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In this post two solutions have been proposed to give a meaning to the integral

$$I=\int_{0}^{+\infty} \frac{x\tan(x)}{1+x^2}\,dx\tag{*}$$

The main problem with this integral is obviously the occurence of poles of the integrand at $x=\pi(n+\frac{1}{2})$, i.e. where the $\cos$ vanishes and the integrand diverges. The solution to get rid of the poles was provided within the concept of the (Cauchy) principal value.

Here we ask for the value of the simpler integral with the same poles

$$i=\int_{0}^{+\infty} \frac{\sec(x)}{1+x^2}\,dx\tag{1}$$

Solution attempts: The first idea is to split the integration region into intervals of length $\pi$ from $\pi n$ to $\pi(n+1)$, $n=0,1,2,...$ take the pricipal value of the integral of each summand

$$i_{n} =PV\int_{\pi n}^{\pi(n+1)} \frac{\sec(x)}{1+x^2}\,dx\tag{2}$$

and sum them up

$$i = \sum_{n=0}^{+\infty} i_{n}\tag{3}$$

Possibly equivalently we could shift the integration contour up from the real axis by a small amount $\epsilon$ and use residues.

Task:

calculate the values of the integral by both ways and show if they are equvalent of not.

Bonus question:

Calculate the integral

$$f=\int_{0}^{+\infty} \frac{x \sec(x)}{1+x^2}\,dx\tag{4}$$

Dr. Wolfgang Hintze
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  • For reasons very similar to the reasons given here, the Cauchy principal value of the integral should be $\frac{\pi}{2 \cosh (1)}$. – Random Variable Jul 07 '21 at 00:06
  • @ Random Variable Thank you. That result corresponds to the second method mentioned in my OP. But why is it the same as that obtained by the first method ($(2)$ and $(3)$)? And, what about the problem $(4)$ where the integrand is not symmetric in $x$? – Dr. Wolfgang Hintze Jul 07 '21 at 09:44
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    Using contour integral technique, we can check that for $b>0$, $$\operatorname{PV}!!\int_{0}^{\infty}\frac{x\sec x}{x^2+b^2},\mathrm{d}x=\frac{\log b}{\cosh b} + \sum_{k=1}^{\infty}(-1)^k\frac{2x_k}{x_k^2+b^2}\log x_k, \qquad x_k=\frac{\pi}{2}(2k-1).$$ I am skeptical that the sum on the right-hand side will simplify into elementary terms. – Sangchul Lee Jul 09 '21 at 08:26
  • Thank you, very interesting. For b=1 I get using Mathematica numerically for the principal value the positive value 0.0627108, while your sum gives the negative result -0.0625179, which is about the same, except for the sign. Could you please elaborate a bit more on the contour integral technique you mentioned? – Dr. Wolfgang Hintze Jul 09 '21 at 21:31
  • Indeed I was worried about signs, as I often make sign mistakes. As for the technique, the idea is to write $$\operatorname{PV}!!\int_{0}^{\infty}\frac{x\sec x}{x^2+b^2},\mathrm{d}x=\frac{1}{2}\operatorname{PV}!!\int_{-\infty}^{\infty}\frac{x\sec x}{x^2+b^2}\left(1 - \frac{2}{\pi i}\log x\right),\mathrm{d}x$$ and then apply the usual contour integration technique as mentioned in other comments. – Sangchul Lee Jul 09 '21 at 23:05

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