In a recent question, it was stated in a comment, without proof, that
PV∫∞−∞tanxxdx=π
What is the easiest way to prove this? I was able to show that
PV∫∞−∞tanxxdx=−PV∫∞−∞1xtanxdxPV∫∞−∞1xsinxdx=0∫∞−∞sinxxdx=π
but failed to compute the original integral from this.
Answer
The
question is : What would be a right way to define the principal value
of this integral, knowing that it has infinitely many singularities at
the points π2+πZ ? I will propose the following
PV∫∞0tanxxdx def= lim
Next I will show that the limit in this definition does exist and that
its value is \frac{\pi}{2}.
First, note that the convergence of the integral
\int_0^{\infty}\frac{\sin x\cos x}{x(\cos^2 x+\lambda^2)}dx is easy to prove
using integration by parts. Now
\eqalign{ \int_0^{\infty}\frac{\sin x\cos x}{x(\cos^2 x+\lambda^2)}dx &=\frac{1}{2}\lim_{n\to\infty}\int_{-\pi n}^{\pi (n+1)}\frac{\sin x\cos x}{x(\cos^2 x+\lambda^2)}dx\cr &=\frac{1}{2}\lim_{n\to\infty}\sum_{k=-n}^{n}\int_{\pi k}^{\pi(k+1)}\frac{\sin x\cos x}{x(\cos^2 x+\lambda^2)}dx\cr &=\frac{1}{2}\lim_{n\to\infty}\sum_{k=-n}^{n}\int_{0}^{\pi}\frac{\sin x\cos x}{(x+ \pi k)(\cos^2 x+\lambda^2)}dx\cr &=\frac{1}{2}\lim_{n\to\infty}\int_{0}^{ \pi}\left(\sum_{k=-n}^{n}\frac{1}{x+ \pi k}\right) \frac{\sin x\cos x}{ \cos^2x+\lambda^2}dx\cr &=\frac{1}{2}\lim_{n\to\infty}\int_{0}^{ \pi}U_n(x) \frac{ \cos^2 x}{ \cos^2x+\lambda^2}dx\cr }
where
U_n(x)=\tan(x)\left(\sum_{k=-n}^{n}\frac{1}{x+ \pi k}\right)
But using the well-known expansion of the cotangent function, it is easy to see that
\{U_n \}_n converges point-wise to 1, and that this sequence is bounded uniformely on the interval [0,\pi]. Thus, we can interchange the signs of integral and limit in the above formula to get
\int_0^{\infty}\frac{\sin x\cos x}{x(\cos^2 x+\lambda^2)}dx =\frac{1}{2} \int_{0}^{ \pi} \frac{ \cos^2 x}{ \cos^2x+\lambda^2}dx = \int_{0}^{ \pi/2} \frac{ \cos^2 x}{ \cos^2x+\lambda^2}dx
Finally, taking the limit as \lambda\to0 we get
PV\int_0^\infty\frac{\tan x}{x}dx~=~ \lim_{\lambda\to0}\int_0^{\infty}\frac{\sin x\cos x}{x(\cos^2 x+\lambda^2)}dx=\frac{\pi}{2}.
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