Ngô Quốc Anh

April 5, 2008

More integral involving Gamma function

Filed under: Giải Tích 2 — Ngô Quốc Anh @ 14:58

Show that

\int_0^1\;e^{ - \alpha\,x}\;\ln\,x\;\textbf dx\; = \;\boxed{ - \dfrac{1}{\alpha}\left(\ln\,\alpha + \gamma + \Gamma(0,\alpha)\right)}

where represents the incomplete gamma function.

Solution. With , we have
\mathcal{I} = \frac {1}{\alpha} \int_0^{\alpha} e^{ - t} \ln t \, dt - \frac {\ln \alpha}{\alpha} \int_0^{\alpha} e^{ - t} \, dt =
= \frac {1}{\alpha} \left\{\left( \int_0^{ + \infty} - \int_{\alpha}^{ + \infty} \right) e^{ - t} \ln t dt - (1 - e^{ - \alpha})\ln \alpha \right\}
= \frac {1}{\alpha} \left\{ - \gamma - \ln \alpha + e^{ - \alpha} \ln \alpha - \underbrace{\int_{\alpha}^{ + \infty}e^{ - t}\ln t \, dt}_{\mathcal{J}} \right\}
To compute J we integrate by parts:
\mathcal{J} = \left[ - e^{ - t} \ln t \right]_{\alpha}^{ + \infty} + \int_{\alpha}^{ + \infty} \frac {e^{ - t}}{t} dt = e^{ - \alpha} \ln \alpha + \underbrace{\int_{ - \infty}^{ - \alpha} \frac {e^{x}}{x} \, dx}_{\mbox{Ei}( - \alpha) = - \Gamma (0, \alpha)}
And Finally:
\mathcal{I} = - \frac {1}{\alpha} \left\{\gamma + \ln \alpha + \Gamma (0, \alpha) \right\}.

Done.

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