Problem 9:  Integration by substitution ($✓$) 1998 Paper I

Show, by means of a change of variable or otherwise, that

${\int }_{0}^{\infty }\mathrm{f}\left(\right{\left({x}^{2}+1\right)}^{\frac{1}{2}}+x\left)\right\phantom{\rule{0.3em}{0ex}}\mathrm{d}x=\frac{1}{2}{\int }_{1}^{\infty }\left(1+{t}^{-2}\right)\mathrm{f}\left(t\right)\phantom{\rule{0.3em}{0ex}}\mathrm{d}t\phantom{\rule{2.77695pt}{0ex}},$

for any given function $\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}\phantom{\rule{0.3em}{0ex}}$.

Hence, or otherwise, show that

${\int }_{0}^{\infty }\left(\right{\left({x}^{2}+1\right)}^{\frac{1}{2}}+x{\left)\right}^{-3}\phantom{\rule{0.3em}{0ex}}\mathrm{d}x=\frac{3}{8}\phantom{\rule{2.77695pt}{0ex}}.$

Note that ‘by change of variable’ means the same as ‘by substitution’.

There are two things to worry about when you are trying to ﬁnd a change of variable to convert one integral to another: you need to make the integrands match up and you need to make the limits match up. Sometimes, the limits give the clue to the change of variable. (For example, if the limits on the original integral were 0 and 1 and the limits on the transformed integral were 0 and $\frac{1}{4}\pi$, then an obvious possibility would be to make the substitution $t=tanx\phantom{\rule{0.3em}{0ex}}$). Here, the change of variable is determined by the integrand, since it must work for all choices of f.

Perhaps you are worried about the inﬁnite upper limit of the integrals. If you are trying to prove some rigorous result about inﬁnite integrals, you might use the deﬁnition

${\int }_{0}^{\infty }\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}\left(x\right)\phantom{\rule{0.3em}{0ex}}\mathrm{d}x=\underset{a\to \infty }{lim}{\int }_{0}^{a}\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}\left(x\right)\phantom{\rule{0.3em}{0ex}}\mathrm{d}x\phantom{\rule{2.77695pt}{0ex}},$

but for present purposes you just do the integral and put in the limits. The inﬁnite limit will not normally present problems. For example,

${\int }_{1}^{\infty }\left({x}^{-2}+{\mathrm{e}}^{-x}\right)\mathrm{d}x=\left(-{x}^{-1}-{\mathrm{e}}^{-x}\right){\left|\right}_{1}^{\infty }=-\frac{1}{\infty }-{\mathrm{e}}^{-\infty }+\frac{1}{1}+{\mathrm{e}}^{-1}=1+{\mathrm{e}}^{-1}\phantom{\rule{2.77695pt}{0ex}}.$

Don’t be afraid of writing $1∕\infty =0\phantom{\rule{0.3em}{0ex}}$. It is perfectly OK to use this as shorthand for $\underset{x\to \infty }{lim}1∕x=0\phantom{\rule{2.77695pt}{0ex}}$; but $\infty ∕\infty$, $\infty -\infty$ and $0∕0$ are deﬁnitely not OK, because of their ambiguity.

Solution to problem 9

Clearly, to get the argument of f right, we must set

$t={\left({x}^{2}+1\right)}^{\frac{1}{2}}+x\phantom{\rule{2.77695pt}{0ex}}.$

We must check that the new limits are correct (if not, we are completely stuck). When $x=0$, $t=1$ as required. Also, ${\left({x}^{2}+1\right)}^{\frac{1}{2}}\to \infty$ as $x\to \infty$, so $t\to \infty$ as $x\to \infty$. Thus the upper limit is still $\infty \phantom{\rule{0.3em}{0ex}}$, again as required.

Using the standard method of changing variable in an integral (basically the chain rule), the integral becomes

${\int }_{1}^{\infty }\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}\left(t\right)\frac{\mathrm{d}x}{\mathrm{d}t}\phantom{\rule{0.3em}{0ex}}\mathrm{d}t$

so the next task is to ﬁnd $\frac{\mathrm{d}x}{\mathrm{d}t}\phantom{\rule{0.3em}{0ex}}$. This we can do in two ways: we ﬁnd $x$ in terms of $t$ and differentiate it; or we could ﬁnd $\frac{\mathrm{d}t}{\mathrm{d}x}\phantom{\rule{0.3em}{0ex}}$ and turn it upside down. The snag with the second method is that the answer will be in terms of $x$, so we will have to express $x$ in terms of $t$ anyway — in which case, we may as well use the ﬁrst method.

We start by ﬁnding $x$ in terms of $t$:

$t={\left({x}^{2}+1\right)}^{\frac{1}{2}}+x⇒{\left(t-x\right)}^{2}=\left({x}^{2}+1\right)⇒x=\frac{{t}^{2}-1}{2t}=\frac{t}{2}-\frac{1}{2t}\phantom{\rule{2.77695pt}{0ex}}.$

Then we differentiate:

$\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{1}{2}+\frac{1}{2{t}^{2}}=\frac{1}{2}\left(1+{t}^{-2}\right)$

which is exactly the factor that we require for the transformed integrand given in the question.

For the last part, we take $\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}\left(t\right)={t}^{-3}$. Thus

$\begin{array}{rcll}{\int }_{0}^{\infty }\left(\right{\left({x}^{2}+1\right)}^{\frac{1}{2}}+x{\left)\right}^{-3}\phantom{\rule{0.3em}{0ex}}\mathrm{d}x& =& \frac{1}{2}{\int }_{1}^{\infty }\left({t}^{-3}+{t}^{-5}\right)\phantom{\rule{0.3em}{0ex}}\mathrm{d}t& \text{}\\ & =& \frac{1}{2}{\left(\frac{{t}^{-2}}{-2}+\frac{{t}^{-4}}{-4}\right)|}_{1}^{\infty }& \text{}\\ & =& \frac{1}{2}\left(\frac{1}{2}+\frac{1}{4}\right)=\frac{3}{8}\phantom{\rule{0.3em}{0ex}}.& \text{}\end{array}$

as required.

Post-mortem

This is not a difficult question conceptually once you realise the signiﬁcance of the fact that the change of variable must work whatever function $\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}$ is in the integrand.

There was a useful point connected with calculating $\frac{\mathrm{d}x}{\mathrm{d}t}\phantom{\rule{0.3em}{0ex}}$. It was a good idea not to plunge into the algebra without ﬁrst thinking about alternative methods; in particular, we might have used (but didn’t)

$\frac{\mathrm{d}x}{\mathrm{d}t}=1/\frac{\mathrm{d}t}{\mathrm{d}x}\phantom{\rule{2.77695pt}{0ex}}.$

Finally, there was the inﬁnite limit of the integrand, which I hope you saw was not something to worry about (even though inﬁnite limits are excluded from the A-level core). If you were setting up a formal deﬁnition of what an integral is, you would have to use ﬁnite limits, but if you are merely calculating the value of an integral, you just go ahead and do it, with whatever limits you are given.