Evaluate an integral involving $\arctan$ and a square root

Solution

Set

$$h(x)=\frac{\arctan\!\left(\sqrt{2-x^2}\right)}{1+x^2}.$$

Since $h$ is even, we have

$$I = \int_0^1 h(x)\,dx + \int_0^{\sqrt2} h(x)\,dx.$$

Let

$$K=\int_0^1 h(x)\,dx, \qquad J=\int_0^{\sqrt2} h(x)\,dx.$$

Thus

$$I=K+J.$$

We first evaluate $J$. More generally, define

$$F(a) = \int_0^a \frac{\arctan\!\left(\sqrt{a^2-x^2}\right)} {1+x^2}\,dx, \qquad a>0.$$

Differentiating with respect to $a$, the boundary term vanishes since $\arctan(0)=0$, and therefore

$$F'(a) = a\int_0^a \frac{dx} {(1+x^2)(1+a^2-x^2)\sqrt{a^2-x^2}}.$$

Using

$$x=a\sin\theta, \qquad 0\leq \theta\leq \frac{\pi}{2},$$

we obtain

$$F'(a) = a\int_0^{\pi/2} \frac{d\theta} {(1+a^2\sin^2\theta)(1+a^2\cos^2\theta)}.$$

Now

$$(1+a^2\sin^2\theta)(1+a^2\cos^2\theta) = 1+a^2+\frac{a^4}{4}\sin^2(2\theta).$$

Hence, using the standard integral

$$\int_0^{\pi/2}\frac{du}{A+B\sin^2u} = \frac{\pi}{2\sqrt{A(A+B)}},$$

we get

$$F'(a) = \frac{\pi a}{(a^2+2)\sqrt{1+a^2}}.$$

Since $F(0)=0$,

$$F(a) = \pi\int_0^a \frac{t\,dt}{(t^2+2)\sqrt{1+t^2}}.$$

Put

$$u=\sqrt{1+t^2}.$$

Then

$$du=\frac{t\,dt}{\sqrt{1+t^2}}, \qquad t^2+2=u^2+1,$$

so

$$F(a) = \pi\int_1^{\sqrt{1+a^2}} \frac{du}{1+u^2}.$$

Therefore

$$F(a) = \pi\left( \arctan\sqrt{1+a^2}-\frac{\pi}{4} \right).$$

Taking $a=\sqrt2$, we find

$$J=F(\sqrt2) = \pi\left(\frac{\pi}{3}-\frac{\pi}{4}\right) = \frac{\pi^2}{12}.$$

It remains to determine $K$.

Using

$$\arctan y=\int_0^y\frac{dt}{1+t^2},$$

we may write

$$J = \iint_Q \frac{dx\,dy}{(1+x^2)(1+y^2)},$$

where

$$Q= \left\{ (x,y)\in\mathbb R^2: x\geq0,\ y\geq0,\ x^2+y^2\leq2 \right\}.$$

Inside $Q$, consider the square

$$C=[0,1]\times[0,1].$$

Its weighted area is

$$\iint_C \frac{dx\,dy}{(1+x^2)(1+y^2)} = \left(\int_0^1\frac{dx}{1+x^2}\right)^2 = \left(\frac{\pi}{4}\right)^2 = \frac{\pi^2}{16}.$$

The two remaining parts of $Q$,

$$\{x>1\} \qquad\text{and}\qquad \{y>1\},$$

have equal weighted areas by the symmetry $x\leftrightarrow y$.

If $L$ denotes either of these two equal contributions, then

$$J=\frac{\pi^2}{16}+2L.$$

On the other hand, $K$ corresponds to the part of $Q$ with $0\leq x\leq1$, hence

$$K=\frac{\pi^2}{16}+L.$$

Consequently,

$$K = \frac12\left( J+\frac{\pi^2}{16} \right).$$

Thus

$$I=J+K = \frac12\left( 3J+\frac{\pi^2}{16} \right).$$

Since

$$J=\frac{\pi^2}{12},$$

we finally obtain

$$I = \frac12 \left( \frac{\pi^2}{4} + \frac{\pi^2}{16} \right) = \boxed{\frac{5\pi^2}{32}}.$$

