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How do I evaluate $$\int_0^\infty e^{-u^2}\cos(xu)du$$

Any hints?

Ben Sheller
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math31
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  • According to Wolfram Alpha this is equal to $\frac{1}{2} \sqrt{\pi} e^{-\frac{x^2}{4}}$. – mxian Apr 11 '17 at 19:00
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    It is even so look at $\int_{-\infty}^\infty e^{-u^2} \cos(xu)du$. Note that for $a \in \mathbb{R}$ : $\int_{-\infty}^\infty e^{-u^2-2au}du=\int_{-\infty}^\infty e^{-(u+a)^2+a^2}du=e^{a^2}\int_{-\infty}^\infty e^{-u^2}du = e^{a^2} \sqrt{\pi} $. Then extend to $a \in \mathbb{C}$ by analytic continuation, and conclude with $a = ix$ – reuns Apr 11 '17 at 19:08
  • Let $I(x)= \int_{0}^{\infty} e^{-u^2} \cos(xu) du$ ... calculate $\frac{dI}{dx}$ ... also integrate the original integral by part (integrate 1) ... I am still working out the detail :-) – Donald Splutterwit Apr 11 '17 at 19:11
  • @BenS. Can you show the OP how to derive this? – unseen_rider Apr 11 '17 at 19:13
  • You might like to consider $\displaystyle \int_0^\infty \exp\left(-(u-\tfrac{ix}{2})^2\right),du$ and some multiple of its real part – Henry Apr 11 '17 at 19:16
  • https://math.stackexchange.com/q/317249/321264 – StubbornAtom Jan 14 '22 at 16:43

2 Answers2

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One may differentiate under the integral sign: let $J(x) = \int_0^{\infty} e^{-u^2}\cos{xu} \, du$. We have $$ J'(x) = -\int_0^{\infty} ue^{-u^2}\sin{xu} \, du $$ Integrating by parts, $$ J'(x) = \left[\frac{1}{2} e^{-u^2}\sin{xu} \right]_0^{\infty} - \frac{x}{2}\int_0^{\infty} e^{-u^2}\cos{xu} \, du, $$ so $J'(x) = -\frac{x}{2}J(x)$. We also know $J(0) = \sqrt{\pi}/2$, and solving the differential equation gives $$ J(x) = \frac{\sqrt{\pi}}{2} e^{-x^2/4}. $$

Chappers
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We have $$\int_0^\infty\!du\,e^{-u^2} \cos(xu) = \frac12 \mathop{\rm Re}\int_{-\infty}^\infty\!du\,e^{-u^2} e^{ixu}$$ shifting the contour to $ u= v+ i x/2$ with $v\in\mathbb{R}$ and noting that the contributions at $|u| \to \infty$ vanish, we obtain $$\int_0^\infty\!du\,e^{-u^2} \cos(xu)= \frac12 \mathop{\rm Re}\int_{-\infty}^\infty\!dv\, e^{-v^2 -x^2/4} = \frac{\sqrt{\pi}}2 e^{-x^2/4}. $$

Fabian
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  • You should warn first the OP that this solution assumes knowledge of contour integration. It is likely that he/she hasn't had lectures on this subject. – Jean Marie Apr 11 '17 at 21:55