Question

Sa se calculeze integrala:
$\displaystyle\int_{-\infty}^{\infty}e^{-x^2}dx$
Am reusit sa o aduc pana aici
$\displaystyle\int_0^{\infty}e^{-t}t^{\frac{1}{2}-1}dt = \Big\Gamma\Big(\frac{1}{2}\Big)$​

Answer 1

Răspuns:

Salut, gamma de 1/2 este radical din pi, munca grea deja ai realizat-o. E o integrala improprie aceea de tip Gamma.

Explicație pas cu pas:

Answer 2

$\displaystyle I = \int_{-\infty}^{\infty}e^{-x^2}\, dx\\ \\\\ I\cdot I =\int_{-\infty}^{\infty}e^{-x^2}\, dx\cdot \int_{-\infty}^{\infty}e^{-y^2}\, dy =\\ \\\\=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x^2} e^{-y^2}\, dx\, dy =\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x^2-y^2}\, dx\,dy\\\\\\\Rightarrow I^2=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x^2-y^2}\, dx\, dy$

Se observă că regiunea de integrare este întregul plan xOy.

Voi schimba variabilele în coordonate polare.

x² + y² = r² ⇒ -x² - y² = -r²

$\displaystyle I^2 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}\, dx\, dy\\\\\\x = r\cos \Theta \\ y = r\sin \Theta\\ r^2 = x^2+y^2\\ dx\,dy =r\,dr\,d\Theta\\ \\r:\,\, 0\to \infty \\ \Theta:\,0\to 2\pi\\ \\\\I^2 = \int_{r=0}^{\infty}\int_{\Theta = 0}^{2\pi}e^{-r^2}r\,d\Theta\, dr =\int_{\Theta=0}^{2\pi}\, d\Theta\cdot \int_{r=0}^{\infty}e^{-r^2}r\, dr = \\ \\= 2\pi\int_{r=0}^{\infty}e^{-r^2}r\, dr$

$\displaystyle -r^2 = u \Rightarrow -2r\, dr = du \Rightarrow r\, dr= -\dfrac{1}{2}\,du\\r\to 0\Rightarrow u\to 0\\ r\to \infty\Rightarrow u\to -\infty\\ \\ I^2 = 2\pi\int_{0}^{-\infty}-e^{u}\cdot \dfrac{1}{2}\, du \\ \\ I^2 = -\pi\int_{0}^{-\infty}e^u\, du \\ \\ I^2 = -\pi e^{u}\Big|_{0}^{-\infty}\\ \\ I^2 = -\pi\cdot 0 -(-\pi \cdot 1)\\ \\ I^2 = \pi \\ \\ \boxed{\Rightarrow I = \sqrt{\pi}}$