A How to solve this definite integral?

Ad VanderVen
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How to solve the integral ##\int_{0}^{x}\!-{\frac {\lambda\,{{\rm e}^{-\lambda\,t}}{\beta}^{\alpha} \left( -\lambda+\beta \right) ^{-\alpha} \left( -\Gamma \left( \alpha \right) +\Gamma \left( \alpha, \left( -\lambda+\beta \right) t \right) \right)}{\Gamma \left( \alpha \right) }}\,{\rm d}t##
I would like to solve the integral underneath:

$$\displaystyle \int_{0}^{x}\!-{\frac {\lambda\,{{\rm e}^{-\lambda\,t}}{\beta}^{\alpha} \left( -\lambda+\beta \right) ^{-\alpha} \left( -\Gamma \left( \alpha \right) +\Gamma \left( \alpha, \left( -\lambda+\beta \right) t \right) \right)}{\Gamma \left( \alpha \right) }}\,{\rm d}t$$

I tried it before using Integration by Substitution, but that didn't help either.
 
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The integrand is written as
Ae^{-\lambda t}+Be^{-\lambda t}\Gamma (\alpha ,Ct)
The first term is easy to integrate. What is two variable function of ##\Gamma(\alpha, Ct)## you wrote ?

[EDIT] It is an incomplete gamma function.
 
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The only non-trivial part is <br /> \int_0^x e^{-\lambda t} \Gamma(\alpha, kt)\,dt<br /> = \frac 1k \int_0^{kx} e^{-\mu s}\Gamma(\alpha, s)\,ds<br /> where k = \beta - \lambda and \mu = \lambda/k. Now <br /> \Gamma(\alpha, s) = \int_s^\infty t^{\alpha - 1} e^{-t}\,dt = \Gamma(\alpha) - \int_0^s t^{\alpha - 1} e^{-t}\,dt so <br /> \int_0^{X} e^{-\mu s} \Gamma(\alpha, s)\,ds = \int_0^X e^{-\mu s} \Gamma(\alpha)\,ds -<br /> \int_0^X \int_0^s e^{-\mu s - t} t^{\alpha - 1}\,dt\,ds. The first integral is trivial; in the second we can set (u,v) = (\mu s + t, t) and we find that <br /> \int_0^X \int_0^s e^{-\mu s - t} t^{\alpha - 1}\,dt\,ds = \frac{1}{\mu}\int_0^{X}<br /> \int_{u/(1+ \mu)}^u e^{-u} v^{\alpha - 1}\,dv\,du <br /> + \frac{1}{\mu} \int_X^{(1 + \mu)X} \int_{u/(1 + \mu)}^X e^{-u} v^{\alpha - 1}\,dv\,du and after doing the inner integral over v you will be left with integrals of e^{-u} (trivial) and e^{-u} u^\alpha (expressible in terms of incomplete gamma functions).

EDIT: The limits should be 0 \leq u \leq (1 + \mu)X and \max\{0, u - \mu X\} \leq v \leq u/(1 + \mu). That introduces a term of the form \int_{\mu X}^{(1 + \mu)X} e^{-u} (u - \mu X)^\alpha \,du = e^{-\mu X} \int_0^{X} e^{-z}z^\alpha\,dz.
 
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Hi folks, I'm trying to respond to your answers, but I can't read them because of the right column, which prevents me from seeing the whole page. I don't know how to click away that column.
 
pasmith said:
The only non-trivial part is <br /> \int_0^x e^{-\lambda t} \Gamma(\alpha, kt)\,dt<br /> = \frac 1k \int_0^{kx} e^{-\mu s}\Gamma(\alpha, s)\,ds<br /> where k = \beta - \lambda and \mu = \lambda/k.

This can be tidied up considerably: <br /> \begin{align*}<br /> \frac{1}{k}\int_0^{kx} e^{-\mu t} \Gamma(\alpha, t)\,dt &amp;= \frac 1k \int_0^{kx} \int_t^\infty e^{-\mu t-s}s^{\alpha - 1}\,ds\,dt \\<br /> &amp;= \frac 1{\mu k} \int_0^\infty v^{\alpha - 1} \int_v^{U(v)} e^{-u}\,du\,dv \\<br /> &amp;= \frac 1{\mu k} \int_0^\infty v^{\alpha - 1} (e^{-v} - e^{-U(v)})\,dv<br /> \end{align*} using (u,v) = (\mu t+s,s) where U(v) = \min\{(1 + \mu)v, v + \mu kx\}. Hence <br /> \begin{align*}<br /> \frac{1}{k} \int_0^{kx} e^{-\mu t} \Gamma(\alpha, t)\,dt &amp;= \frac{\Gamma(\alpha)}{\mu k}<br /> - \frac{1}{\mu k}\int_0^{kx} v^{\alpha - 1}e^{-(1 + \mu)v}\,dv <br /> - \frac{e^{-\mu kx}}{\mu k} \int_{kx}^\infty v^{\alpha-1} e^{-v}\,dv \\<br /> &amp;= \frac{\Gamma(\alpha)}{\mu k} - \frac{\gamma(\alpha, (1+\mu)kx)}{\mu k(1 + \mu)^\alpha} <br /> - \frac{e^{-\mu kx} \Gamma(\alpha,kx)}{\mu k}.<br /> \end{align*}
 
pasmith

For me the non-trivial part is:

$$\displaystyle \int_{0}^{x}\!{\frac {\Gamma \left( \alpha,\beta\,t-\lambda\,t \right) }{{{\rm e}^{\lambda\,t}}}}\,{\rm d}t$$

Please, would you be so kind to start from this equation.
 
pasmith

Sorry for my earlier reply. Now that I've looked again, I've seen that you've actually started from the equation I've suggested.
 
pasmith

With your help I got the final solution for the definite integral:

$$\displaystyle {\frac {{\beta}^{\alpha} \left( {{\rm e}^{\lambda\,x}}-1 \right) {{\rm e}^{-\lambda\,x}}}{ \left( -\lambda+\beta \right) ^{\alpha}}}-{\frac {{\beta}^{\alpha}}{\alpha\, \left( 1+\alpha \right) \left( -\lambda+\beta \right) ^{\alpha}\Gamma \left( \alpha \right) } \left( -{{\rm e}^{-x\beta}} \left( {\frac {\beta}{-\lambda+\beta}} \right) ^{-\alpha} \left( x\beta \right) ^{\alpha}\alpha- \left( x\beta \right) ^{\alpha/2}{{\rm e}^{-1/2\,x\beta}}{{WhittakerM}_{\alpha/2,\,\alpha/2+1/2}\left(x\beta\right)} \left( {\frac {\beta}{-\lambda+\beta}} \right) ^{-\alpha}-{{\rm e}^{-x\beta}} \left( {\frac {\beta}{-\lambda+\beta}} \right) ^{-\alpha} \left( x\beta \right) ^{\alpha}-{{\rm e}^{-\lambda\,x}}\Gamma \left( \alpha, \left( -\lambda+\beta \right) x \right) {\alpha}^{2}+\Gamma \left( \alpha \right) {\alpha}^{2}-{{\rm e}^{-\lambda\,x}}\Gamma \left( \alpha, \left( -\lambda+\beta \right) x \right) \alpha+\Gamma \left( \alpha \right) \alpha \right) }$$
 
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