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While studying Dirichlet's proof of an infinity of primes in any AP with first term and common difference coprime, the formula below involving the gamma function was quoted as being well known.

$$\int_{0}^{1} x^{k-1} \log ^{\varrho } \left ( \frac{1}{x} \right ) dx = \frac{\Gamma \left ( 1 + \varrho \right )}{k + \varrho }$$

where k is a positive constant and $\varrho$ is complex.

I had not seen this formula before and wanted to derive it (after first looking around in the literature and not finding it). My question concerns only the above formula, can someone please show how this was derived ?

I tried to derive it from :

$$\Gamma \left ( \varrho + 1 \right ) = \int_{0}^{\infty } t^{\varrho }\ e^{-t} dt$$

using the simple substitution $$x = e^{-t} \Rightarrow t = \log \frac{1}{x}$$

which yields $$\Gamma \left ( \varrho + 1 \right )=\int_{0}^{1} \log ^{\varrho }\left ( \frac{1}{x} \right )dx$$

From this I derived a reduction formula and expanded it but the above form did not come out ... can someone show the derivation or point to a reference book ?

Thanks, Aethelred the Unwashed.

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  • $\begingroup$ Something is off: the $k=1$ case of the formula at top doesn't match your integral at the bottom. $\endgroup$ – Semiclassical Jul 21 '14 at 20:00
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    $\begingroup$ Mathematica gives the correct formula as having a denominator of $k^{1+\rho}$. $\endgroup$ – Semiclassical Jul 21 '14 at 20:04
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The substitution $t = \log \frac{1}{x}$ is indeed the natural candidate. So let's start from the given integral and see where it takes us

$$\begin{align} \int_0^1 x^{k-1} \log^\varrho \left(\frac{1}{x}\right)\,dx &= \int_0^\infty t^\varrho e^{-kt}\,dt\\ &= \frac{1}{k^{\varrho+1}} \int_0^\infty (kt)^\varrho e^{-kt}\,d(kt)\\ &= \frac{1}{k^{\varrho+1}}\int_0^\infty u^\varrho e^{-u}\,du\\ &= \frac{\Gamma(1+\varrho)}{k^{\varrho+1}}. \end{align}$$

Now, $k^{\varrho+1}$ and $k+\varrho$ are rarely the same, so the purported formula is incorrect. Possibly a typo?

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  • $\begingroup$ Heh. I posted my comment above just before reading your new answer! $\endgroup$ – Semiclassical Jul 21 '14 at 20:04
  • $\begingroup$ Eleven seconds difference. It's a short answer, but that's too short to read it nevertheless. $\endgroup$ – Daniel Fischer Jul 21 '14 at 20:06
  • $\begingroup$ Indeed the above formula had a typo, should have been $k^{\varrho + 1}$ not as shown. Excellent thanks ! $\endgroup$ – aethelred_the_unwashed Jul 21 '14 at 20:10

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