This is my work for solving the improper integralI=∞∫0dxlog2xx2+a2
I feel like I did everything write, but when I substitute values into a, it doesn’t match up with Wolfram Alpha.
First substitute x=a2u soI=−0∫∞du2log2a−log2ux2+a2=∞∫0du2log2aa2+x2−∞∫0dulog2ua2+x2
HenceI=∞∫0dulog2aa2+x2=log2aa2π/2∫0dtasec2t1+tan2t=πlog2a2a
However, when a=e Wolfram Alpha evaluates the integral numerically asI≈2.00369
however the input that I arrived at evaluates numericallyπ2e≈0.5778
Where did I go wrong? And how would you go about solving this integral?
Answer
Note
log2(a2u)≠2log2a−log2u.
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