Write your $\Lambda$ as $$\Lambda(t)=\frac{1}{2}\sum_{-\infty}^\infty f(n),$$ where $$f(x)=\pi(x+1/2)\sin\pi(x+1/2)\exp\left(-\pi^2(x+1/2)^2t/2\right)=y\sin y\,e^{-ty^2/2},$$ where $y=\pi(x+1/2),$ and $t>0$. Then use Poisson's summation formula $$\sum_{-\infty}^\infty f(n)=\sum_{-\infty}^\infty \hat{f}(2\pi n),$$ where $$\hat{f}(s)=\int_{-\infty}^\infty f(x)e^{-isx}dx$$ is the Fourier transform. This Fourier transform can be explicitly computed: $$\hat{f}(s)=\frac{1}{\sqrt{2\pi}}t^{-3/2}e^{is/2}\left((s/\pi+1)e^{-(s/\pi+1)^2/(2t)}-(s/\pi-1)e^{-(s/\pi-1)^2/(2t)}\right).$$ The trick is that in the Fourier transform your parameter $t$ will stand in the denominator of the exponent, so the series $\sum\hat{f}(n)$ will be asymptotic to the sum of $3$ terms (with $n=0,\pm1$), z when $t\to 0$, $$\sum_{-\infty}^\infty\hat{f}(2\pi n)\sim \hat{f}(0)+\hat{f(2\pi)}+\hat{f}(-2\pi)\sim 2\hat{f}(0)=2\sqrt{\frac{2}{\pi t^3}}e^{-1/(2t)},\quad t\to 0+,$$ and since $\Lambda(t)$ is $1/2$ of this sum, we obtain the answer that wrote.
(In the computation of Fourier transform, I started with the $e^{-y^2/(2t)}$, and then used the transformation rules of Fourier transform: multiplied my function on $y$, then on $\sin y$, and then scaled by $\pi$ and added $1/2$ to the argument).