0

$$ \int_0^\infty e^{-x^2}\cos(x) dx$$

I got to the following

$$\Re[\int_0^\infty e^{-(x-{i\over2})^2}.e^{1\over4}dx]$$

How I proceed further ?

RKK
  • 418
  • Start by writing

    $$\int_0^\infty e^{-x^2}\cos(x),dx=\frac12 \int_{-\infty}^\infty e^{-x^2}\cos(x),dx$$

    Then, applying Euler's formula we have

    $$\int_0^\infty e^{-x^2}\cos(x),dx=\frac12 \text{Re}\int_{-\infty}^\infty e^{-1/4}e^{-(x-i/2)^2},dx$$

    Finally, exploiting Cauchy's integral theorem, we find that

    $$\int_0^\infty e^{-x^2}\cos(x),dx=\frac12 \sqrt\pi e^{-1/4}$$

     – Mark Viola Oct 10 '21 at 17:18
  • Thanks I got it – RKK Oct 10 '21 at 19:18
  • You're welcome. My pleasure. – Mark Viola Oct 10 '21 at 21:52

0 Answers0