Show $\int_0^\infty \frac{\cos ax}{1+x^2}dx$ is continuous function of $a$ by definition of Riemann integration.

Question

One could show the integral is uniformly convergent by definition of Riemann integration.

Let division $D(n)=x_0, x_1,\dots, x_n\dots$, where$x_i=\frac1a(\frac{i}{2n}\pi)=\frac1a(\lambda_i+2k\pi)$, $\lambda_i=\frac{i\ \text{mod}( 4n)}{2n}\pi, k =[\frac i{4n}]$.

The upper sum $S_n=\lim_{n\to\infty}\sum_{i=0}^\infty \frac1n M_i =\lim_{n\to\infty}\frac1n[\dots+\sum_{i=-2n}^{-n-1} \frac{\cos ax_{i+1}}{1+{x_{i+1}}^2} +\sum_{i=-n}^{-1} \frac{\cos ax_{i+1}}{1+{x_i}^2} +\sum_{i=0}^{n-1} \frac{\cos ax_i}{1+{x_i}^2} +\sum_{i=n}^{2n-1} \frac{\cos ax_i}{1+{x_{i+1}}^2}+\dots].$

Similarly we have the lower sum $s_n$. Obviously, $S_n-s_n$ has terms such as

$\frac{\cos ax_{i+1}}{1+{x_{i+1}}^2}- \frac{\cos ax_{i}}{1+{x_{i}}^2} \quad \text{(using Taylor expansion)}\\ =\frac{\left\{\left[\cos ax_i(1-\frac{(\frac{\pi}{2n})^2}{2!}+O(\frac1{n^4}))-\sin ax_i ((\frac{\pi}{2n})-\frac{(\frac{\pi}{2n})^3}{3!}+O(\frac1{n^5}))\right](1+{x_{i}}^2) -\cos ax_i\left(1+{x_i}^2+2x_i\frac\pi{2an}+(\frac\pi{2an})^2\right)\right\} (1-\frac{2x_i}{(1+{x_i}^2)}\frac\pi{2an}+O(\frac1{n^2}))} {(1+{x_i}^2)(1+{x_{i}}^2)}\\ =\frac{\frac{\pi}{2n}\left\{|\sin ax_i| (1+{x_{i}}^2)+|\cos ax_i|2x_i\frac1a\right\}+O(\frac1{n^2})} {(1+{x_{i}}^2)^2}.$

With similar calculation, we have $S_n-s_n= \frac{\pi}{2n^2}\sum_{i=0}^{\infty} \frac{|\sin ax_i|} {(1+{x_{i}}^2)} +\frac{|\cos ax_i|2x_i\frac1a} {(1+{x_{i}}^2)^2}+O(\frac1{n})\\ =\frac{\pi}{2n^2}\sum_{k=0}^\infty\sum_{i=0}^{4n-1} \frac{|\sin \lambda_i|} {(1+{[\frac1a(\lambda_i+2k\pi)]}^2)} +\frac{|\cos \lambda_i|2[\frac1a(\lambda_i+2k\pi)]\frac1a} {(1+{[\frac1a(\lambda_i+2k\pi)]}^2)^2}+O(\frac1{n})\\ <\frac1n\sum_{k=0}^\infty\sum_{i=0}^{4n-1} \frac{\frac{\pi}{2n}|\sin \lambda_i|} {(1+{[\frac1a(2k\pi)]}^2)} +\frac{\frac{\pi}{2n}|\cos \lambda_i|2[\frac1a(2\pi+2k\pi)]\frac1a} {(1+{[\frac1a(2k\pi)]}^2)^2}+O(\frac1{n}) \quad (\text{using defintion of integration})\\ <\frac1n\sum_{k=0}^\infty \frac{4} {(1+{[\frac1a(2k\pi)]}^2)} +\frac{4\cdot2[\frac1a(2\pi+2k\pi)]\frac1a} {(1+{[\frac1a(2k\pi)]}^2)^2}+O(1),$

How does one then prove the uniform convergence?

