Thursday, 25 January 2018

integration - Finding intlimitsi1nftyfracsin4(logx)x2logxmathrmdx




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



0dueusin4uu



This is very computable by introducing a parameter and differentiating under the integral. In this case, consider




F(k)=0duekusin4uu



F(k)=0duekusin4u



F(k) is relatively easy to compute using the fact that sin4u=3812cos2u+18cos4u, and that



0duekucosmu=kk2+m2



Thus




F(k)=38k+12kk2+418kk2+16



and



F(k)=116log[k6(k2+16)(k2+4)4]+C



To evaluate C, we must consider limkF(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


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...