2019牛客多校第一场

$$已知int_0^{infty}frac{1}{1+x^2}dx=frac{pi}{2}.给定数组a,求frac{1}{pi}int_0^{infty}frac{1}{prod{left( a_i^2+x^2 right) }}dx的值. 假设结果为frac{p}{q},输出p cdot q^{-1}.$$

先把所给公式分母下面的公式拿出来化简

$$frac{1}{prod_{i=1}^{n}left( a_{i}^2+x^2right )}$$ $$begin{align}&frac{1}{prod_{i=1}^{n}left( a_{i}^2+x^2right )}\ =&frac{1}{left(a_1^2+x^2right) left(a_2^2+x^2right) }times frac{1}{prod_{i=3}^{n}left( a_{i}^2+x^2right )} \ =& frac{1}{ a_2^2-a_1^2 }times left( frac{1}{a_1^2+x^2}- frac{1}{a_2^2+x^2 }right)times frac{1} {prod_{i=3}^{n}left( a_{i}^2+x^2 right)}\ =& frac{1}{ a_2^2-a_1^2 }times left( frac{1}{a_1^2+x^2}timesfrac{1}{a_3^2+x^2}- frac{1}{a_2^2+x^2 }timesfrac{1}{a_3^2+x^2}right)times frac{1} {prod_{i=4}^{n}left( a_{i}^2+x^2 right)}\ =& left(frac{1}{(a_2^2-a_1^2)(a_3^2-a_1^2)}frac{1}{(a_1^2+x^2)}+frac{1}{(a_1^2-a_2^2)(a_3^2-a_2^2)}frac{1}{(a_2^2+x^2)}+frac{1}{(a_1^2-a_3^2)(a_2^2-a_3^2)}frac{1}{(a_3^2+x^2)} right )times frac{1} {prod_{i=4}^{n}left( a_{i}^2+x^2 right)} end{align}$$

然后总结公式继续化简

$$begin{aligned} &sumlimits_{i=1}^{n}frac{1}{prodlimits_{j=1,jnot=i}^{n}(a_j^2-a_i^2)}int_0^{infty}frac{1}{a_i^2+x^2}dx&\ =&sumlimits_{i=1}^{n}frac{1}{prodlimits_{j=1,jnot=i}^{n}(a_j^2-a_i^2)times a_i^2}int_0^{infty}frac{1}{1+(frac{x}{a_i})^2}dx&\ =&sumlimits_{i=1}^{n}frac{1}{prodlimits_{j=1,jnot=i}^{n}(a_j^2-a_i^2)times a_i}int_0^{infty}frac{1}{1+(frac{x}{a_i})^2}dleft(frac{x}{a_i}right)\ = &sumlimits_{i=1}^{n}frac{1}{2timesprodlimits_{j=1,jnot=i}^{n}(a_j^2-a_i^2)times a_i}end{aligned}$$