Value of $\int_0^\infty e^{-x^2} \cos\alpha x \ dx$
(i) $\frac {\sqrt \pi}{2}e^{-\frac{\alpha^2}{4}}$
(ii)$\frac { \pi}{2}e^{-\frac{\alpha^2}{4}}$
(iii) $\frac {\sqrt \pi}{2}$
(iv) $e^{-\frac{\alpha^2}{4}}$
Explain the answer.
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Realistically the best idea is to note that the above is $\textrm{Re} \int_0^\infty e^{-x^2} e^{i \alpha x} dx$, then complete the square. – Rellek Feb 17 '18 at 19:37
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2I hoped this was a question about the applications for the gamma function... – Yuriy S Feb 17 '18 at 19:52
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https://math.stackexchange.com/q/317249/321264 – StubbornAtom Jan 14 '22 at 16:50
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Let us write $$I(a)=\int_0^\infty e^{-x^2} \cos(ax)dx$$. It is known that $I(0) = \frac{\sqrt{\pi}}{2}$. Taking a derivative wrt $a$ and per partes we get $$I'(a) = -\frac{a}{2}I(a)$$ Solution of this differential equation is $$I(a) = A e^{-\frac{a^2}{4}}$$ Substituting the known value at zero for our integral we get desired result.
Machinato
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Note : This method I found in Nahin's Inside interesting integrals p. 77 – Machinato Feb 22 '18 at 18:08