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Write your $\Lambda$ as $$\Lambda(t)=\frac{1}{2}\sum_{-\infty}^\infty f(n),$$ where $$f(x)=\pi(x+1/2)\sin\pi(x+1/2)\exp\left(-\pi^2(x+1/2)^2t/2\right).$$ 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 its zero-th term, $$\hat{f}(0)=\sqrt{\frac{2}{\pi t^3}}e^{-1/(2t)},$$ as you wrote.