Encontrar un integral con fracciones

Question

Cómo encontrar el % integral $$\int_0^\infty \frac{e^{-x^2}}{(x^2+1/2)^2}dx?$$

Me parece que es difícil de hacer si integrar por partes... ¿Cuál es el truco?

Answer 1

$$\int\dfrac{e^{-x^2}}{\left(x^2+\dfrac{1}{2}\right)^2}dx=-\int\dfrac{e^{-x^2}}{2x}d\left(\dfrac{1}{x^2+\dfrac{1}{2}}\right)=-\dfrac{e^{-x^2}}{2x\left(x^2+\dfrac{1}{2}\right)}-\int\dfrac{e^{-x^2}}{x^2}$$ and $$-\int\dfrac{e^{-x^2}}{x^2}=\int e^{x^2}d\dfrac{1}{x}=\dfrac{e^{-x^2}}{x}+2\int e^{-x^2}dx$$

tan $$me = \int_ {0} ^ {+ \infty} \dfrac {e ^ {-x ^ 2}} {\left (x ^ 2 + \dfrac {1} {2} \right) ^ 2} dx = \dfrac{xe^{-x^2}}{x^2+\dfrac{1}{2}}|_{0}^{+\infty}+2\int_{0}^{+\infty}e^{-x^2}dx=\sqrt{\pi}$$

Solution2: desde $$e^{-x^2}=\int_{x}^{+\infty}2te^{-t^2}dt$ $ lo$$ me = \int_ {0} ^ {+ \infty} \left [\int_ {x} ^ {+ \infty} \dfrac {2te ^ {-t ^ 2}} {\left (x ^ 2 + \dfrac {1} {2} \right) ^ 2} dt\right] dx = \int_ {0} ^ {+ \infty} 2te ^ {-t ^ 2} \left (\sqrt {2} \arctan {\sqrt {2} t} + \dfrac {t} {t ^ 2 + 1/2} \right) dt así $$I=\int_{0}^{+\infty}\dfrac{2t^2}{t^2+1/2}e^{-t^2}-\int_{0}^{+\infty}\sqrt{2}\arctan{\sqrt{2}t}de^{-t^2}=2\int_{0}^{+\infty}e^{-t^2}dt=\sqrt{\pi}