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How can one evaluate the following integral?

$\int_{-c}^{c} e^{-ax^2}\cos^2(bx) \,\mathrm{d}x$

Wolfram Alpha gives this. Is there not a more compact form?

If $\int_{-c}^{c} e^{-ax^2} \, \mathrm{d}x=k$, then can we express the first integral in terms of $k$? Thanks.

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    $\text{erfi}(x) = -i \text{erf}(ix)$2011-10-16

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How can one integrate $\int_{-c}^{c} e^{-ax^2}\cos^2(bx) \,\mathrm{d}x$?

Write $\cos^2(bx)=\frac12(1+\cos(2bx))=\frac12+\frac14 e^{2b i x}+\frac14 e^{-2b i x}\tag1$ Then $e^{-ax^2}\cos^2(bx) =\frac12e^{-ax^2}+ \frac14 e^{-ax^2+2b i x}+\frac14 e^{-ax^2-2b i x}\tag2$ With appropriate substitutions, the matter reduces to integration of $\exp(-t^2)$, which yields the error function. Of course, this is what WA already did for you.

If $\int_{-c}^{c} e^{-ax^2} \, \mathrm{d}x=k$, then can we express the first integral in terms of $k$?

No, because the integral of a product is not the product of integrals... But it is true that $ \lim_{b\to\infty} \int_{-c}^{c} e^{-ax^2}\cos^2(bx) \,\mathrm{d}x =\frac{k}{2}\tag3$ because the highly oscillatory terms $e^{\pm 2b i x} $ in (1) contribute little when $b$ is large. See the Riemann-Lebesgue lemma.