MHB Integrating $\frac{x^2}{(1+x^2)^3}$ Over the Real Line

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The integral of the function $\frac{x^2}{(1+x^2)^3}$ over the real line can be evaluated using complex analysis, specifically by identifying a pole of order 3 at z=i in the upper half plane. The result of the integral is calculated as $\frac{\pi}{8}$. An alternative method suggested involves using a tangent substitution for integration. There is a question raised about whether this is a homework problem, indicating a potential academic context. The discussion emphasizes the use of both complex analysis and substitution techniques for solving the integral.
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integrate $\frac{x^2}{(1+x^2)^3}$ over the real line
 
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Fermat said:
integrate $\frac{x^2}{(1+x^2)^3}$ over the real line

[sp]The function has in the upper half plane a pole of order 3 in z=i, so that is...

$\displaystyle \int_{- \infty}^{+ \infty} \frac{x^{2}}{(1 + x^{2})^{3}}\ d x = 2\ \pi i\ \lim_{z \rightarrow i} \frac{1}{2}\ \frac {d^{2}}{d z^{2}}\ \frac{z^{2}}{(z+i)^{3}} = \frac{\pi}{8}$[/sp]


Kind regards

$\chi$ $\sigma$
 
chisigma said:
[sp]The function has in the upper half plane a pole of order 3 in z=i, so that is...

$\displaystyle \int_{- \infty}^{+ \infty} \frac{x^{2}}{(1 + x^{2})^{3}}\ d x = 2\ \pi i\ \lim_{z \rightarrow i} \frac{1}{2}\ \frac {d^{2}}{d z^{2}}\ \frac{z^{2}}{(z+i)^{3}} = \frac{\pi}{8}$[/sp]


Kind regards

$\chi$ $\sigma$

that is a way I had no really expected (although it is correct). What about a tan substitution
 
Fermat said:
integrate $\frac{x^2}{(1+x^2)^3}$ over the real line

Is this a homework problem?
Suspecting that this is a homework problem, I leave a hint, use $x=\tan\theta$. The resulting integral is straightforward.
 
Thread 'Erroneously finding discrepancy in transpose rule'

Thread 'Erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
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