Write $\sum_{k=K}^{K+p}$ for $\sum_{k=0}^\infty$, then uniform convergence implies -- for integration on a tail is smaller than $\epsilon$, the tail ($[K, +\infty)$) defined independently of $a$; and so are upper and lower sums $S_n, s_n$ over this tail (and $N$ is independent of $a$ too), right?, -- that, $\forall \epsilon, \exists K, N \text{, both independent of $a$, such that }\forall p>0, n>N$, the above expression is less than $\epsilon$.

It seems the O(1) term makes it divergent. We could also find for however large $K, a=K^2$ such that one term in $\sum_{k=K}^{K+p}>\frac4{1+4\pi^2}$ which is not arbitrarily small, and so the series doesn't converge uniformly, right?

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Edited to add: Below I gave a simpler proof inspired by an answer and tried to answer part of my questions above.

Another Q: Perhaps the integral is not uniform convergent but still a continuous function of $a$?

Answer 1

This is a sketch of solution based on the notion of Riemann improper integrals, which seems to be what the OP is referring to (the interval of integration is infinite).

Let $\varepsilon>0$. Fix $a_0$.

Since $\frac{1}{1+x^2}$ is integrable (in the sense that $\lim_{c\rightarrow\infty}\int^c_0\frac{dx}{1+x^2}$), there is $C$ such that $$\int^{c'}_c\frac{dx}{1+x^2}<\frac{\varepsilon}{2}$$ for all $c'>c>C$.

Since $\cos$ is periodic and continuous, it is also uniformly continuous; hence, there is $\delta>0$ such that if $|b-b'|<\delta$, then $|\cos b - \cos b'|<\frac{\varepsilon}{2\pi}$

Putting things together, if $c>C$ and $|a-a_0|C<\delta$, then

$$ \Big|\int^c_0\frac{\cos(ax)-\cos(a_0x)}{1+x^2}\,dx\Big|\leq\int^C_0\frac{|\cos(ax)-\cos(a_0x)|}{1+x^2}+\int^c_C\frac{dx}{1+x^2}<2\varepsilon $$

Edit: Just to complete the argument a little further:

Letting $c\rightarrow\infty$ implies that

$$ \Big|\int^\infty_0\frac{\cos(ax)-\cos(a_0x)}{1+x^2}\,dx\Big|\leq2\varepsilon $$ whenever $|a-a_0|<\frac{\delta}{C}$, and the conclusion follows.

Answer 2

Regarding continuity of the function.

This uses the idea of the answer @Oliver Diaz.

We need to show that For each $a_0, \forall \epsilon>0, \exists\delta>0,$ such that when $|a-a_0|<\delta,$ $|\int_{0}^{\infty} \frac{\cos ax}{1+x^2}dx-\int_{0}^{\infty} \frac{\cos a_0x}{1+x^2}dx| <\epsilon$.

$\forall \epsilon_0, \exists X,$ independent of $a$, such that $|\int_{X}^{\infty} \frac{\cos ax}{1+x^2}dx-\int_{X}^{\infty} \frac{\cos a_0x}{1+x^2}dx|<\int_{X}^{\infty} \frac{2}{1+x^2}dx <\epsilon_0$, for this integral is convergent.

