Graduate How to solve this definite integral?

[Ad VanderVen]

TL;DR

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.

 

[anuttarasammyak]

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.

 

[pasmith]

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.

 

[Ad VanderVen]

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]

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*}

 

[Ad VanderVen]

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.

 

[Ad VanderVen]

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.

 

[Ad VanderVen]

pasmith

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

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