I need help to solve the next improper integral using complex analysis:
∫∞−∞x6(4+x4)2dx
I have problems when I try to find residues for the function f=1(z4+4)2.
This is what I tried.
res(f,√2ei(π4+kπ2))=limz→√2ei(π4+kπ2)((z−√2ei(π4+kπ2))2(z4−4)2)′
with k∈{0,1,2,3}. What do you think about it?
I know there is a little more general problem involving this integral; for all a>0
∫∞−∞x6(a4+x4)2dx=3π√28a
Edit.
I've had an idea: from the integration by parts
∫udv=uv−∫vdu
and if we let
dv=4x3(4+x4)2,u=x34
with
dv=−ddx14+x4=4x3(4+x4)2
we get finally
∫∞−∞x6(4+x4)2dx=0+34∫∞−∞x21+x4dx
which I think is more easy to solve.
Anyway, if you know another idea or how to complete my first try will be welcome.
Answer
If you are right, then ∫∞−∞x21+x4dx can be easily done by a semi-circle contour, computing the residues at (1+i)/√2 and (−1+i)/√2. It is easy to check that the integral of z21+z4 on the arc of the semi-circle goes to 0.
z21+z4=z2(z−eiπ/4)(z−ei3π/4)(z−ei5π/4)(z−ei7π/4) shows that it is a simple pole at those residues.
Res(z21+z4,eiπ/4)=limz→eiπ/4z2(z−eiπ/4)1+z4=limz→eiπ/43z2−2zeiπ/44z3=(34−12)e−iπ/4.
Res(z21+z4,ei3π/4)=limz→ei3π/4z2(z−ei3π/4)1+z4=limz→ei3π/43z2−2zei3π/44z3=(34−12)e−i3π/4.
I used L'Hospital's rule above because it seems simpler.
So the answer is ∫∞−∞x21+x4dx=2πi14(1−i√2+−1−i√2)=√22π.
Edit: Something's wrong with the integration by parts; we do not seem to get the right answer. It is a 4 in the bottom and not 1. But we can do a further substitution to get it right.
∫∞−∞x24+x4dx=1√2∫∞−∞1√2(x√2)21+(x√2)4dx=1√2∫∞−∞x21+x4dx.
So the final answer is ∫∞−∞x6(4+x4)2dx=34∫∞−∞x24+x4dx=38π. You also have to argue that you are limiting r→∞ in [x34(4+x4)]r−r which is why it is 0.
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