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Write your $\Lambda$ as $$\Lambda(t)=\frac{1}{2}\sum_{-\infty}^\infty f(n),$$ where $$f(x)=\pi(x+1/2)\sin\pi(x+1/2)\exp\left(-\pi^2(x+1/2)^2t/2\right)=y\sin y\,e^{-ty^2/2},$$ 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).

Remark. This approximation is related to the famous computation of the age of Earth by Lord Kelvin. Roughly speaking, the series corresponds to the exact solution for the heat equation inside the spherical Earth, while the asymptotics corresponds to the flat Earth approximation.

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).

Remark. This approximation is related to the famous computation of the age of Earth by Lord Kelvin. Roughly speaking, the series corresponds to the exact solution for the heat equation inside the spherical Earth, while the asymptotics corresponds to the flat Earth approximation. It is unclear from his papers on the subject, whether Kelvin knew the exact solution and Poisson's formula.

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).

Remark. This approximation is related to the famous computation of the age of Earth by Lord Kelvin. Roughly speaking, the series corresponds to the exact solution for the heat equation inside the Earth, while the asymptotics corresponds to the flat Earth approximation. See age of earth, and age of Earth 2

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).

Remark. This approximation is related to the famous computation of the age of Earth by Lord Kelvin. Roughly speaking, the series corresponds to the exact solution for the heat equation inside the spherical Earth, while the asymptotics corresponds to the  flat Earth approximation.

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).

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).

Remark. This approximation is related to the famous computation of the age of Earth by Lord Kelvin. Roughly speaking, the series corresponds to the exact solution for the heat equation inside the Earth, while the asymptotics corresponds to the flat Earth approximation. See age of earth, and age of Earth 2