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Consider the function $$ I(a,x) = x^{-a} \int_{1}^x y^{a-1} \exp(-y a) dy $$ where $x \geq 1$, and $a \geq 0$.

I am not really interested in the parameter $x$, so define $$ I(a) = \sup_{x \geq 1} I(a,x) $$

Now what is the asymptotic behavior of $I(a)$ as $a \rightarrow 0$?

I can show the bound $I(a) = \log(1/a) + O(1)$. This is by noting that $I(a,x) \leq \int_{1}^x \exp(-y a)/y \ dy \leq \int_{1}^{\infty} \exp(-y a)/y \ dy = \Gamma(0, a) = \log(1/a) + O(1)$.

The asymptotic behavior of $I(a)$ appears to be smaller than this, though, something like $I(a) \sim 0.9 \log(1/a)$

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Shouldn't your inequality go the other way? The supremum $\ge$ the limit as $x \to \infty$, not $\le$. – Robert Israel Apr 3 '12 at 21:30
up vote 5 down vote accepted

In a first step, we use the variable transform $y=(y-1)/(x-1)$ and obtain $$I(a,x)= \int_0^1\!dt\, x^{-a} (x-1) [1+ t(x-1)]^{a-1} e^{-a [1+ t (x-1)]}.$$ As we are interested in $a\to0$, we expand the integrand to first order in $a$ and obtain $$\begin{align}I(a,x) &= \int_0^1\!dt\, \left[ \frac{x-1}{1+t(x-1)} + a \frac{(x-1)(\log[1+t (x-1)] - \log x -1 - t(x-1)}{1+ t (x-1)} + O(a^2) \right] \\ &=\log x +a \left(1-x - \tfrac12 \log^2 x\right) + O(a^2) \end{align}.$$

The function $I(a,x)$ assumes its maximum at $x^* = a^{-1} + O(\log a) $. Thus, we have $$I(a) = I(a,x^*) = \log a^{-1} -1 + o(a). $$

Edit: As oenamen pointed out the answer is not self consistent. In expanding the exponent to get from the first expression to the second, I assumed that $ax\ll1$. Then I found that $x \simeq a^{-1}$ which of course outside the scope of the first assumption. However, it is not difficult to convince oneself that the $\log a^{-1}$ scaling (with unit prefactor and not like the OP assumed with a different pre factor) is indeed the correct asymptotic expression. Thus $$I(a) = \log a^{-1} + O(1).$$

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The parameter $a x$ is not small, so the expansion $e^{-a x t} = 1 - a x t + O(a^2)$ is no good. Thus, there is a problem with your location of the maximum $x^*$ and therefore with the finite part of $I(a)$. The integral does go like $\ln a^{-1}$ however. This is a more interesting problem than it appears! – user26872 Apr 5 '12 at 21:48
Thanks for spotting the error. I edited the answer to make it correct. – Fabian Apr 6 '12 at 13:16
Glad to help. Cheers! – user26872 Apr 6 '12 at 13:59

We must be careful when expanding in small $a$, we do not know a priori that $a x$ is small. It is in fact necessary to treat $a$ as small, and $a x$ to be at least of order one. While the maximum value of the integral goes like $\ln a^{-1}$, there is an error of order one in the accepted answer.

Change variables. Let $z = a y$. The integral can then be written as $$I(a,x) = (a x)^{-a}\left[\Gamma(a,a) - \Gamma(a,a x)\right],$$ where $\Gamma(s,x) = \int_x^\infty dt \ t^{s-1} e^{-t}$ is the incomplete gamma function.

Look for an extremum, $X$. (The reader can verify the extremum found below is a maximum.) We find ${\partial I}/{\partial x}|_{x=X} = 0$ implies \begin{equation*} a\Gamma(a,a X) + (a X)^a e^{-a X} = a \Gamma(a,a) \tag{1} \end{equation*} so that \begin{equation*} I(a,X) = \frac{1}{a} e^{-a X}. \end{equation*} Integrating $a\Gamma(a,a X)$ in (1) by parts we find the condition on $X$ is $$\frac{1}{a}\Gamma(a+1,a X) = \Gamma(a,a).$$ So far no expansion has been made. Expanding the left hand side in small $a$ (but not small $a x$), we find $$\begin{eqnarray} \frac{1}{a}\Gamma(a+1,a X) &=& \frac{1}{a}e^{-a X} + \mathrm{h.o.} \\ &=& I(a,X) + \mathrm{h.o.}. \end{eqnarray}$$ We find the higher order terms go like $e^{-a X} \ln a X$. We assume $a X$ is large enough so these terms are suppressed. We will find this assumption to be self consistent. Therefore, $I(a,X) = \Gamma(a,a) + \mathrm{h.o.}$. Expanding $\Gamma(a,a)$ in small $a$ we find $$\begin{eqnarray} I(a,X) &=& -\mathrm{Ei}(-a) + \mathrm{h.o.} \\ &=& -\gamma + \ln a^{-1} + \mathrm{h.o.}, \end{eqnarray}$$ where $\mathrm{Ei}(x) = \int_{-\infty}^x d t \ t^{-1} e^t$ is the exponential integral and $\gamma$ is the Euler-Mascheroni constant. The location of the maximum is $$X = -\frac{1}{a} \ln\left[a(\ln a^{-1}-\gamma)\right] + \mathrm{h.o.}$$ For $a = e^{-k}$ we find $a X \approx k$ so the higher order terms discussed above go like $e^{-k}\ln k$. Thus, since $k$ is large the expansion is valid.

Below is a plot of $I(a,x)$ for $a=10^{-4}$. The predicted maximum is $I(a,X) \approx 8.63$ at $X \approx 7.05\times 10^4$. Clearly $\ln a^{-1} -1 \approx 8.21$ underestimates the maximum value of the integral.

Plot of I(a,x) vs x for a = 10^{-4}

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