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Numerical calculation of a Duhamel-Integral coming up considering a unsteady state diffusion in a thin film electrode with zero initial concentration leads to the following strange identity:

$$ \int_0^t \theta_3(e^{-\pi^2 (t-\tau)}) \, \theta_2(e^{-\pi^2 \tau}) \ d \tau = 1$$

Question: Has anybody an idea to prove that.

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    Sorry, this was my first question here, the formula in the body is the correct one! – 2017-01-03
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    \pi is just the number. – 2017-01-03
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    Let me explain a little bit more. – 2017-01-03
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    The integral may be proven by the convolution theorem. – 2017-01-03
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    H(s)=F(s) G(s), now, I found a physical problem showing that $H(s)=Coth(\sqrt{s})/\sqrt{s} Tanh(\sqrt{s})/\sqrt{s}=1/s$. The Laplace-Transform of the first term is the first term of the integral in the body $\theta_3$ the second term is the second term of the integral in the body $\theta_2$ – 2017-01-03
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    If you already have a method of proof, then what exactly are you looking for? – 2017-01-03
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    I try to understand, whether this identity is trivial or useful for other problems. The aim is to use a Laplace transform on a geometrical sequence $Sum (Coth(\sqrt{s}/\sqrt(s)))^k to solve the diffusion equation of a spherical electrode particle with a new method, never published. – 2017-01-03

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