Question

求一道高数反常积分的计算题，谢谢了

Answer 1

【答案】B

【解析】

原式=∫[0~a]lnx/(x²+a²)·dx+∫[a~+∞]lnx/(x²+a²)·dx

先计算 P=∫[0~a]lnx/(x²+a²)·dx
设x=a·tant，
则dx=a·sec²t·dt
x=0时，t=0
x=a时，t=π/4

P=∫[0~π/4]ln(a·tant)/(a²sec²t)·a·sec²t·dt

=∫[0~π/4]ln(a·tant)/a·dt

=1/a·∫[0~π/4](lna+lntant)·dt

=πlna/(4a)+1/a·∫[0~π/4]lntant·dt

再计算 Q=∫[a~+∞]lnx/(x²+a²)·dx
设x=a·cotu，
则dx=-a·csc²u·du
x=a时，u=π/4
x→+∞时，u→0

Q=∫[π/4~0]ln(a·cotu)/(a²csc²u)·(-a·csc²u·du)

=∫[0~π/4]ln(a·cotu)/a·du

=1/a·∫[0~π/4](lna+lncotu)·du

=πlna/(4a)+1/a·∫[0~π/4]lncotu·du

综上所述，

原式=P+Q
=πlna/(2a)+1/a·∫[0~π/4]lntant·dt+1/a·∫[0~π/4]lncotu·du

=πlna/(2a)+1/a·∫[0~π/4]lntanu·du+1/a·∫[0~π/4]lncotu·du

=πlna/(2a)+1/a·∫[0~π/4](lntanu+lncotu)·du

=πlna/(2a)+1/a·∫[0~π/4]ln(tanu·cotu)·du

=πlna/(2a)+1/a·∫[0~π/4]ln1·du

=πlna/(2a)