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Question: I have the following "Analysis 1"-limit: $$\lim_{t \rightarrow 0}\: \frac{e^{-\frac{t}{4}}}{t} \left(\frac{1}{4t^{\frac{3}{2}}} \int_0 ^\infty \frac{x^3 e^{-\frac{x^2}{4t}}}{\sinh(\frac{x}{2})}dx - \frac{1}{2t^{\frac{1}{2}}} \int_0 ^\infty \frac{x e^{-\frac{x^2}{4t}}}{\sinh(\frac{x}{2})}dx\right).$$

The result should be $-\frac{\sqrt{\pi}}{6}$. How to compute it?

Motivation: McKean, in page 242 of his article "Selberg's Trace Formula as Applied to a Compact Riemann Surface" states the formal power series expansion: $$\frac{e^{-\frac{t}{4}}}{(4\pi t)^{\frac{3}{2}}} \int_0^\infty \frac{x e^{-\frac{x^2}{4t}}}{\sinh(\frac{x}{2})}dx = \frac{1}{4 \pi t} \left(1 -\frac{t}{3} + O(t^2) \right).$$

He gives no proof for it, so I decided to do my homework and compute it. My approach consisted in defining: $$f(t):= \frac{e^{-\frac{t}{4}}}{(4\pi t)^{\frac{1}{2}}} \int_0^\infty \frac{x e^{-\frac{x^2}{4t}}}{\sinh(\frac{x}{2})}dx.$$

Then in computing its Taylor expansion in $t=0$. The first term of the Taylor expansion follows easily (see below in "My progresses"). To compute the second term of the expansion I need:

$$ \partial_t(f(t))_{t=0} = \lim_{t \rightarrow 0} \partial_t(f(t))= \lim_{t\rightarrow 0 } -\frac{f(t)}{4} + \frac{1}{2\sqrt{\pi}}\cdot\{\text{Limit above}\}.$$

This is why I'm interested in the limit above. The conjectured result follows by McKean statement and $\underset{t \rightarrow 0}{\lim} f(t) =1$.

My progresses: A useful partial result I have got is the limit, for $n$ odd: $$\lim_{t\rightarrow 0}\: \frac{1}{t^{\frac{n}{2}}} \int_0^\infty \frac{x^n e^{-\frac{x^2}{4t}}}{\sinh(\frac{x}{2})}dx=\left(\prod_{i=1}^{\frac{n-1}{2}}\frac{2i-1}{2}\right)2^n \sqrt{\pi}.$$

Which I use to compute the first term of the Taylor expansion. Applying it to the "main" limit we see that it is of the form $\frac{0}{0}$. But applying L'Hopital rule we erase the $\frac{1}{t}$ from the denominator just to get it again from the derivative of the numerator, and we come back to the $\frac{0}{0}$ situation. I iterated the rule for some steps but it doesn't seem to bring anywhere.

Remark: I'm actually interested in the expansion as in McKean, so if you have any way to compute it avoiding the limit above for me it would be a completely satisfying answer. Moreover if you see any error I made to get to the limit I would be very grateful if you could point it out. Thank you very much in advance!

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You correctly observed that you obtain $0/0$ form but notice also that since in the denominator you have only $t$ it's clear that in order for the limit to exist and be finite (which we assume), the numerator must also behave as $t$ near $0$. Therefore, it's enough to keep the cubic terms in the Taylor expansion of $\sinh(\sqrt t x / 2)$. The linear terms (which are the only ones you've kept in the OP) cancel but the cubic terms will give you the answer. Let me know if this comment is too sketchy and unclear. I'll post an answer when I'll have more time. –  Marek Jan 10 '13 at 11:35
    
@Marek Thank you very much! I unzipped your comment and you`re totally right (there is still a factor 2 missing, but I'm confident I'll find it once reviewing the computation)! If you want to put it as an answer I will accept and upvote it as soon as possible! :) –  Giovanni De Gaetano Jan 10 '13 at 12:23
    
@Giovanni: no problem, you're welcome :) As for the answer: I'll post it later (if nobody does it first). It takes time to write out everything clearly in full, so when I'm busy I prefer to leave comments instead. Cheers. –  Marek Jan 10 '13 at 12:38

2 Answers 2

up vote 6 down vote accepted

About McKean's expansion:

It can be computed directly. First, notice that $$ \int_0^\infty\frac{xe^{-x^2/4t}}{\sinh(x/2)}\,dx = 4t\int_0^\infty \frac{x e^{-x^2}}{\sinh(x\sqrt{t})}\,dx. $$

For $|t| \leq 1$, you can use the expansion (with uniform big O) $$ \frac{xe^{- x^2}}{\sinh(x\sqrt{t})} = \frac{xe^{-x^2}}{x\sqrt{t}}\left(1 - x^2\frac{t}{6} + O(t^2e^{2x})\right) $$ to get $$ \int_0^\infty \frac{x e^{-x^2}}{\sinh(x\sqrt{t})}\,dx = \sqrt{\frac{\pi}{4t}}\left(1 - \frac{t}{12} + O(t^2)\right). $$

On the other hand, $$ e^{-t/4} = 1 - \frac{t}{4} + O(t^2) $$

as $t \to 0$, so that $$ \frac{e^{-t/4}}{(4\pi t)^{3/2}}\int_0^\infty\frac{xe^{-x^2/4t}}{\sinh(x/2)}\,dx = \frac{1}{4\pi t}\left(1 - \frac{t}{3} + O(t^2)\right). $$

About your limit: Same techniques apply.

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Use

$$ \int_{0}^{\infty}{\rm f}\left(x\right){\rm e}^{-x^{2}/\left(4t\right)}\,{\rm d}x = 2t^{1/2}\int_{0}^{\infty}{\rm f}\left(2t^{1/2}x\right){\rm e}^{-x^{2}}\,{\rm d}x \approx 2t^{1/2}\left\lbrack\lim_{x \to 0}{\rm f}\left(x\right)\right\rbrack\ \overbrace{\quad\int_{0}^{\infty}{\rm e}^{-x^{2}}\,{\rm d}x\quad}^{\sqrt{\pi}\,/\,2} $$

$$ \mbox{Then}\quad \int_{0}^{\infty}{\rm f}\left(x\right){\rm e}^{-x^{2}/\left(4t\right)}\,{\rm d}x \approx \sqrt{\pi}\,t^{1/2}\lim_{x \to 0}{\rm f}\left(x\right) \quad\mbox{when}\quad t \gtrsim 0 $$

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