Alternative solution

We use the identity

$$\arctan u=\int_0^u \frac{dy}{1+y^2}.$$

Hence

$$I = \int_{-1}^{\sqrt2} \int_0^{\sqrt{2-x^2}} \frac{dy\,dx}{(1+x^2)(1+y^2)}.$$

Thus $I$ is the weighted area of the region

$$R= \left\{ (x,y):x^2+y^2\le 2,\quad y\ge0,\quad x\ge-1 \right\},$$

with weight

$$w(x,y)=\frac1{(1+x^2)(1+y^2)}.$$

Let

$$Q=\{(x,y):x\ge0,\ y\ge0,\ x^2+y^2\le2\}$$

be the quarter disk, and denote

$$J=\iint_Q w(x,y)\,dx\,dy.$$

We first compute $J$.

Using polar coordinates

$$x=r\cos\theta,\qquad y=r\sin\theta,$$

we obtain

$$J= \int_0^{\sqrt2} \int_0^{\pi/2} \frac{r\,d\theta\,dr} {(1+r^2\cos^2\theta)(1+r^2\sin^2\theta)}.$$

Now observe that

$$\frac{1} {(1+r^2\cos^2\theta)(1+r^2\sin^2\theta)}$$

can be written as

$$\frac1{2+r^2} \left( \frac1{1+r^2\cos^2\theta} + \frac1{1+r^2\sin^2\theta} \right),$$

because

$$(1+r^2\cos^2\theta)+(1+r^2\sin^2\theta) =2+r^2.$$

Therefore

$$\int_0^{\pi/2} \frac{d\theta} {(1+r^2\cos^2\theta)(1+r^2\sin^2\theta)} = \frac{2}{2+r^2} \int_0^{\pi/2} \frac{d\theta}{1+r^2\sin^2\theta}.$$

Using the classical formula

$$\int_0^{\pi/2} \frac{d\theta}{1+r^2\sin^2\theta} = \frac{\pi}{2\sqrt{1+r^2}},$$

we get

$$J = \pi \int_0^{\sqrt2} \frac{r\,dr}{(r^2+2)\sqrt{1+r^2}}.$$

Set

$$u=\sqrt{1+r^2}.$$

Then

$$du=\frac{r\,dr}{\sqrt{1+r^2}}, \qquad r^2+2=u^2+1.$$

Thus

$$J = \pi\int_1^{\sqrt3}\frac{du}{1+u^2}.$$

Hence

$$J = \pi\left( \arctan\sqrt3-\arctan1 \right) = \pi\left( \frac{\pi}{3}-\frac{\pi}{4} \right),$$

so

$$\boxed{J=\frac{\pi^2}{12}}.$$

It remains to recover $I$ from $J$.

Inside the quarter disk $Q$, consider the unit square

$$S=[0,1]\times[0,1].$$

Its weighted area is

$$A = \iint_S w(x,y)\,dx\,dy = \left( \int_0^1\frac{dx}{1+x^2} \right)^2.$$

Therefore

$$A= \left(\frac{\pi}{4}\right)^2 = \frac{\pi^2}{16}.$$

The remaining part $Q\setminus S$ consists of two regions which are interchanged by

$$(x,y)\longmapsto(y,x).$$

Since the weight $w(x,y)$ is also symmetric in $x$ and $y$, these two regions have the same weighted area.

Call this common value $C$. Then

$$J=A+2C,$$

and hence

$$C=\frac{J-A}{2}.$$

Now the part of $R$ with $-1\le x\le0$ is, after the reflection $x\mapsto -x$, exactly the part of $Q$ with $0\le x\le1$.

Its weighted area is therefore

$$A+C.$$

Consequently

$$I=J+A+C.$$

Using

$$C=\frac{J-A}{2},$$

we find

$$I = J+A+\frac{J-A}{2} = \frac{3J+A}{2}.$$

Finally,

$$I = \frac12 \left( 3\frac{\pi^2}{12} + \frac{\pi^2}{16} \right).$$

Therefore

$$\boxed{ I=\frac{5\pi^2}{32} }.$$