I'm looking for the close form of the integral $\int_u^{\infty}(x+a)^ve^{-bx}dx$, where $u,a,b$ are positive real numbers (b could be integer in special case), $v$ is a complex number. Are there anyone aware of it? Thanks in advance!
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Do you consider the incomplete gamma function a closed form solution? – Fabian Oct 25 '16 at 15:13
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yes, the incomplete gamma function is also viewed as a close form. – kawofengche Oct 25 '16 at 15:18
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$$ \int_u^\infty(x+a)^{\nu}\mathrm{e}^{-bx}dx $$ lets make the sub $t = x+ a$ we find $$ \mathrm{e}^{ab}\int_{u+a}^\infty t^\nu\mathrm{e}^{-b(t-a)}dt = \mathrm{e}^{ab}\int_{u+a}^\infty t^\nu\mathrm{e}^{-bt}dt =\frac{\mathrm{e}^{ab}}{b^{\nu+1}}\int_{b(u+a)}^\infty s^{\nu}\mathrm{e}^{-s}ds $$ If we map $\nu \to \alpha - 1$ we find $$ \frac{\mathrm{e}^{ab}}{b^{\nu+1}}\int_{b(u+a)}^\infty s^{\nu}\mathrm{e}^{-s}ds = \frac{\mathrm{e}^{ab}}{b^{\nu+1}}\int_{b(u+a)}^\infty s^{\alpha-1}\mathrm{e}^{-s}ds $$ The latter is the form of the incomplete gamma function $$ \Gamma(s,x) = \int_x^\infty t^{s-1}\mathrm{e}^{-t}dt $$ so we have $$ \frac{\mathrm{e}^{ab}}{b^{\nu+1}}\int_{b(u+a)}^\infty s^{\alpha-1}\mathrm{e}^{-s}ds = \frac{\mathrm{e}^{ab}}{b^{\nu+1}}\Gamma(\alpha,b(u+a)) = \frac{\mathrm{e}^{ab}}{b^{\nu+1}}\Gamma(\nu+1,b(u+a)) $$ We have suitable conditions on $\alpha$ and thus $\nu + 1$.
Chinny84
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The derivation is complementary to Fabian's answer - So no need to re-assign the tick. – Chinny84 Oct 25 '16 at 16:04
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Wolfram alpha shows that a antiderivative is given by $$\int (x+a)^\nu e^{-b x} dx= -b^{-(\nu+1)} e^{a b} \Gamma(\nu+1,b(a+x)). $$
Evaluating the antiderivative at the boundary values, we obtain $$\int_u^\infty (x+a)^\nu e^{-b x} dx = b^{-(\nu+1)} e^{a b} \Gamma(\nu+1,b(a+u))$$ with $\Gamma$ the incomplete Gamma function.
Fabian
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