Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am interested in the evaluation of

$$\int^1_0 \left(1-x^2\right)^{n/2} \sin(kx)\, dx$$

Really $n$ is an odd integer here (the even case is easy).

Ooops, I think the posting

Integral $\int_{-1}^1 \frac{e^x}{\sqrt{1-x^2}}dx$

pretty much covers it, esp. the answer from J.M.

share|cite|improve this question
    
Did yoy try trigonometric substitutions? like $1-x^2 = \cos^2(\theta)$? – kjetil b halvorsen Oct 22 '12 at 18:32

Set $x = \cos(\theta)$. This gives us $$I = \int_0^{\pi/2} \sin^{n/2}(\theta) \sin(k \cos(\theta)) \sin(\theta) d\theta = \int_0^{\pi/2} \sin(k \cos(\theta)) \sin^{n/2+1}(\theta) d\theta$$ The integral can be expressed in terms of the Struve function.

$$H_{\alpha}(k) = \dfrac{2(k/2)^{\alpha}}{\sqrt{\pi} \Gamma(\alpha + 1/2)} \int_0^{\pi/2} \sin(k\cos(\theta)) \sin^{2 \alpha}(\theta) d \theta$$ In our case, $\alpha = \dfrac{n}4 + \dfrac12$.

Hence, $$I = \int_0^{\pi/2} \sin(k \cos(\theta)) \sin^{n/2+1}(\theta) d\theta = \dfrac{\sqrt{\pi} \Gamma(n/4+1)}{2(k/2)^{n/4+1/2}} H_{n/4+1/2}(k)$$

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.