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I am trying to compute the following contour integration but am quite stuck I have to evaluate it analytically, by extending it to the complex plane and solving an appropriate integral involving a complex variable $z= x + iy$:

$$\int_0^\infty \frac{x^2}{(x^2-4)(x^2-9)}\,\text dx$$

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    $\begingroup$ What does the foggy rectangle symbol mean? $\endgroup$ – Nishant Apr 6 '15 at 17:49
  • $\begingroup$ @Nishant: i'm going to go with a guess that it is a mistaken character that doesn't need to be there and hope that the OP will double-check to make sure that the resulting edit is correct... $\endgroup$ – abiessu Apr 6 '15 at 17:52
  • $\begingroup$ The integral is divergent. If there was a symbol there, maybe it was meant to convey that. $\endgroup$ – Ron Gordon Apr 6 '15 at 18:00
  • $\begingroup$ Are you sure the denominator is not $(x^2+4)(x^2+9)$? $\endgroup$ – N. S. Apr 6 '15 at 18:04
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HINT:

W/O using Complex Calculus,

$$\frac{x^2}{(x^2-4)(x^2-9)}=\frac15\cdot\frac{9(x^2-4)-4(x^2-9)}{(x^2-4)(x^2-9)}=\cdots$$

Now use this

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    $\begingroup$ You do realize that the integral is divergent, right...? $\endgroup$ – Ron Gordon Apr 6 '15 at 17:59
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by using the partial fraction $$\int_{0}^{\infty }\frac{x^2}{(x^2-4)(x^2-9)}dx=(\int_{0}^{\infty }\frac{1}{5(x-2)}-\frac{1}{5(x-2)}+\frac{3}{10(x-3)}-\frac{3}{10(x+3)})dx$$ $$=\frac{1}{5}\log(\frac{x-2}{x+2})+\frac{3}{10}\log(\frac{x+3}{x-3}) $$ take the limit when $x=\infty$ to get $$\lim_{x\rightarrow \infty }\log(\frac{x-2}{x+2})=0$$ $$\lim_{x\rightarrow \infty }\log(\frac{x+3}{x-3})=0$$

$$I=-\frac{1}{5}\log(-1)-\frac{3}{10}\log(-1)=-\frac{i\pi }{2}$$

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    $\begingroup$ How does it make sense to get a complex number out of a limit of sums of real numbers...? Even the Cauchy PV will never do this. You have picked the wrong branch cuts. $\endgroup$ – Ian Apr 6 '15 at 20:27
  • $\begingroup$ I used a direct substitute of the boundary in the integral $\endgroup$ – E.H.E Apr 6 '15 at 20:30
  • $\begingroup$ What about the poles on the axis? Any thought as to why you think the Fundamental theorem of Calculus applies there? $\endgroup$ – Ron Gordon Apr 7 '15 at 17:02

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