# $\Gamma [\frac {(m+1)}{(n-r+1)}]$ while it could be $\Gamma [(\frac {m+1}{n})-r+1]$which one is correct?

$$\int_0^\infty \frac{1}{(x^2+1)^{3/2}}\,dx$$

point:

$$\int_0^\infty \frac{x^m \, dx}{({x^n+a^n)}^r}=\frac{(-1)^{r-1}\pi a^{m+1-nr}\Gamma [(m+1)/n]}{n\sin[\pi (m+1)/n](r-1)!\Gamma[(m+1)/n-r+1]}, \quad 0<m+1<nr$$

$\Gamma [\frac {(m+1)}{(n-r+1)}]$ while it could be $\Gamma [(\frac {m+1}{n})-r+1]$ which one is correct?

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What do you mean by invalid value? –  anon271828 Feb 7 '13 at 19:22
This integral converges. –  Ron Gordon Feb 7 '13 at 19:23
@anon271828 The answer should be defined, but there is no answer for this one! –  Neo Feb 7 '13 at 19:24
@Neo: Many problems are easier to solve without a calculator. And if you do need a numerical answer, often it is best if the calculator is only used at the end. –  André Nicolas Feb 7 '13 at 19:40
The formula is for integer $r$, not for values like $r=3/2$. The $((m+1)/n)-r+1$ is the correct one. –  GEdgar Feb 7 '13 at 20:40

Let $x=\tan{\theta}$, $dx=\sec^2{\theta} \, d \theta$:

\begin{align}\int_0^\infty dx \: \frac{1}{(x^2+1)^{3/2}} &= \int_0^{\pi/2} d \theta \frac{\sec^2{\theta}}{\sec^3{\theta}} \\ &= \int_0^{\pi/2} d \theta \: \cos{\theta} \\&= 1 \end{align}

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The integral converges. It is even easy to evaluate, make the substitution $x=\tan\theta$. After a short while, you will be integrating $\cos\theta$.

One can see that the the integral converges without evaluating it. Break it up as the integral from $0$ to $1$, where the function is well-behaved, and the integral from $1$ to $\infty$. On the second interval, our function is less than $\dfrac{1}{x^3}$. So by the Comparison Test the integral from $1$ to $\infty$ converges.

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It is well-defined.

By definition, the improper integral is equal to $$\lim_{a\to \infty} \int_0^a \frac{1}{(1+x^2)^{3/2}} dx,$$ if this limit exists. And it does:

$$\lim_{a\to \infty} \int_0^a \frac{1}{(1+x^2)^{3/2}} dx = \lim_{a\to\ \infty} \frac{a}{\sqrt{1+a^2}}-\frac{0}{\sqrt{1+0^2}} = 1 - 0.$$

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Letting $\sinh(u)=x$, we get

$$\int_{0}^{\infty}\frac{1}{\cosh^2(u)}\mathrm{du}=\left[\tanh u\right]_0^{\infty}=1$$

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