Solution
We use du=d(arctanx)=1/(1+x²)dx,
We have that the integral is
∫ (_0^(π/2))(tanx)^pdx,-1<p<1
By using the beta function,we have that the intergal is equal to the under
∫(_0^(π/2))(tanx)^pdx
Noticed that
∫(_0^(π/2))(tanx)^pdx
=∫ (_0^(π/2))(sinx/cosx)^pdx
=(B(1/2+p/2,1/2-p/2))/2
Using the
relationship between the gamma function and the beta function,
B(p,q)=Γ(p)Γ(q)/Γ(p+q)
it is easy to get the result:
B(1/2+p/2,1/2-p/2)
=Γ(1/2+p/2)Γ(1/2-p/2)/Γ(1/2+p/2+1/2-p/2)
=Γ(1/2+p/2)Γ(1/2-p/2)/Γ(1)
=Γ(1/2+p/2)Γ(1/2-p/2)
As for the formula,in fact, the name of the formula is Remainder formula:
Γ(x)Γ(1-x)=π/sinπx,0<x<1
We have that:
π/(2sinπ(1/2+p/2)). 口
这个题是以前做过的,老师说题不难但看卷子比较严格……
比如,我用了余元却没有证明,扣分5分......
然后没有证明beta函数Γ函数关系,扣分5分
这样,20分的题只有10分了........
