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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.

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

1 Answer 1

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

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