For $\cos x$ is uniformly continuous, $\forall e',\exists \delta',$ such that when $|x_1-x_2|<\delta', |\cos ax-\cos a_0x|<\epsilon'$. Let $|a-a_0|X<\delta'$, then when $x<X, |(a-a_0)x|<\delta'$, and so $|\int_{0}^{X} \frac{\cos ax}{1+x^2}dx-\int_{0}^{X} \frac{\cos a_0x}{1+x^2}dx|<\int_{0}^{X} \frac{|\cos ax-\cos a_0x|}{1+x^2}dx<\epsilon'\int_{0}^{X} \frac1{1+x^2}dx=\epsilon'\eta,$

Therefore, $\exists \delta=\frac {\delta'}X$ independent of $a$, $|\int_{0}^{\infty} \frac{\cos ax}{1+x^2}dx-\int_{0}^{\infty} \frac{\cos a_0x}{1+x^2}dx|<|\int_{0}^{X} \frac{\cos ax}{1+x^2}dx-\int_{0}^{X} \frac{\cos a_0x}{1+x^2}dx|+|\int_{X}^{\infty} \frac{\cos ax}{1+x^2}dx-\int_{X}^{\infty} \frac{\cos a_0x}{1+x^2}dx|\\ <\epsilon'\eta+\epsilon_0,$

which can be arbitrarily small (for $\eta$ is independent of $a$.)

In a word, for the tail, we use convergence of integral, for the rest, we use uniform continuity. (It seems even non-uniform continuity suffices.)

Though I'm still looking for answers to the questions above.

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Regarding logic of proof and dependence of $n$ on $a$.

In the proof.

I. There is a convolution of logic. The following could make things clearer.

To prove uniform convergence one needs to show

$\exists K \text{ independent of a, such that } |I_{K,p}=S_{n;K,p}-(S_{n;K,p}-I_{K,p})|=|S_{n;K,p}-\epsilon_{n;K,p}|<\epsilon, \quad (1)$

where $I_{K,p}=\int_{x_{4nK}}^{x_{4n(K+p+1)}}\frac{\cos(ax)}{1+x^2}dx,x_{4nK}=\frac{2K\pi}{a}, x_{4n(K+p+1)}=\frac{2(K+p+1)\pi}{a},$ (both unaffected by the choice of $D(n)$,) $S_{n;K,p}=\sum_{i=4nK}^{4n(K+p+1)}\frac1n M_i$.

We need to i. consider only one $n$ (more exactly, one for each $a$) that satisfies (1), that is, ii. we needn't consider many, or all, sufficiently large $n$; iii. by far, n's value is irrelevant, for n's effect on function $S_{n;K,p}$ and error $\epsilon_{n;K,p}$ cancel out; it's but used to calculate $I_{K,p}$.

Then it suffices (not necessary) to show $|S_{n;K,p}-\epsilon_{n;K,p}|<|S_{n;K,p}|+|(S_{n;K,p}-s_{n;K,p})| <\epsilon. \quad (2)$

For that it suffices to show $|S_{n;K,p}|<\frac \epsilon 2$ and $|(S_{n;K,p}-s_{n;K,p})|<\frac \epsilon 2$. (We proved above the latter, which is necessary for (2) (not for (1)) but insufficient; notice the inconsistent use there of sufficiency and necessity leads to no valid proof.) (2) doesn't affect (i, ii) but it constrains $n$ to be sufficiently large,

that is, if there exists K independent of a, then it doesn't matter whether n depends on a; for example, if we can find an n dependent on a to allow K independent of a, then the proof is good, right?

In other words, to show $I_{K,p} < \epsilon$, it suffices--actually equals--to prove $S_{\infty;K,p}<\frac\epsilon2,\epsilon_{\infty;K,p}<\frac\epsilon2$.If $\exists n, \text{ such that } S_{n;K,p} < \frac\epsilon2$,(even if n depends on a, e.g. $n=[\frac1a]$) then $S_{\infty;K,p}<S_{n;K,p} < \frac\epsilon2$. So does $\epsilon_{\infty;K,p}$. (though we possibly can't use $S_{\infty}$ arbitrarily, and it's why we use $\epsilon-N$ expression.) Obviously, existence of such n independent of a is sufficent (though possibly not necessary) condition for (2).

Not sure this understanding is accurate. At least it shows it helps to lay out logical sequence of steps.

II. The calculation of $O(\frac1n)$ is perhaps incomplete.