How do we prove that ∫∞1sin4(logx)x2logxdx=log(62517)16
I tried substitutions like logx=arcsint, but it doesn't seem to work out. Please help me out. Thank you.
Answer
The integral screams for a sub x=eu; the result is
∫∞0due−usin4uu
This is very computable by introducing a parameter and differentiating under the integral. In this case, consider
F(k)=∫∞0due−kusin4uu
F′(k)=−∫∞0due−kusin4u
F′(k) is relatively easy to compute using the fact that sin4u=38−12cos2u+18cos4u, and that
∫∞0due−kucosmu=kk2+m2
Thus
F′(k)=−38k+12kk2+4−18kk2+16
and
F(k)=−116log[k6(k2+16)(k2+4)4]+C
To evaluate C, we must consider limk→∞F(k) because F(0) represents a non convergent integral. Because the limit is zero, we must have C=0. The integral we seek is then
F(1)=116log62517
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