On December 20th 2020, I created a post titled Feynman Integration in which I used Feynman's technique to show that: $$\int_\infty^\infty \frac{\cos x}{x^2+1} \mathrm{d}x=\frac{\pi}{\mathrm{e}}$$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:$$\int_0^1 \frac{\sin(\ln x)}{\ln x} \mathrm{d}x=\frac{\pi}{4}$$So let's get started. Firstly, we need to remember that:$$\begin{align}
\sin z &= \frac{e^{iz}-e^{-iz}}{2i}\\
\therefore \sin( \ln x)&=\frac{e^{i \ln x}-e^{-i \ln x}}{2 i}\\
&=\frac{e^{(\ln x)^{i}}-e^{(\ln x )^{-i}}}{2i}\\
&=\frac{x^i-x^{-i}}{2i}
\end{align}$$Thus our original integral changes:$$\begin{align} \int_0^1 \frac{\sin(\ln x)}{\ln x} \mathrm{d}x&=\int_0^1 \frac{x^i-x^{-i}}{2i \ln x} \mathrm{d}x\\
\text{ let }I(b) &=\int_0^1 \frac{x^{bi}-x^{-i}}{2i \ln x} \mathrm{d}x \text{ where }b \text{ is a parameter}\\
\frac{\mathrm{d}}{\mathrm{d}b}I(b) &=\frac{\mathrm{d}}{\mathrm{d}b}\int_0^1 \frac{x^{bi}-x^{-i}}{2i \ln x} \mathrm{d}x\\
I'(b)&=\int_0^1 \frac{\partial{}}{\partial{b}} \left (\frac{x^{bi}-x^{-i}}{2i \ln x} \right ) \mathrm{d}x \\
&=\int_0^1 \frac{\ln x \cdot x^{bi} \cdot i}{2i \ln x}\\
&=\frac{1}{2} \int_0^1 x^{bi} \mathrm{d} x\\
&=\frac{1}{2}\left[\frac{x^{bi+1}}{bi+1}\right]_{x=0}^{x=1}\\
&=\frac{1}{2(bi+1)}\\
\therefore I(b)&=\frac{\ln(bi+1)}{2i}+C\\
\text{but } I(-1)&=\int_0^1 \frac{x^{-1}-x^{-1}}{2i \ln x} \mathrm{d} x \text{ when }b=-1\\
&=0\\
\therefore C&=-\frac{\ln(1-i)}{2i}\\
\text{and } I(b)&=\frac{\ln(bi+1)}{2i}-\frac{\ln(1-i)}{2i}\\
I(1)&=\frac{\ln(1+i)}{2i}-\frac{\ln(1-i)}{2i}\\
&=\frac{1}{2i} \left (\ln(1+i)-\ln(1-i) \right )\\
&=\frac{\ln \left (\dfrac{1+i}{1-i} \right )}{2i}\\
&=\frac{1}{2i} \cdot \ln i\\
&=\frac{1}{2i} \cdot \frac{\pi}{2} \cdot i\\
&=\frac{\pi}{4}\\
\text{thus }\int_0^1 \frac{\sin(\ln x)}{\ln x} \mathrm{d}x&=\frac{\pi}{4}
\end{align}$$
\sin z &= \frac{e^{iz}-e^{-iz}}{2i}\\
\therefore \sin( \ln x)&=\frac{e^{i \ln x}-e^{-i \ln x}}{2 i}\\
&=\frac{e^{(\ln x)^{i}}-e^{(\ln x )^{-i}}}{2i}\\
&=\frac{x^i-x^{-i}}{2i}
\end{align}$$Thus our original integral changes:$$\begin{align} \int_0^1 \frac{\sin(\ln x)}{\ln x} \mathrm{d}x&=\int_0^1 \frac{x^i-x^{-i}}{2i \ln x} \mathrm{d}x\\
\text{ let }I(b) &=\int_0^1 \frac{x^{bi}-x^{-i}}{2i \ln x} \mathrm{d}x \text{ where }b \text{ is a parameter}\\
\frac{\mathrm{d}}{\mathrm{d}b}I(b) &=\frac{\mathrm{d}}{\mathrm{d}b}\int_0^1 \frac{x^{bi}-x^{-i}}{2i \ln x} \mathrm{d}x\\
I'(b)&=\int_0^1 \frac{\partial{}}{\partial{b}} \left (\frac{x^{bi}-x^{-i}}{2i \ln x} \right ) \mathrm{d}x \\
&=\int_0^1 \frac{\ln x \cdot x^{bi} \cdot i}{2i \ln x}\\
&=\frac{1}{2} \int_0^1 x^{bi} \mathrm{d} x\\
&=\frac{1}{2}\left[\frac{x^{bi+1}}{bi+1}\right]_{x=0}^{x=1}\\
&=\frac{1}{2(bi+1)}\\
\therefore I(b)&=\frac{\ln(bi+1)}{2i}+C\\
\text{but } I(-1)&=\int_0^1 \frac{x^{-1}-x^{-1}}{2i \ln x} \mathrm{d} x \text{ when }b=-1\\
&=0\\
\therefore C&=-\frac{\ln(1-i)}{2i}\\
\text{and } I(b)&=\frac{\ln(bi+1)}{2i}-\frac{\ln(1-i)}{2i}\\
I(1)&=\frac{\ln(1+i)}{2i}-\frac{\ln(1-i)}{2i}\\
&=\frac{1}{2i} \left (\ln(1+i)-\ln(1-i) \right )\\
&=\frac{\ln \left (\dfrac{1+i}{1-i} \right )}{2i}\\
&=\frac{1}{2i} \cdot \ln i\\
&=\frac{1}{2i} \cdot \frac{\pi}{2} \cdot i\\
&=\frac{\pi}{4}\\
\text{thus }\int_0^1 \frac{\sin(\ln x)}{\ln x} \mathrm{d}x&=\frac{\pi}{4}
\end{align}$$
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 \( \dfrac{\sin(\ln x)}{\ln x}\) look like? This is what my next post will examine.
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