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$ \sqrt {(1+a^2/x^2)} =>\frac1 x\sqrt{(x^2+a^2)}$
The first expression is even (i.e, remains same when we put $-x$ in place of $x$), while the second one is odd. What am I doing wrong while going from the first to the second expression? The expression has came as a result of integration of $\int\frac{dx}{(x^2+a^2)^{3/2}}$ from $-L$ to $L$. If I take the second expression the integration is non-zero, which I think is correct from the plot of $\frac{1}{(x^2+a^2)^{3/2}}$. For the other expression it is zero.
I am not asking for the result of the integration. My questions are 1. How both the expressions are not same? 2. In general if there is a function which is discontinuous at the origin, how to find whether it is even or odd. Thank you for any help.

Martin Sleziak
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Siddhartha
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    $\sqrt {1+\frac{a^2}{x^2}}=\frac{1}{|x|}\sqrt{x^2+a^2}$, which is even either way. – Shubham Johri Nov 29 '18 at 19:03
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    Being even/odd means that $f(t)=\pm f(-t)$ for $t>0$. How does this have anything to do with a discontinuity at $0$? – Federico Nov 29 '18 at 19:03
  • @Federico yes, there is no relation between being discontinuous at the origin and being even/odd. I understand, thanks. I got confused seeing the plots... for even and odd both cases f(0) give same value, if f(0) is finite. But in 1/x or 1/x^2 cases for even and for odd the plots shows different things at the origin.. but still then, even and odd, can be easily defined as you mentioned. Thanks! – Siddhartha Nov 29 '18 at 19:17
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    @ShubhamJohri Thanks! I got it. $\sqrt{x^2}=\vert x \vert $ always. – Siddhartha Nov 29 '18 at 19:29

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