Integral: Investigating Convergence I

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    Convergence Integral
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SUMMARY

The discussion focuses on the convergence of the integral $\int_{0}^{\infty} \frac{-\ln \left[ \frac{1}{a} x ^{ \frac{2}{a-1} } e ^{- x^{\frac{2}{a} }} \right] }{1+x^2} \mbox{d}x$ for values of $a \ge 2$. It is established that the integral converges for $a > 2$ and diverges when $a = 2$. The analysis involves rewriting the integral as $I = \int_{0}^{\infty} f(x)\ d x = \int_{0}^{\infty} \frac{\ln a - \frac{2}{a-1}\ \ln x + x^{\frac{2}{a}}}{1 + x^{2}}\ dx$, leading to the asymptotic behavior of $f(x)$ as $x$ approaches infinity.

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Investigate the convergence of the integral
$\int_{0}^{\infty} \frac{-\ln \left[ \frac{1}{a} x ^{ \frac{2}{a-1} } e ^{- x^{\frac{2}{a} }} \right] }{1+x^2} \mbox{d}x$ for $a \ge 2$
 
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maxkor said:
Investigate the convergence of the integral
$\int_{0}^{\infty} \frac{-\ln \left[ \frac{1}{a} x ^{ \frac{2}{a-1} } e ^{- x^{\frac{2}{a} }} \right] }{1+x^2} \mbox{d}x$ for $a \ge 2$

If You write the integral as ...

$\displaystyle I = \int_{0}^{\infty} f(x)\ d x = \int_{0}^{\infty} \frac{\ln a - \frac{2}{a-1}\ \ln x + x^{\frac{2}{a}}}{1 + x^{2}}\ dx\ (1)$

... in x tends to $\infty$ You have $\displaystyle f(x) \sim \frac{x^{\frac{2}{a}}}{1 + x^{2}}$, so that the integral converges for a>2 and diverges for a=2...

Kind regards

$\chi$ $\sigma$
 
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