- #1

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$$\int_{0}^{\infty}{dk \frac{k^2 e^{-k^2 x^2}}{(k^2 + s)^{3/2}}}$$

Thank you for your help.

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- Thread starter fermi
- Start date

- #1

- 76

- 5

$$\int_{0}^{\infty}{dk \frac{k^2 e^{-k^2 x^2}}{(k^2 + s)^{3/2}}}$$

Thank you for your help.

- #2

ShayanJ

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You should show some attempt at solving it.

- #3

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- #4

- 94

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- #5

FactChecker

Science Advisor

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Do you have a reason to think that it has a closed form solution?

- #6

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- #7

ShayanJ

Gold Member

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$$

\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{n!} \int_{0}^\infty \frac{k^{2n+2}}{(k^2+s)^{\frac 3 2}} dk

$$

But I'm not sure how to deal with the integrals!

As you can see, its an even function of k so maybe it can be contour integrated but I have problem with it.

- #8

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- #9

FactChecker

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I think the e^{-k2x2} needs to stay inside the integration to make it converge.

- #10

FactChecker

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But I think ## \exp(-|z|^2 x^2) ## will not be an analytic function, so the contour integration would not be valid.

- #11

- 94

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Probably. But maybe a better analytic function can be found.

- #12

- 76

- 5

$$

\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{n!} \int_{0}^\infty \frac{k^{2n+2}}{(k^2+s)^{\frac 3 2}} dk

$$

But I'm not sure how to deal with the integrals!

As you can see, its an even function of k so maybe it can be contour integrated but I have problem with it.

I am happy to get an answer in terms of a power series if a closed form answer cannot be found. However, the power series you suggested has a problem for n=0 term in the series. For n=0, the integral:

$$

\int_{0}^\infty \frac{k^{2n+2}}{(k^2+s)^{\frac 3 2}} dk

$$

does not converge. This sort of divergence can happen when you expand the integrand in power series, even when the whole integral is strongly convergent. The integral I am trying to evaluate is strongly convergent for s > 0 (That's easy to prove). In fact, it is likely to be convergent for all Real s, by allowing the variable 's' to acquire a small positive imaginary part, and taking the limit that the imaginary part goes to zero after integration. (But that's much harder to prove. It feels intuitively right, but I have no proof yet for s<=0.)

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