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Let $c,d>0$. I do not know how to integrate the following:

$$\int_{-d}^d \frac{z}{2\sqrt{z+c}\ \cdot \sqrt{d^2-z^2}}\ \text{d}z.$$


Sorry~~ it should be $z^2$.

I think it can be simplified to the following: for $c\ge d>0$

$$\frac{\sqrt{d}}{2} \int_{-\pi/2}^{\pi/2} \frac{\sin y}{\sqrt{\sin y+c/d}}\ \text{d}y.$$

Thanks again!

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you should start by simplifying the denominator. –  lampShade Mar 11 '11 at 2:50
    
$z^{z}$ i don't think we can solve this. –  anonymous Mar 11 '11 at 4:01
    
@moron: is it $z^{z}$ or $z^{2}$ –  anonymous Mar 11 '11 at 4:02
    
@Chandru: Looks like $z^z$ to me. –  Aryabhata Mar 11 '11 at 5:13

3 Answers 3

up vote 3 down vote accepted

$$\frac12\int_{-d}^d\frac{z}{\sqrt{(d^2-z^2)(c+z)}}\mathrm{d}z$$

is equivalent to (by depressing the cubic within the square root):

$$=-\int_{-d-c/3}^{d-c/3}\frac{z+c/3}{\sqrt{\frac{8cd^2}{3}-\frac{8c^3}{27}-\left(\frac{4c^2}{3}+4d^2\right)z+4z^3}}\mathrm{d}z$$

Using the substitution

$$z=\wp\left(u;\frac{4c^2}{3}+4d^2,\frac{8c^3}{27}-\frac{8cd^2}{3}\right)$$

where $\wp(u;g_2,g_3)$ is a Weierstrass elliptic function, and making use of the differential relation for the Weierstrass function:

${\wp^{\prime}}^2=4\wp^3-g_2\wp-g_3$

we have

$$-\int_{\wp^{(-1)}(-d-c/3)}^{\wp^{(-1)}(d-c/3)}\left(\wp\left(u;\frac{4c^2}{3}+4d^2,\frac{8c^3}{27}-\frac{8cd^2}{3}\right)+\frac{c}{3}\right)\mathrm du$$

whose integration I'll leave to the interested reader.

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Chapeau, monsieur. –  Gunnar Magnusson Jun 28 '11 at 20:32

I see somebody did it Weierstrass-style; I'll try out the Jacobi approach here for completeness.

With

$$\frac12\int_{-d}^d\frac{x}{\sqrt{(x+c)(d-x)(d+x)}}\mathrm dx$$

and the assumption $c \geq d > 0$, we use the substitution

$$x=d\left(1-2\mathrm{sn}^2\left(u\mid\frac{2d}{c+d}\right)\right)$$

which turns the integral into

$$\frac{d}{\sqrt{c+d}}\int_0^{K\left(\frac{2d}{c+d}\right)}\left(1-2\mathrm{sn}^2\left(u\mid\frac{2d}{c+d}\right)\right)\mathrm du$$

or, simplifying,

$$\frac{d}{\sqrt{c+d}}\left(K\left(\frac{2d}{c+d}\right)-2\int_0^{K\left(\frac{2d}{c+d}\right)}\mathrm{sn}^2\left(u\mid\frac{2d}{c+d}\right)\mathrm du\right)$$

We can use formula 22.16.15 in the DLMF to handle the remaining integral. Simplification yields the expression

$$\sqrt{c+d}E\left(\frac{2d}{c+d}\right)-\frac{c}{\sqrt{c+d}}K\left(\frac{2d}{c+d}\right)$$

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I wouldn't know how to do it manually, but contrarily to the case of yunone, Wolfram does yield an answer to me:

$$-\frac{\pi d^2 \left(2\frac{d^2}{c^2}\right)}{8 c^{3/2}}$$

Under the conditions that $$\Re(c)>0\land d>0\land (c\notin \mathbb{R}\lor d\leq \Re(c))$$

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hi, just use copy - Latex option in mathematica. –  Kerry Mar 11 '11 at 7:22
    
Hm, that resulted in as it stands now.... –  JBSnorro Mar 11 '11 at 7:25
    
I don't think this is the correct result. Why would it be negative? As long as $d^d< d^2$ and $c>d$ you are actually integrating a positive function and the result has to be positive. –  Fabian Mar 11 '11 at 9:34

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