# Evaluation of improper integrals

Let $f(x)$ be a real-valued function of real variable $x$, whose anti-derivative is difficult to obtain. Suppose we wish to compute the definite integral $$I=\int_a^\infty f(x)\text{d}x,$$ where $a$ is finite.

Assuming all standard attempts to evaluate the integral have failed, is the following method plausible and/or known?:

Compute the power series of $f(x)$ at $\infty$ (so that we obtain a power series in terms of powers of $1/x$). Integrate the the resulting power series, possibly (?) over an interval, say $[0,y]$ or $[1,y]$, so as to avoid constants of integration, and assuming convergence, let $y\longrightarrow 0$. This should (?) give the limit of the anti-derivative of $f(x)$ at $\infty$.

Compute the lower limit in a similar way. Subtract the two limits. The result should be the real number $I$.

## 1 Answer

While I don't really understand the strategy which you describe in the last two paragraphs of the question, the general method of integrating the Taylor series will work provided that

1. $a>0$,
2. $f$ is the restriction to $(a,\infty)$ of a complex function which is analytic in a neighborhood of $\infty$ which contains the complement of a disk of the form $\{|z|\leq r\}$ where $r<a$
3. $f=O(1/|z|^2)$ at $\infty$ (so in other words there is no constant or $1/z$ term in the Taylor series).

The assumption of analyticity is important, for otherwise you could take a positive smooth bump function supported in $(a,\infty)$ and the Taylor series at $\infty$ would have all zeros for coefficients - which would give an integral of zero by integration of series, but the integral is certainly not zero since the function is nonnegative and nonzero. Also, the assumption that $f=O(1/|z|^2)$ at $\infty$ is important, as demonstrated by various elementary examples like $1/z$ (which is not integrable on $(a,\infty)$).

For the given hypotheses, the proof would go as follows: First, $\int_a^\infty |f(x)|dx<\infty$ by hypothesis 3 and continuity on $[a,\infty)$.

If $f(x)=\sum_{j=2}^\infty c_j(1/x)^j$ is the Taylor series then by hypothesis 2, $\sum_{j=2}^\infty |c_j|(1/a)^j<\infty$ so \begin{align} \sum_{j=2}^\infty\int_a^\infty |c_j|(1/x)^jdx&= \sum_{j=2}^\infty |c_j|a^{1-j}/(j-1)\\ &=a\sum_{j=2}^\infty |c_j|(1/a)^j/(j-1)\\ &\leq a\sum_{j=2}^\infty |c_j|(1/a)^j\\ &<\infty \end{align} therefore $[x\mapsto \sum_{j=2}^\infty |c_j(1/x)^j|]\in L^1(a,\infty)$ by monotone convergence. Thus \begin{align} \int_a^\infty f(x)dx&=\int_a^\infty\lim_{n\to \infty}\sum_{j=2}^nc_j(1/x)^jdx \\ &=\lim_{n\to \infty}\int_a^\infty \sum_{j=2}^nc_j(1/x)^jdx \\ &=\lim_{n\to \infty}\sum_{j=2}^nc_j\int_a^\infty (1/x)^jdx \\ &=\sum_{j=2}^\infty c_ja^{1-j}/(j-1). \end{align} The first equality is the pointwise convergence of the series to the function $f$ which is valid by hypothesis 2, the second equality is by dominated convergence, which is justified by the monotone convergence argument given above.

Example: $$f(x)=\frac{1}{1+x^2}=\frac{1}{x^2}\frac{1}{1-(-1/x^2)}=\frac{1}{x^2}\sum_{k=0}^\infty (-1)^k(1/x^2)^k=\sum_{k=0}^\infty(-1)^k(1/x^{2k+2})$$ the third equality being valid only if $x>1$, in which case $$\int_a^\infty f(x)dx=\sum_{k=0}^\infty(-1)^ka^{-1-2k}/(1+2k)$$ provided $a>1$. If $a<1$ then the series does not even converge because, despite the real analyticity of $1/(1+x^2)$ on the entire real line, the expansion at $\infty$ can't be continued to the unit disk since there are singularities at $\pm i$ on the unit circle. This is why hypothesis 2 is crucial.

To check this, observe that $1/(1+x^2)$ has a nice antiderivative $\arctan(x)$... so you can compare the series answer to the usual answer $$\int_a^\infty 1/(1+x^2)dx=\pi/2-\arctan(a).$$ Since $a\mapsto \pi/2-\arctan(a)$ and $a\mapsto \sum_{k=0}^\infty(-1)^ka^{-1-2k}/(1+2k)$ have the same derivative on $(1,\infty)$ and the same limit as $a\rightarrow \infty$, these answers are the same.