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I am trying to evaluate the Integral:

$$\mathrm\int_{-\infty}^{\infty}\frac{x^2+1}{[x^2-2x\sin\alpha+1][x^2+2x\sin\alpha+1]}dx\\$$

Where $\alpha$ is an arbitrary constant

My book says : Since this is an even function, adding an odd function to the numerator will not change the integral

Therefore this integral can be written as:

$$\int_{-\infty}^{\infty}\frac{x^2-2x\sin\alpha+1}{[x^2-2x\sin\alpha+1][x^2+2x\sin\alpha+1]}dx\\$$

What I understand: I know and understand that the integrand of the integral I am trying to evaluate is Even.

What I don't understand is that why adding an odd function to the numerator won't change the integral, and how is this valid?

My thoughts:

I know for an odd function: $$\int_{-a}^{a} f_o(x) dx=0$$

And I can write what the book says as :

$$\int_{-\infty}^{\infty}\frac{x^2+1}{[x^2-2x\sin\alpha+1][x^2+2x\sin\alpha+1]}dx$$

$$=\int_{-\infty}^{\infty}\frac{x^2+1}{[x^2-2x\sin\alpha+1][x^2+2x\sin\alpha+1]}dx -\int_{-\infty}^{\infty}\frac{2x\sin\alpha}{[x^2-2x\sin\alpha+1][x^2+2x\sin\alpha+1]}dx$$


But , how do we know that:

$$\int_{-\infty}^{\infty}\frac{2x\sin\alpha}{[x^2-2x\sin\alpha+1][x^2+2x\sin\alpha+1]}dx$$

is ODD? ,

because this integrand is of the form: $$\frac{f_o(x)}{g_e(x)}$$

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  • $\begingroup$ An odd function divided by an even function is again odd, for if $f$ is odd and $g$ is even then $(f/g)(-x) = f(-x)/g(-x) = (-f(x))/g(x) = - (f/g)(x)$. $\endgroup$ – Joshua Mundinger Dec 2 '14 at 4:31
  • $\begingroup$ @Alqatrkapa Thank you so much $\endgroup$ – M.S.E Dec 2 '14 at 4:35
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If $\displaystyle H(x)=\frac{2x\sin\alpha}{[x^2-2x\sin\alpha+1][x^2+2x\sin\alpha+1]},$

$$H(-x)=\frac{2(-x)\sin\alpha}{[(-x)^2-2(-x)\sin\alpha+1][(-x)^2+2(-x)\sin\alpha+1]}$$

$$=\frac{-2x\sin\alpha}{[x^2+2x\sin\alpha+1][x^2-2x\sin\alpha+1]}=-H(x)$$

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