Prove $ \frac1{\sqrt{(z+a)(z+b)}}=\frac1\pi \int_a^b \frac1{\sqrt{(t-a)(b-t)}}\frac1{t+z}dt $

Question

Reading papers online i found the following definition

$$ \frac1{\sqrt{(z+a)(z+b)}}=\frac1\pi \int_a^b \frac1{\sqrt{(t-a)(b-t)}}\frac1{t+z}dt $$

for $$ b>a \,\, and \,\,\,z \in \Bbb C |(-\infty,-a] $$

Question : How can this be proven? would the proof for this above definition be similar for another expression of type:

$$\frac{f(x)}{\sqrt{(z+a)(z+b)}}$$

Thank you kindly for your help and time.

Answer 1

Let $t = \frac{a+b}{2}+\frac{b-a}{2}\sin\theta$. Then

$$\pi I(z) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{d\theta}{\frac{a+b}{2}+\frac{b-a}{2}\sin\theta+z} = \int_0^\pi \frac{d\theta}{\frac{a+b}{2}+\frac{b-a}{2}\cos\theta+z}$$

Then denoting $\frac{a+b}{2}+z \equiv k_1$ and $\frac{b-a}{2} \equiv k_2$ we get that

$$\pi I(z) = \int_0^\pi \frac{d\theta}{(k_1+k_2)\cos^2\left(\frac{\theta}{2}\right)+(k_1-k_2)\sin^2\left(\frac{\theta}{2}\right)}$$

$$= \frac{2}{k_1-k_2}\int_0^{\pi}\frac{\frac{1}{2}\sec^2\left(\frac{\theta}{2}\right) \:d\theta}{\frac{k_1+k_2}{k_1-k_2}+\tan^2\left(\frac{\theta}{2}\right)} = \frac{2}{\sqrt{k_1^2-k_2^2}}\tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\sqrt{\frac{k_1+k_2}{k_1-k_2}}\right)\Biggr|_0^\pi = \frac{\pi}{\sqrt{k_1^2-k_2^2}}$$

Then plugging in for $k_1$ and $k_2$ gives us that

$$I(z) = \frac{1}{\sqrt{(k_1-k_2)(k_1+k_2)}} = \frac{1}{\sqrt{(z+a)(z+b)}}$$

Answer 2

Substitute $x= \frac1{t+z}$, along with the shorthands $p = b+z$, $q=a+z$,

\begin{align} I &=\int_a^b \frac{1}{\sqrt{(t-a)(b-t)}}\frac{1}{t+z} dt \\ &=-\frac{1}{\sqrt{pq}}\int_{\frac1q}^{\frac1p}\frac{dx}{\sqrt{(x-\frac1p)(\frac1q-x)}} =\frac{1}{\sqrt{pq}}\int_{\frac1q}^{\frac1p}\frac{-\frac{2pq}{p-q}dx}{\sqrt{1-\left(\frac{p+q-2pqx}{p-q}\right)^2}}\\ \end{align}

Then, let $u = \frac{p+q-2pqx}{p-q}$, $$I =\frac{1}{\sqrt{pq}}\int_{-1}^{1}\frac{du}{\sqrt{1-u^2}}=\frac{1}{\sqrt{pq}}\sin^{-1}u|_{-1}^{1} = \frac{\pi}{\sqrt{pq}} = \frac{\pi}{\sqrt{(z+b)(z+a)}} $$