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This integral $$\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}=a\pi, ~ a, b \in \Re$$ looks suspiciously interesting as it is independent of the parameter $b$. The question is: What is the best way of proving or disproving this?

Z Ahmed
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    Looks like you could use Glasser's Master Theorem here. – clathratus Jun 19 '19 at 03:03
  • The first way can be using the contour integration with a semicircular contour centered at origin. It also seems that with some proper substitution or manipulation, the Glasser's Master Theorem can be used. Using Beta function could also be an appropriate choice, if you could somehow create the required form in the integrand. I am currently in a journey , I will rely to post the complete solution when I will be back home. – Rohan Shinde Jun 19 '19 at 03:04
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    The integral is crying for some Glasser's Master Theorem... – Sangchul Lee Jun 19 '19 at 03:27
  • Thanks, I did not know GMT! – Z Ahmed Jun 19 '19 at 04:12

3 Answers3

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One way:

Denote the integral as $$I=\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}.$$ $$(x^2-b^2)^2+a^2x^2=(x^2+p^2)(x^2+q^2) \Rightarrow p=(a+c)/2,q=(a-c)/2, c=\sqrt{a^2-4b^2}.$$ Then $$I=\frac{2a^2}{p^2-q^2} \int_{0}^{\infty} \left( \frac{p^2}{x^2+p^2}-\frac{q^2}{x^2+q^2} \right)dx= \frac{2a^2}{p^2-q^2}[p\tan^{-1}(x/p)-q \tan^{-1}(x/q)]_{0}^{\infty}=\frac{a^2\pi}{p+q}=a\pi.$$

Z Ahmed
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Glasser's master theorem states that for arbitrary constants $\alpha$, $(\alpha_n)_{n=1}^{N}$, $(\beta_n)_{n=1}^N$, the function $$\phi(x)=|\alpha|x-\sum_{n=1}^{N}\frac{|\alpha_n|}{x-\beta_n},$$ and any integrable function $F(x)$, $$\mathrm{PV}\int_{-\infty}^{\infty}F(\phi(x))dx=\mathrm{PV}\int_{-\infty}^\infty F(x)dx.$$ For your integral, set $$\phi(x)=x-\frac{b^2}{x}$$ and $$F(\phi(x))=\frac{a^2x^2}{(x^2-b^2)^2-a^2x^2}=\frac{1}{(\frac{x}{a}-\frac{b^2}{ax})^2+1}$$ to immediately yield the desired result.

clathratus
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    If you are citing the version stated in Wolfram MathWorld, then there are some error in the statement. The correct statement is that, if you are using $\phi$ as in your answer, then $$\text{PV}!!\int_{-\infty}^{\infty}F(\phi(x)),\mathrm{d}x=\frac{1}{|\alpha|}\text{PV}!!\int_{-\infty}^{\infty}F(x),\mathrm{d}x.$$ Just consider the trivial case $N = 0$ to see the necessity of the factor $1/|\alpha|$ on the RHS. – Sangchul Lee Jun 19 '19 at 06:33
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How about this, please check it critically

Denote the integral as $$I=\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}.$$ $$(x^2-b^2)^2+a^2x^2=(x^2-r^2)(x^2-s^2) \Rightarrow r=(d+ia)/2,s=(d-ia)/2, d=\sqrt{4b^2-a^2}.$$ $$I=\frac{a^2}{r^2-s^2} \int_{-\infty}^{\infty} \left( \frac{r^2}{x^2-r^2}-\frac{s^2}{x^2-s^2} \right)dx.$$ By considering a semi-circle contour in the upper half plane and applying the residue theorem, we get $$I=\frac{a^2}{r^2-s^2} 2i\pi \left( r^2 Res \left (\frac{1}{x^2-r^2} \right)_{x=r}-s^2 Res \left ( \frac{1}{x^2-s^2} \right)_{x=-s}\right)=\frac{i\pi a^2}{r-s}=a\pi.$$

Z Ahmed
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