Multivariable calculus

Double Integral of y/(1 + x^2) Over [0,1] by [0,2]

Q: What if the numerator were xy instead of y?

Then the x integral becomes \int_{0}^{1}\frac{x}{1+x^2}\,dx = \frac{\ln 2}{2}, and the double integral is \ln 2 \approx 0.693. The single factor x converts an arctangent into a logarithm.

The double integral of y divided by 1 plus x squared, over x from 0 to 1 and y from 0 to 2, equals pi/2, about 1.5708. The integrand factors, so the answer is the integral of 1 over 1 plus x squared from 0 to 1, which is pi/4, times the integral of y from 0 to 2, which is 2.

f(x,y) = \frac{y}{1 + x^2}

\int_{0}^{1}\int_{0}^{2} \frac{y}{1+x^2}\,dy\,dx = \frac{\pi}{2}

Numerically 1.570796, confirmed by quadrature on every build.

Factor, then recognise the arctangent

The integrand is $\frac{1}{1+x^2}$ times $y$ , one factor per variable, and both limits are constants. That is the separable case.

\int_{0}^{1}\int_{0}^{2} \frac{y}{1+x^2}\,dy\,dx = \left(\int_{0}^{1} \frac{1}{1+x^2}\,dx\right)\left(\int_{0}^{2} y\,dy\right)

= \left[\arctan x\right]_{0}^{1}\left[\frac{y^2}{2}\right]_{0}^{2} = \frac{\pi}{4}\cdot 2 = \frac{\pi}{2}

The $\frac{\pi}{4}$ is $\arctan 1$ , the angle whose tangent is 1. Every $\pi$ in the final answer traces back to that single evaluation.

The same thing one layer at a time

If you would rather not use the shortcut, freeze $x$ and note that $\frac{1}{1+x^2}$ is then a constant multiplier.

\int_{0}^{2} \frac{y}{1+x^2}\,dy = \frac{1}{1+x^2}\left[\frac{y^2}{2}\right]_{0}^{2} = \frac{2}{1+x^2}

\int_{0}^{1} \frac{2}{1+x^2}\,dx = 2\left[\arctan x\right]_{0}^{1} = \frac{\pi}{2}

A size check before you trust it: on this rectangle the integrand runs from 0 to 2 and the area is 2, so the answer has to lie between 0 and 4. The value 1.5708 passes.

The mistake: the logarithm that is not there

Writing $\int \frac{1}{1+x^2}\,dx = \ln\left(1+x^2\right)$ is the standard wrong turn. The substitution $u = 1 + x^2$ needs $du = 2x\,dx$ , and there is no $x$ in this numerator to supply it.

$\int \frac{1}{1+x^2}\,dx = \arctan x$ , an inverse trigonometric function.
$\int \frac{x}{1+x^2}\,dx = \frac{1}{2}\ln\left(1+x^2\right)$ , a logarithm, and note the $\frac{1}{2}$ .
One factor of $x$ in the numerator is the whole difference between the two.

That extra $x$ changes the answer, not just its form. With integrand $\frac{xy}{1+x^2}$ over the same rectangle you get $\left(\frac{\ln 2}{2}\right)(2) = \ln 2 \approx 0.693$ rather than $\frac{\pi}{2} \approx 1.571$ .

Frequently asked questions

Why does the y integral not disturb the arctangent?

Because the two variables live in separate factors. Integrating in $y$ multiplies $\frac{1}{1+x^2}$ by the number $\int_{0}^{2} y\,dy = 2$ , and a constant multiple passes straight through the $x$ integral.

What if the numerator were xy instead of y?

Then the $x$ integral becomes $\int_{0}^{1}\frac{x}{1+x^2}\,dx = \frac{\ln 2}{2}$ , and the double integral is $\ln 2 \approx 0.693$ . The single factor $x$ converts an arctangent into a logarithm.