Study the convergence and calculate the improper integral

Question

Hoping someone can help me with this one.

Study the convergence and, in case it's affirmative, calculate the improper integral:

$\int_a^b{\frac{x dx}{\sqrt{(x-a)(b-x)}}} $ , where $ b>a $

Answer 1

Hints.

The indefinite integral exists and it's pretty trivial to evaluate:

$$\int \frac{dx}{\sqrt{(x-a)(b-x)}} = -\frac{\sqrt{x-a} \sqrt{b-x} \tan ^{-1}\left(\frac{a+b-2 x}{2 \sqrt{x-a} \sqrt{b-x}}\right)}{\sqrt{(x-a) (b-x)}}$$

Where as your definite integral is simply $\pi$.

You have a square root of a second degree equation in the denominator. Start from it.

You can also come to the final result by taking limits.

Answer 2

substitute $$\sqrt{x^2-x(a+b)+ab}=t+x$$

Answer 3

$I=\displaystyle\int _{a}^{b} \dfrac{x}{\sqrt{(x-a)(b-x)}} dx$

use property $\displaystyle\int_{a}^{b} f(a+b-x)dx=\displaystyle\int_{a}^{b}f(x)dx$ and add both I's together then you will get;

$I=\dfrac{a+b}{2}\displaystyle\int _{a}^{b} \dfrac{1}{\sqrt{(x-a)(b-x)}} dx$

substitute $(x-a)=t(b-x)\implies x=\dfrac{a+tb}{t+1}\implies dx=\dfrac{b-a}{(t+1)^2}dt$

$$I=\dfrac{a+b}{2}\displaystyle\int _{0}^{\infty} \dfrac{t^\frac{-1}{2}}{t+1} dt$$

$$I=\dfrac{a+b}{2}\displaystyle\int _{0}^{\infty} \dfrac{t^\frac{-1}{2}}{t+1} dt$$

substitute $t=x^2\implies 2xdx=dt$

$$I={(a+b)}.\displaystyle\int _{0}^{\infty} \dfrac{1}{x^2+1} dx$$

$$I=\dfrac{\pi(a+b)}{2}$$