Consider the Confluent hypergeometric function $U(a,b,z)$, which is a solution of the Kummer's Equation : $$zw''+(b-z)w'-aw=0$$

it has the following integral representation when $- \pi/2 < \arg z< \pi/2$ $$U(a,z)= \frac{1}{\Gamma(a)} \int^{\infty}_0 t^{a-1}(1+t)^{b-a-1} e^{-tz} dt$$

I know that there is the following asymptotic series for $U(a,1,z)$ :

$$\mathop{U}\nolimits\!\left(a,1,z\right)=-\frac{1}{\mathop{\Gamma}\nolimits% \!\left(a\right)}\left(\mathop{\ln}\nolimits z+\mathop{\psi}\nolimits\!% \left(a\right)+2\gamma\right)+\mathop{O}\nolimits\!\left(z\mathop{\ln}% \nolimits z\right) \ \ \ \ (*) \ \ \ \ z \to 0$$


Where $\gamma$ is the Euler constant and $\psi(z) = \displaystyle \frac{\Gamma'(z)}{\Gamma(z)}$ is the digamma function.

I would be very thankful if one can explain how should I start deriving asymptotic series $(*)$


It might be easier to start with one of the other integral representations, but here's one way. We'll begin with some manipulations to strip away the lower order terms:

$$ \begin{align} &\int_0^\infty t^{a-1} (1+t)^{-a} e^{-zt}\,dt \\ &\qquad = \left(\int_0^1 + \int_1^\infty\right) t^{a-1} (1+t)^{-a} e^{-zt}\,dt \\ &\qquad = \int_0^1 t^{a-1} (1+t)^{-a}\,dt + O(z) + \int_1^\infty t^{a-1} (1+t)^{-a} e^{-zt}\,dt \\ &\qquad = \int_0^1 t^{a-1} (1+t)^{-a}\,dt + O(z) + \int_1^\infty \left[t^{a-1} (1+t)^{-a} - t^{-1} + at^{-2}\right]e^{-zt}\,dt \\ &\qquad \qquad - a\int_1^\infty t^{-2}e^{-zt}\,dt + \int_1^\infty t^{-1} e^{-zt}\,dt. \tag{1} \end{align} $$

The last integral in the last line diverges logarithmically as $z \to 0^+$ while the other integrals converge. Thus the leading-order behavior comes from the last integral. To extract this behavior let's first make the change of variables $zt=s$, then integrate by parts:

$$ \begin{align} \int_1^\infty t^{-1} e^{-zt}\,dt &= \int_z^\infty s^{-1} e^{-s}\,ds \\ &= e^{-s}\log s \Bigr|_z^\infty + \int_z^\infty e^{-s}\log s\,ds \\ &= -e^{-z}\log z + \int_0^\infty e^{-s}\log s\,ds - \int_0^z e^{-s}\log s\,ds. \end{align} $$

To get the last line we split the integral like $\int_z^\infty = \int_0^\infty - \int_0^z$. Note that we could repeatedly integrate the last integral by parts to obtain further terms of the asymptotic expansion if we desire. We only need the terms up to $O(z\log z)$, so it's enough to notice that

  • $e^{-z}\log z = \log z + O(z\log z)$,

  • $\int_0^\infty e^{-s}\log s\,ds = -\gamma$, and

  • $\int_0^z e^{-s}\log s\,ds = O(z\log z)$ by L'Hopital's rule.


$$ \int_1^\infty t^{-1} e^{-zt}\,dt = -\log z - \gamma + O(z\log z). \tag{2} $$

It remains to estimate the remaining integrals in $(1)$, namely

$$ \int_1^\infty \left[t^{a-1} (1+t)^{-a} - t^{-1} + at^{-2}\right]e^{-zt}\,dt \qquad \text{and} \qquad \int_1^\infty t^{-2}e^{-zt}\,dt. $$

The integrand in the first of these is $O(t^{-3})e^{-zt}$, so we can apply the mean value theorem (i.e. $e^{-zt} = 1 -zte^{-zc(t)}$ with $0 < c(t) < t$) to conclude that

$$ \int_1^\infty \left[t^{a-1} (1+t)^{-a} - t^{-1} + at^{-2}\right]e^{-zt}\,dt = \int_1^\infty \left[t^{a-1} (1+t)^{-a} - t^{-1} + at^{-2}\right]dt + O(z).$$ $$ \tag{3} $$

For the second integral we proceed as we did with the singular integral in $(2)$:

$$ \begin{align} \int_1^\infty t^{-2}e^{-zt}\,dt &= \int_1^\infty t^{-2}\,dt + \int_1^\infty t^{-2}(e^{-zt}-1)\,dt \\ &= \int_1^\infty t^{-2}\,dt + z\int_z^\infty s^{-2}(e^{-s}-1)\,ds \\ &= \int_1^\infty t^{-2}\,dt + z\int_1^\infty s^{-2}(e^{-s}-1)\,ds - z\int_1^z s^{-2} (e^{-s}-1)\,ds. \end{align} $$


$$ \int_1^z s^{-2} (e^{-s}-1)\,ds = O(\log z) $$

by L'Hopital's rule, so

$$ \int_1^\infty t^{-2}e^{-zt}\,dt = \int_1^\infty t^{-2}\,dt + O(z\log z). \tag{4} $$

Combining $(3)$ and $(4)$ we get

$$ \begin{align} &\int_1^\infty \left[t^{a-1} (1+t)^{-a} - t^{-1} + at^{-2}\right]e^{-zt}\,dt - a\int_1^\infty t^{-2}e^{-zt}\,dt \\ &\qquad = \int_1^\infty \left[t^{a-1} (1+t)^{-a} - t^{-1}\right]dt + O(z\log z). \end{align} $$

Substituting this and $(2)$ into the first equation we got, we conclude that

$$ \int_0^\infty t^{a-1} (1+t)^{-a} e^{-zt}\,dt = -\log z - \gamma + f(a) + O(z\log z), $$ where $$ f(a) = \int_0^1 t^{a-1} (1+t)^{-a}\,dt + \int_1^\infty \left[t^{a-1} (1+t)^{-a} - t^{-1}\right]dt. $$

It shouldn't be hard to show that

$$ f(a) = \int_0^1 \frac{t^{a-1}-1}{1-t}\,dt = -\psi(a)-\gamma, $$

from which the result would follow.

  • $\begingroup$ Thank you so much for your answer ! I will try to follow every single step, I hope I can get it all $\endgroup$ – the8thone Apr 14 '15 at 2:16
  • $\begingroup$ Sure thing. I hope it's helpful. This is the usual method I use for these types of problems, though I suspect there is a more elegant approach out there... $\endgroup$ – Antonio Vargas Apr 14 '15 at 6:41

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