Question-133454

Question Number 133454 by rs4089 last updated on 22/Feb/21

Answered by TheSupreme last updated on 22/Feb/21

Ω = {(x, y) ∣ x > 0; 0 < y < x}

Ω = {(x, y) ∣ y > 0, x > y}

I = \int_{0}^{\infty} \int_{0}^{x} e^{- x y} y d y d x = {\int_{0}}^{\infty} {\int_{y}}^{\infty} e^{- x y} y d y d x

\int_{0}^{\infty} \int_{y}^{\infty} e^{- x y} y d x d y = \int_{0}^{\infty} {[- \frac{e^{- x y}}{y^{2}}]}_{y}^{\infty} d y

\int_{0}^{\infty} \frac{e^{- y^{2}}}{y^{2}} d y

f (x) = e^{- y^{2}} \to f^{'} (y) = - 2 {y e}^{y^{2}}

g^{'} (x) = \frac{1}{y^{2}} \to g (x) = - \frac{1}{y}

\int f (x) g^{'} (x) = f (x) g (x) - \int f^{'} (x) g (x)

\int_{0}^{\infty} e^{- y^{2}} / y^{2} d y = - \frac{e^{- y^{2}}}{y} - \int 2 e^{- y^{2}} d y

I = \frac{e^{- y^{2}}}{y} - 2 \frac{\sqrt{π}}{2}

I = \frac{e^{- y^{2}}}{y} - \sqrt{π}