Question

please solve this integration

Answer 1

Answer:

$I=\sqrt{\Pi}$

Step-by-step explanation:

$I=\int_{-\infty }^{\infty }e^{-y^{2}}dy\\I^{2}=[\int_{-\infty }^{\infty }e^{-y^{2}}dy]^{2}\\I^{2}=\int_{-\infty }^{\infty }e^{-y^{2}}dy\cdot \int_{-\infty }^{\infty }e^{-y^{2}}dy\\I^{2}=\int_{-\infty }^{\infty }e^{-x^{2}}dx\cdot \int_{-\infty }^{\infty }e^{-y^{2}}dy\\I^{2}=\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }e^{-(x^{2}+y^{2})}dxdy$

Now since in polar coordinate system r² = x² + y² and dxdy = rdrdθ

Substituting,

$I^{2}=\int_{0}^{2\pi}\int_{0}^{\infty }r\cdot e^{-r^{2}}drd\Theta \\\\Take\thinspace u=-r^{2},du=\frac{-1}{2}r\cdot dr\\\\\\I^{2}=\frac{-1}{2}\int_{0}^{2\pi }{\left ( e^{-r^{2}} \right )_{0}}^{-\infty }d\Theta \\\\I^{2}=\frac{-1}{2}\int_{0}^{2\pi }(0-1)d\Theta \\I^{2}=\frac{1}{2}\int_{0}^{2\pi }d\Theta \\I^{2}=\Pi\\I=\sqrt{\Pi}$

Answer 2

Answer:

sorry dear......................