On December 20th 2020, I created a post titled Feynman Integration in which I used Feynman's technique to show that: ∫∞∞cosxx2+1dx=πe
Lately my daily number analysis has not generated any material for a post but I have been practising my LaTeX skills and I thought that this blackpenredpen video that involved the use of Feynman's technique provided a good opportunity in this regard. Here is the video:
In this post, I'll be showing that:∫10sin(lnx)lnxdx=π4
So let's get started. Firstly, we need to remember that:sinz=eiz−e−iz2i∴sin(lnx)=eilnx−e−ilnx2i=e(lnx)i−e(lnx)−i2i=xi−x−i2i
Thus our original integral changes:∫10sin(lnx)lnxdx=∫10xi−x−i2ilnxdx let I(b)=∫10xbi−x−i2ilnxdx where b is a parameterddbI(b)=ddb∫10xbi−x−i2ilnxdxI′(b)=∫10∂∂b(xbi−x−i2ilnx)dx=∫10lnx⋅xbi⋅i2ilnx=12∫10xbidx=12[xbi+1bi+1]x=1x=0=12(bi+1)∴I(b)=ln(bi+1)2i+Cbut I(−1)=∫10x−1−x−12ilnxdx when b=−1=0∴C=−ln(1−i)2iand I(b)=ln(bi+1)2i−ln(1−i)2iI(1)=ln(1+i)2i−ln(1−i)2i=12i(ln(1+i)−ln(1−i))=ln(1+i1−i)2i=12i⋅lni=12i⋅π2⋅i=π4thus ∫10sin(lnx)lnxdx=π4
Even though complex numbers were used in the solution, the final result for our real integral yielded a real solution. The typesetting for this post took quite some time but it provided valuable practice. So what does the graph of the function sin(lnx)lnx look like? This is what my next post will examine.
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