For $a^2
∫∞0dxx2+2ax+b
What about:
∫∞0dx(x2+2ax+b)2
My attempt is using partial fractions and completing the square, but I still failed to obtain a nice result.
Answer
Let the first integral be I and the second one be J, then by putting x=y−a and y=z√b−a2 we have
I(a,b)=∫∞0dx(x+a)2+b−a2=∫∞adyy2+b−a2=1√b−a2∫∞a√b−a2dzz2+1=1√b−a2(π2−arctan(a√b−a2))
and the 2nd integral is just a first derivative of I with respect to b times −1
J(a,b)=−∂I∂b=∂∂b[1√b−a2(arctan(a√b−a2)−π2)]
Can you take it from here?
No comments:
Post a Comment