# Another simple integral

What follows is a recent generalization of the technique of computing some definite integrals discussed some time ago. The discussion here is a compilation of an article __Lazy Student Integrals__ by Gregory Galperin and Gregory Ronsse (*Math Magazine*, v 81, n 2, April 2008, pp.152-154).

Assume you are asked to evaluate the integral

$\displaystyle I(\alpha )=\int_{0}^{\infty}\frac{dx}{(1+x^{\alpha})(1+x^{2})}$

as a function of $\alpha.$ You may observe that the substitution $x = \tan\theta$ leads to

(1)

$\displaystyle I(\alpha )=\int_{0}^{\pi /2}\frac{\cos^{\alpha}\theta\,d\theta}{\cos^{\alpha}\theta +\sin^{\alpha}\theta }$

an integral of the sort we have considered previously. Thus we know that the latter integral is just plain $\pi /4,$ independent of $\alpha !$ Being a curiosity, we'll obtain the same result in a different, but still a simple manner, after the substitution $u = 1/x.$ This substitution yields

$\displaystyle I(\alpha )=\int_{0}^{1}\frac{dx}{(1+x^{\alpha})(1+x^{2})}+\int_{0}^{1}\frac{u^{\alpha}du}{(1+u^{\alpha})(1+u^{2})}.$

Adding the two expressions for $I(\alpha )$ gives

$\displaystyle I(\alpha )=\int_{0}^{1}\frac{dx}{1+x^{2}}=\frac{\pi}{4},$

as expected.

On the whole, the independence of the integral in (1) of $\alpha$ suggests that there may be more to the integral than meets the eye. Indeed, the integral in (1) falls in the same class as, say, integrals

$\displaystyle\int_{-3}^{3}\frac{\sin^{3}x\,dx}{\sqrt{1+x^{2}}+\cos 2x}=0$

or

(2)

$\displaystyle\int_{0}^{10}\frac{(3+x^{\sqrt{7}})dx}{6+x^{\sqrt{7}}+(10-x)^{\sqrt{7}}}=5.$

Do you see why one is $0$ and the other $5?$ As a matter of fact, the two are almost immediate with a little piece of theory.

Let $f:\,[0, a] \rightarrow \mathbb{R}$ be any continuous function. Substitute $u = x - a$ to obtain

(3)

$\displaystyle\int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-u)du$

which becomes obvious if you note that the substitution just reflects function $f$ in $x = a/2.$ Now suppose that $f$ satisfies the following **symmetry condition**

$f(x) + f(a - x) = 1.$

Then adding the integrals in (3) shows that

$\displaystyle\int_{0}^{a}f(x)dx=\frac{a}{2}.$

To make the approach general we only need a way of finding more functions $f$ that obey the symmetry condition. This too is simple. All such function can be obtained from

$\displaystyle f(x)=\frac{g(x)}{g(x)+g(a-x)}$

where $g$ is any continuous function on $[0, a]$ for which the denominator above does not vanish. Any function $f$ that satisfies the symmetry condition comes in this form with $g = f !$ In particular, (2) is obtained with $g(x) = 3 + x^{\sqrt{7}}$ while (1) follows from $\sin (x) = \cos(\pi /2 - x).$ But we see that even more generally

$\displaystyle\int_{0}^{\pi /2}\frac{f(\cos x)}{f(\sin x)+f(\cos x)}=\frac{\pi}{4}.$

for any continuous function $f$ defined on $[0, 1]$ such that the denominator of the integrand above does not vanish.

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