let-a-gt-0-calculate-0-dx-x-2-a-2-2-2-calculate-x-2-x-2-a-2-2-dx-

Question Number 30773 by abdo imad last updated on 25/Feb/18

l e t a > 0 c a l c u l a t e \int_{0}^{\infty} \frac{d x}{{(x^{2} + a^{2})}^{2}}

2) c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{x^{2}}{{(x^{2} + a^{2})}^{2}} d x .

Commented by abdo imad last updated on 26/Feb/18

1) t h e c h . x = a t a n t g i v e

I = \int_{0}^{\infty} \frac{d x}{{(x^{2} + a^{2})}^{2}} = \int_{0}^{\frac{π}{2}} \frac{1}{a^{4} {(1 + {t a n}^{2} t)}^{2}} a (1 + {t a n}^{2} t) d t

= \frac{1}{a^{3}} \int_{0}^{\frac{π}{2}} \frac{d t}{1 + {t a n}^{2} t} \Rightarrow a^{3} I = \int_{0}^{\frac{π}{2}} {c o s}^{2} t d t

= \int_{0}^{\frac{π}{2}} \frac{1 + c o s (2 t)}{2} d t = \frac{π}{4} + {[\frac{1}{4} s i n (2 t)]}_{0}^{\frac{π}{2}} = \frac{π}{4} \Rightarrow

I = \frac{π}{4 a^{3}}

2) l e t p u t J = \int_{- \infty}^{+ \infty} \frac{x^{2}}{{(x^{2} + a^{2})}^{2}} d x

J = 2 \int_{0}^{\infty} \frac{x^{2} + a^{2} - a^{2}}{{(x^{2} + a^{2})}^{2}} d x = 2 \int_{0}^{\infty} \frac{d x}{(x^{2} + a^{2})} d x - 2 a^{2} \int_{0}^{\infty} \frac{d x}{{(x^{2} + a^{2})}^{2}} d x

c h x = a t \Rightarrow \int_{0}^{\infty} \frac{d x}{(x^{2} + a^{2})} = \int_{0}^{\infty} \frac{a d t}{a^{2} (1 + t^{2})} = \frac{1}{a} \frac{π}{2} = \frac{π}{2 a}

J = \frac{π}{a} - 2 a^{2} I = \frac{π}{a} - 2 a^{2} \frac{π}{4 a^{3}} = \frac{π}{a} - \frac{π}{2 a} \Rightarrow J = \frac{π}{2 a} .