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I want to compute the following integral,

$\int_0^\infty\frac{dx}{(x^2 + p)^{n}}$

for any $n \in \mathbb{N}$

I've tried to add a parameter, obtaining its integral and then taking derivatives with respect to the extra parameter and so on, but wasn't able to generalize for any $n$.

For example, I've tried integrating

$\frac{dx}{(x^2 + p + t)^{-1}}$

and that gives $\int_0^\infty\frac{dx}{(x^2 + p + t)^{-1}} = \frac{\pi}{2(t+p)^{0.5}}$.

Then I differentiate both sides and get $\int_0^\infty\frac{dx}{(x^2 + p + t)^{-2}} = \frac{\pi}{4(t+p)^{1.5}}$.

If I use t = 0 using the last expression I'll obtain the integral for $n = 2$ but I cannot generalize this approach for any $n \in \mathbb{N}$.

Thanks

EDIT: If I do this for $n = 3$, I obtain $\int_0^\infty\frac{dx}{(x^2 + p + t)^{-3}} = \frac{3\pi}{16(t+p)^{2.5}}$ and so on. I can answer it for every case through differentiating again and again. Let me rephrase my question, is it possible to obtain some pattern from the RHS of the results of the integrals such that I can have a function that under $t = 0$ only depends on $n$? [That is, given any $n$ I can obtain automatically get the desired integral]. I'm not sure if this is possible or trivial but I cannot see it.

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  • $\begingroup$ Why can't you generalize it? It seems like you just have to keep taking derivatives (do not compute the integral new each time, but use that the integral for $n+1$ is (up to constant factors) the derivative of the integral for $n$). $\endgroup$ – PhoemueX Oct 27 '14 at 5:47
  • $\begingroup$ Thanks for the reply. I add more detail to my question. Besides that can you give an example of how can you generalize it? $\endgroup$ – user3777221 Oct 27 '14 at 15:34
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First, a straightforward induction shows that

$$ \frac{d^n}{d p^n} (p + x^2)^{-1} = (-1)^n \cdot n! \cdot (p+x^2)^{-(n+1)}. $$

Using differentiation under the integral sign, we see that the function

$$ F(p) := \int_0^\infty \frac{1}{x^2 + p}\, dx $$

has $n$-th derivative

$$ F^{(n)} (p) = (-1)^n \cdot n! \cdot \int_0^\infty \frac{1}{(x^2 + p)^{n+1}}\, dx. $$

But as you already calculated, we also have

$$ F(p) = \frac{\pi}{2} p^{-1/2}. $$

Another straightforward induction shows that

$$ \frac{d^n}{d p^n} p^{-1/2} = (-1)^n \frac{\prod_{j=1}^n (2j-1)}{2^n} p^{-(2n+1)/2}. $$

Rearranging everything, we finally arrive at

$$ \int_0^\infty \frac{1}{x^2 + p} \,dx = \frac{\pi}{2^{n+1} \cdot n!} \cdot \prod_{j=1}^n (2j-1) \cdot p^{-(2n+1)/2}. $$

Comparison shows that this is identical to your results in the cases that you calculated.

In general, if you have expressions like these, I recommend to try the induction step first in order to see how the derivative changes (multiplication with $n+1$ or with $2n+1$, ...) in order to find the "closed form" for the induction.

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  • $\begingroup$ Also note that the introduction of the parameter $t$ was not actually necessary, as $p$ can be considered as a variable in its own right. $\endgroup$ – PhoemueX Oct 27 '14 at 20:45
  • $\begingroup$ Everything is clear now. Thanks, @PhoemueX ! $\endgroup$ – user3777221 Oct 27 '14 at 21:00

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