charlesreid1

An Integral from Hardy

Here's an integral from G. H. Hardy:

$$ \int \dfrac{dx}{\sqrt{x+a} + \sqrt{x+b}} $$

It doesn't fit any of the standard undergraduate forms. There's no obvious $u$-substitution, integration by parts doesn't help, and a trigonometric substitution turns it into a mess.

The Setup

The trouble is the sum of two square roots in the denominator. With just one square root the integral is a power-rule exercise:

$$ \int \dfrac{dx}{\sqrt{x+a}} = 2\sqrt{x+a} + C $$

But the sum resists the usual substitutions. Its derivative involves both roots in a form that doesn't cancel cleanly against anything you'd put in the numerator.

Multiply by the Conjugate

Multiply numerator and denominator by the conjugate $\sqrt{x+a} - \sqrt{x+b}$:

$$ \dfrac{1}{\sqrt{x+a} + \sqrt{x+b}} \cdot \dfrac{\sqrt{x+a} - \sqrt{x+b}}{\sqrt{x+a} - \sqrt{x+b}} $$

By the difference-of-squares identity $(u+v)(u-v) = u^2 - v^2$, with $u = \sqrt{x+a}$ and $v = \sqrt{x+b}$, the denominator becomes:

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

The $x$'s cancel and the denominator is a constant.

Putting It Together

The integral is now:

$$ \int \dfrac{dx}{\sqrt{x+a} + \sqrt{x+b}} = \dfrac{1}{a - b} \int \left( \sqrt{x+a} - \sqrt{x+b} \right) dx $$

Each piece is a power-rule integral. Using $\int (x+c)^{1/2} \, dx = \tfrac{2}{3}(x+c)^{3/2}$:

$$ \int \sqrt{x+a} \, dx = \dfrac{2}{3}(x+a)^{3/2} $$

$$ \int \sqrt{x+b} \, dx = \dfrac{2}{3}(x+b)^{3/2} $$

So:

$$ \int \dfrac{dx}{\sqrt{x+a} + \sqrt{x+b}} = \dfrac{2}{3(a-b)} \left[ (x+a)^{3/2} - (x+b)^{3/2} \right] + C $$

Sanity Check by Differentiation

Let

$$ F(x) = \dfrac{2}{3(a-b)} \left[ (x+a)^{3/2} - (x+b)^{3/2} \right] $$

Differentiating:

$$ F'(x) = \dfrac{2}{3(a-b)} \cdot \dfrac{3}{2} \left[ (x+a)^{1/2} - (x+b)^{1/2} \right] = \dfrac{\sqrt{x+a} - \sqrt{x+b}}{a-b} $$

To match the original integrand, multiply top and bottom by $\sqrt{x+a} + \sqrt{x+b}$:

$$ \dfrac{\sqrt{x+a} - \sqrt{x+b}}{a - b} \cdot \dfrac{\sqrt{x+a} + \sqrt{x+b}}{\sqrt{x+a} + \sqrt{x+b}} = \dfrac{(x+a) - (x+b)}{(a-b)(\sqrt{x+a} + \sqrt{x+b})} $$

The numerator is $a - b$, which cancels, leaving:

$$ F'(x) = \dfrac{1}{\sqrt{x+a} + \sqrt{x+b}} $$

The $a = b$ Case

The answer has $a - b$ in the denominator, so it's undefined at $a = b$. But the original integral degenerates there: the denominator becomes $2\sqrt{x+a}$ and

$$ \int \dfrac{dx}{2\sqrt{x+a}} = \sqrt{x+a} + C $$

The singularity is removable — taking $b \to a$ in the closed form (via L'Hôpital or a Taylor expansion of $(x+b)^{3/2}$ around $b = a$) recovers $\sqrt{x+a} + C$.

References

G. H. Hardy, A Course of Pure Mathematics, Cambridge University Press. Available on Project Gutenberg.
Wikipedia, Rationalisation (mathematics).

A Hard(y) Math Problem