Computation of a (probably) tricky limit. 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!
 A: 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.
A: Use
$$
\int_{0}^{\infty}{\rm f}\left(x\right){\rm e}^{-x^{2}/\left(4t\right)}\,{\rm d}x
=
2t^{1/2}\int_{0}^{\infty}{\rm f}\left(2t^{1/2}x\right){\rm e}^{-x^{2}}\,{\rm d}x
\approx
2t^{1/2}\left\lbrack\lim_{x \to 0}{\rm f}\left(x\right)\right\rbrack\
\overbrace{\quad\int_{0}^{\infty}{\rm e}^{-x^{2}}\,{\rm d}x\quad}^{\sqrt{\pi}\,/\,2}
$$
$$
\mbox{Then}\quad
\int_{0}^{\infty}{\rm f}\left(x\right){\rm e}^{-x^{2}/\left(4t\right)}\,{\rm d}x
\approx
\sqrt{\pi}\,t^{1/2}\lim_{x \to 0}{\rm f}\left(x\right)
\quad\mbox{when}\quad
t \gtrsim 0
$$
