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I want to solve the following integral by using Leibnitz rule:

$$I(b) = \int_{0}^{\infty} e^{-x} x^{b-1} dx$$

$$I'(b) = \int_{0}^{\infty} \frac{\partial}{\partial b} e^{-x} x^{b-1} dx = \int_{0}^{\infty}ln(x) e^{-x} x^{b-1} dx$$

Honestly, I am not really certain how to solve the final integral form. If i were to first plug in a value for (b), solving the integral would be much easier, but I am not so sure that this is allowed.

My question is, can I plug in a value for (b) before integrating [why/why not], or is there a way to compute the second integral without plugging in a value for (b)?

I know that this integral can be more easily solved using other methods, however I am curious to know if it is solvable using Leibnitz rule.

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The answer is basically no: this integral is, of course, the Gamma function. Since $\Gamma(n+1)=n!$ for positive integer $n$ (easily checked using integration by parts), and the factorial doesn't behave well under differentiation, it's not likely that the Gamma function will either. Worse, if you do put $b=1$, you have the integral $$ \int_0^{\infty} e^{-x}\log{x} \, dx, $$ which is the (negative of) the rather nasty transcendental Euler-Mascheroni constant.

Also, there's no way to deal with the integrand $x^{b-1}e^{-x}\log{x}$ easily: you can't split it up into things with nice antiderivatives, or anything useful like that. The best you've got is probably the integration by parts, $$ I'(b) = [-e^{-x}x^{b-1}\log{x}]_0^{\infty} + \int_0^{\infty} e^{-x} ((b-1)x^{b-2}\log{x} + x^{b-2}) \, dx = (b-1)I'(b-1) + I(b-1). $$

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  • $\begingroup$ Yes, I wanted to avoid an answer that involved the gamma function, but I was hopeful that there could have been more useful forms of the function. Thank you very much for the quick response. $\endgroup$ –  Konstantin Ashkenazy Mar 28 '15 at 20:52
  • $\begingroup$ *I meant differentiating the Gamma function... $\endgroup$ –  Konstantin Ashkenazy Mar 28 '15 at 20:59

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