I'm struggling with the convolution $f \ast g(x)$ where $$f(x)=e^{-x^2}$$ $$g(x)=e^{-5x^2}$$ I integrated $$\int_{-\infty}^{\infty} e^{-t^2}e^{-5(x-t)^2}\ dt = \int_{-\infty}^{\infty} e^{-5x^2+10tx-6t^2}\ dt = \left[\frac{1}{10x-12t}e^{-5x^2+10tx-6t^2}\right]_{-b}^a$$ When I expanded using a and b and let them both go to $\infty$ respectively, both parts went to zero. I'm not sure what I did wrong. $$=\left[\frac{1}{10x-12a}\frac{1}{e^{6a^2-10ax+5x^2}}-\frac{1}{10x+12b}\frac{1}{e^{6b^2+10bx+5x^2}}\right]$$ In every fraction the denominator $\to \infty$ so the whole thing is $0$ but that's incorrect.
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1Your antiderivative is not correct (check it by differentiating). You probably want to make a substitution and transform it to a Gaussian integral. – Matthew Leingang May 07 '23 at 10:29
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Represente the exponent as $a(t+bx)^2+cx^2$ – Ryszard Szwarc May 07 '23 at 10:33
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See https://math.stackexchange.com/questions/9286/evaluation-of-gaussian-integral-int-0-infty-mathrme-x2-dx – Matthew Leingang May 07 '23 at 10:35
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@MatthewLeingang Thank you! I'll try that instead – rose May 07 '23 at 10:41
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We have $$-5x^2+10xt-6t^2=-6\left ( t-{5x \over 6}\right )^2-{5\over 6 }x^2$$ Thus the integral representing the convolution is equal $$e^{-5x^2/6}\int\limits_{-\infty}^\infty e^{-6t^2}\,dt \\ = 6^{-1/2}e^{-5x^2/6}\int\limits_{-\infty}^\infty e^{-t^2}\,dt=\pi^{1/2}6^{-1/2}\,e^{-5x^2/6}$$ On the way we have used the property that the Lebesgue measure is translation invariant.
Ryszard Szwarc
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