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$p(\lambda) = C\frac{\sqrt{(\lambda_{\max}-\lambda)(\lambda-\lambda_{\min})}}{\lambda}$ is defined in the interval $[\lambda_{\min},\lambda_{\max}]$ and zero outside the interval.

By spread, I assume, we mean the variance $\big(\operatorname{Var}(\lambda)\big)^2 = \langle \lambda ^2 \rangle - \langle \lambda \rangle ^2$.

Now, I've tried computing the integral $\displaystyle \langle \lambda ^2 \rangle = \int_{\lambda_{\min}}^{\lambda_{\max}}\lambda^2p(\lambda) \, d\lambda$. But, it doesn't solve that easily for me.

Any guess on an alternative way of finding the so-called spread?

Context.

$p$ describes the distribution of eigenvalues of the Wishart matrix.

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    I don't think this question can admit any reasonable answer until you tell us what function $p$ is. – 2017-01-11
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    It's in the title. – 2017-01-11
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    @MusséRedi: Please refrain from using LateX in the title and move the formula to the body of the question. – 2017-01-11
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    @AlexR. : I don't think there's a reason to avoid MathJax (for which the prevalent casual misnomer is "LaTeX") in titles entirely, but in a case like this, certainly the body of the question should include this information. – 2017-01-11

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First let us complete the square: \begin{align} & (\lambda_\max-\lambda)(\lambda - \lambda_\min) \\[10pt] = {} & -\Big(\lambda^2 -(\lambda_\min+\lambda_\max) \lambda\Big) - \lambda_\min \lambda_\max \\[10pt] = {} & -\left(\lambda^2 -(\lambda_\min+\lambda_\max) \lambda + \left(\frac{\lambda_\min+\lambda_\max} 2 \right)^2 \right) - \lambda_\min \lambda_\max + \left(\frac{\lambda_\min+\lambda_\max} 2 \right)^2 \\[10pt] = {} & -\left( \lambda- \frac{\lambda_\min + \lambda_\max} 2 \right)^2 + \left( \frac{\lambda_\max - \lambda_\min} 2 \right)^2 \end{align}

This suggests a substitution: \begin{align} \left( \frac{\lambda_\max - \lambda_\min} 2 \right) \sin\theta & = \lambda - \frac{\lambda_\min+\lambda_\max} 2 \\[10pt] \left( \frac{\lambda_\max - \lambda_\min} 2 \right) \cos\theta \, d\theta & = d\lambda \end{align} The integral then becomes $$ \int_{\lambda_\min}^{\lambda_\max} \lambda^2 p(\lambda) \, d\lambda $$ $$ = \int_{-\pi/2}^{\pi/2} \Big( (\lambda_\min+\lambda_\max) + (\lambda_\max-\lambda_\min)\sin\theta) \Big) (\cos\theta) (\lambda_\max - \lambda_\min) (\cos\theta \, d\theta). $$

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    How did you arrive at the suggested substitution? – 2017-01-12
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    @MusséRedi : When you have $\Big( \cdots \cdots (a^2 - x^2) \cdots \cdots \Big) \,dx$ then you can try $x= a\sin\theta.$ That's what we have here. $\qquad$ – 2017-01-12