11.1 Infinite Real Integrals

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In this section we will use Cauchy's residue theorem to calculate integrals of the form $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(t)\mathsf{d}t$$ where $f:\mathbb{R}\to \mathbb{R}$ decays suitably at infinity. Generally, the above notation is shorthand for the limit $$\underset{A,B\to \mathrm{\infty }}{lim}\underset{-A}{\overset{B}{\int }}f(t)\mathsf{d}t$$ but in this course we will only consider the special case covered by the following lemma.

Lemma

Fix $f:\mathbb{R}\to \mathbb{R}$ continuous. If there is $C>0$ and $r>1$ with $|f(t)|\le C/|t{|}^r$ for all $t\in \mathbb{R}$ then $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(t)\mathsf{d}t$$ exists and equals $$\underset{T\to \mathrm{\infty }}{lim}\underset{-T}{\overset{T}{\int }}f(t)\mathsf{d}t$$ exists.

Proof:

Fix $0<S<T$ and calculate $$\begin{array}{rl}|\underset{-T}{\overset{T}{\int }}f(t)\mathsf{d}t-\underset{-S}{\overset{S}{\int }}f(t)\mathsf{d}t|& =|\underset{-T}{\overset{-S}{\int }}f(t)\mathsf{d}t+\underset{S}{\overset{T}{\int }}f(t)\mathsf{d}t|\\ & \le \underset{-T}{\overset{-S}{\int }}{\displaystyle \frac{C}{|t{|}^r}}\mathsf{d}t+\underset{S}{\overset{T}{\int }}{\displaystyle \frac{C}{|t{|}^r}}\mathsf{d}t\\ & =2C\underset{S}{\overset{T}{\int }}{\displaystyle \frac{1}{t^r}}\mathsf{d}t\\ & ={\displaystyle \frac{2C}{1-r}}{\left[{\displaystyle \frac{1}{t^{r-1}}}\right]}_S^T={\displaystyle \frac{2C}{r-1}}({\displaystyle \frac{1}{S^{r-1}}}-{\displaystyle \frac{1}{T^{r-1}}})\end{array}$$ which will be at most $ϵ$ whenever $S,T$ are large enough. This imples that, as $T\to \mathrm{\infty }$, the quantity $$\underset{-T}{\overset{T}{\int }}f(t)\mathsf{d}t$$ is Cauchy and therefore converges. A similar line of argument proves the second part of the lemma. $◻$

The connection between infinite real integrals and contour integration is given by the contour $\mathrm{\Gamma }=(\gamma ,\eta )$ where $\gamma (t)=t$ on $[-T,T]$ and $\eta (t)=T{e}^{it}$ on $[0,\pi ]$. Indeed - provided one can extend the definition of $f$ to a domain containing $\mathbb{R}$ - one has $$\underset{\mathrm{\Gamma }}{\int }f=\underset{\gamma }{\int }f+\underset{\eta }{\int }f=\underset{-T}{\overset{T}{\int }}f(t)\mathsf{d}t+\underset{0}{\overset{\pi }{\int }}f(T{e}^{it})iT{e}^{it}\mathsf{d}t$$ and we are faced with two tasks.

1. Computing the integral on the left-hand side using Cauchy's residue theorem.
2. Dispensing with the contour integral over $\eta$.

The first involves the calculation of any residues $f$ may have inside $\mathrm{\Gamma }$. For the second we need $f$ to decay as $T\to \mathrm{\infty }$ in some way.

Figure 1: The contour $\mathrm{\Gamma }=(\gamma ,\eta )$ used to calculate infinite real integrals using Cauchy's residue theorem.

Example

Lets evaluate $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{1}{(t^2+1)(t^2+4)}}\mathsf{d}t$$ using the above approach. First, we estimate that $$\left|{\displaystyle \frac{1}{(t^2+1)(t^2+4)}}\right|\le {\displaystyle \frac{1}{t^4}}$$ for all $t\in \mathbb{R}$ so that the lemma applies, allowing us to follow the strategy outlined above. Define $f$ on $\mathbb{C}\setminus \{i,-i,2i,-2i\}$ by $$f(z)={\displaystyle \frac{1}{(z^2+1)(z^2+4)}}$$ and put $\eta (t)=T{e}^{it}$ on $[0,\pi ]$. We can estimate $$|\underset{0}{\overset{\pi }{\int }}f(T{e}^{it})iT{e}^{it}\mathsf{d}t|$$ if we have an upper bound for $|f(T{e}^{it})|$ on $[0,\pi ]$. But $$|f(z)|={\displaystyle \frac{1}{|z-i||z+i||z-2i||z+2i|}}\le {\displaystyle \frac{1}{(|z|-1{)}^2(|z|-2{)}^2}}$$ using the reverse triangle inequality, so $$|f(T{e}^{it})|\le {\displaystyle \frac{1}{(T-1{)}^2(T-2{)}^2}}\le {\displaystyle \frac{100}{T^4}}$$ for all $T$ large enough. We conclude that $$|\underset{0}{\overset{\pi }{\int }}f(T{e}^{it})iT{e}^{it}\mathsf{d}t|\le \underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{100}{T^4}}T\mathsf{d}t\le {\displaystyle \frac{2\pi }{T^3}}$$ so that the contour integral over $\eta$ goes to zero as $T\to \mathrm{\infty }$. When $T$ is large $\mathrm{\Gamma }$ contains $i$ and $2i$ so $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{1}{t^2+1}}{t}^2+4\mathsf{d}t=2\pi i\mathsf{Res}(f,i)+2\pi i\mathsf{Res}(f,2i)$$ by Cauchy's residue theorem. The residues are $$\begin{array}{rl}\mathsf{Res}(f,i)& =\underset{z\to i}{lim}{\displaystyle \frac{z-i}{(z^2+1)(z^2+4)}}={\displaystyle \frac{1}{6i}}\\ \mathsf{Res}(f,2i)& =\underset{z\to 2i}{lim}{\displaystyle \frac{z-2i}{(z^2+1)(z^2+4)}}=-{\displaystyle \frac{1}{12i}}\end{array}$$ so $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{1}{(t^2+1)(t^2+4)}}\mathsf{d}t={\displaystyle \frac{\pi }{3}}-{\displaystyle \frac{\pi }{6}}={\displaystyle \frac{\pi }{6}}$$ concluding the example.