{5}-\frac{k}{5}\right) \, _4F_6\left(\frac{k}{5}+\frac{3}{5},\frac{k}{5}+\frac{4}{5},\frac{k}{5}+1,\frac{k}{5}+\frac{6}{5};\frac{3}{5},\frac{4}{5},\frac{6}{5},\frac{7}{5},\frac{k}{5}+\frac{11}{15},\frac{k}{5}+\frac{16}{15};\frac{t^3}{27}\right)}{4 \pi }-\frac{3^{-\frac{3 k}{5}-\frac{23}{10}} 5^{k+\frac{5}{2}} t^{\frac{3 k}{5}+\frac{4}{5}} \Gamma \left(-\frac{k}{5}-\frac{4}{15}\right) \Gamma \left(\frac{1}{15}-\frac{k}{5}\right) \Gamma \left(\frac{k}{5}+\frac{4}{5}\right) \Gamma \left(\frac{k}{5}+1\right) \Gamma \left(\frac{k}{5}+\frac{6}{5}\right) \Gamma \left(\frac{k}{5}+\frac{7}{5}\right) \, _4F_6\left(\frac{k}{5}+\frac{4}{5},\frac{k}{5}+1,\frac{k}{5}+\frac{6}{5},\frac{k}{5}+\frac{7}{5};\frac{4}{5},\frac{6}{5},\frac{7}{5},\frac{8}{5},\frac{k}{5}+\frac{14}{15},\frac{k}{5}+\frac{19}{15};\frac{t^3}{27}\right) \sec \left(\frac{\pi k}{5}+\frac{\pi }{10}\right)}{16 \pi ^2 \Gamma (k)}}$$

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The constant c is breaking your neck in this case. Consider the following simplification

Block[{c = 1},
 Integrate[
  c*k*(x^(c - 1)/(1 + x^c)^(k + 1))*Exp[-t*x], {x, 0, Infinity}]
 ]
(* ConditionalExpression[E^t k ExpIntegralE[1 + k, t], Re[t] > 0] *)

Testing some other integer values for c shows, that Mathematica cannot find the integral for them. I'm not sure how helpful this is, but it's maybe a start.

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  • $\begingroup$ For both variables $c$ and $k$ I can fix the values as you can see in the edited question but the result cannot be computed as well. $\endgroup$ – Mattia Mar 6 '17 at 13:18
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    $\begingroup$ Unfortunately, that does not help. Even when we further fix t=1, the value of the Integral cannot be calculated. The deal breaker is really the c constant. Maybe Mathematics is a better place to ask because your integral is not overly complex. Maybe some real mathematician can suggest a way to calculate the it. $\endgroup$ – halirutan Mar 6 '17 at 13:40
  • $\begingroup$ @Mattia If you are going to ask on Mathematics please leave a link here. I find the problem interesting as I had to deal with a similar integral which fortunately could be integrated by Mathematica. $\endgroup$ – halirutan Mar 6 '17 at 13:53
  • $\begingroup$ Here is the link: math.stackexchange.com/questions/2174344/… feel free to edit the question if you think you can make it clearer. Thank you for your help. $\endgroup$ – Mattia Mar 6 '17 at 13:57

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