The question I'm about to ask isn't covered in our lectures, so pardon my ignorance.

Suppose $f:(0,a]\times [c,d]$. How is uniform convergence of the improper integral $\int_0^af(x,a)dx$ defined?

Just to give an insight of what I've (hope so) learnt so far, I'm going to write down some definitions and results (and proofs) I think are relevant for functions $f:[a,+\infty)\times [c,d]$ that I found in the script by prof. Šime Ungar from 2004. It might be available here.

$\underline{\boldsymbol{\text{ definition 1: }}}$

An improper integral $\int_a^\infty f(x)dx$ is defined as $\lim_{b\to\infty}\int_a^b f(x)dx$ under the condition that $f$ is integrable on $[a,b],\forall b>a$ and that the limit exists. In this case we also say the improper integral $\int_a^\infty f(x)dx$ converges. That is equivalent to the following condition: $f$ is integrable on $[a,b],\forall b>a$ and $$(\forall\varepsilon>0)(\exists a_0>a), \left|\int_b^c f(x)dx\right|<\varepsilon,\forall c>b>a_0.$$

Now, suppose we have a function $f:[a,+\infty)\times S\to\Bbb R,$ where $S\subseteq\Bbb R$ is an arbitrary set. Here we can observe convergence of the integral $\displaystyle\int_a^b f(x,y)dx,\forall y\in S.$ In this situation, the following definition makes sense:

$\underline{\boldsymbol{\text{definition 2:}}}$

We say the improper integral $\int_a^\infty f(x,y)dx$ converges uniformly on $S$ if the integral converges $\forall y\in S$ and if $$\lim_{b\to\infty}\left(\sup_{y\in S}\left|\int_b^\infty f(x,y)dx\right|\right)=0.$$

$\underline{\boldsymbol{\text{result 1:}}}$

Suppose $\int_a^b f(x,y)dx$ exists $\forall b>a$ and $\color{red}{\forall y\in S}.$ Then, the improper integral $\int_a^\infty f(x,y)dx$ converges uniformly on $S$ if and only if $$(\forall\varepsilon>0)(\exists a_0>a), \color{red}{\sup_{y\in S}}\left|\int_b^c f(x,y)dx\right|<\varepsilon,\forall c>b\ge a_0.$$

$\boldsymbol{\text{ proof: }}$

$\boxed{\Rightarrow}$ Necessity of the conditions in the theorem is obvious.

$\boxed{\Leftarrow}$ Let's prove thr sufficiency. From the conditions of the theorem, it first follows that, $\forall y\in S,$ the integral $\int_a^\infty f(x,y)dx$ converges. Furthermore, for $y\in S,$ since $\left|\int_b^c\right|<\varepsilon$ whenever $c>b>a_0,$ letting $c\to\infty,$ we have $\left|\int_b^\infty f(x,y)dx\right|\color{red}\le\varepsilon.$ But, as this holds $\color{red}{\forall y\in S},$ it is also true that $\color{red}{\sup_{y\in S}}\left|\int_b^\infty f(x,y)dx\right|\le\varepsilon$ and the claim follows.

Something I believe is important (I'll write down my motivation in the end):

$\underline{\boldsymbol{\text{Weierstrass M-test:}}}$

Suppose $\int_a^b f(x,y)dx$ exists $\forall b>a.$ If there is a function $M:[a,+\infty)\to\Bbb R$ s. t. $|f(x,y)|\le M(x),\forall x\in[a,+\infty)$ and $\forall y\in S$ and if $\int_a^\infty M(x)dx$ converges, then $\int_a^\infty f(x,y)dx$ converges (absolutely and) uniformly on $S$.

I've gone through the proof of the discrete version. If needed, I'll analyze this more carefully.

Last result with the proof, I promise.

$\underline{\boldsymbol{\text{result 2: }}}$

Suppose $f:[a,+\infty)\times [c,d]$ is continuous and that the following statements are true:

  1. $\exists y\in [c,d]$ s. t. the improper integral $\int_a^\infty f(x,y)dx$ converges
  2. $\partial_2f(x,y)$ exists and is continuous on $[a,+\infty)\times [c,d]$
  3. $\int_a^\infty\partial_2f(x,y)dx$ converges uniformly on $[c,d].$

Then, the improper integral $\int_a^\infty f(x,y)dx$ exists and converges uniformly on $[c,d],$ and the function $F(y):=\int_a^\infty f(x,y)dx$ is differentiable on $[c,d]$ and $\color{purple}{F'(y)=\int_a^\infty\partial_2 f(x,y)dx}\forall y\in [c,d].$

$\boldsymbol{\text{ proof: }}$

Suppose that $\int_a^\infty f(x,y_0)dx$ converges. Apart from that, we also know that $\forall b>a$ and $\forall y\in [c,d]$ the integral $\int_a^b f(x,y)dx.$

Because of $2,\forall\varepsilon>0,\exists a_0$ s. t. $a_0<b<g\implies \left|\int_b^g\partial_2f(x,t)dx\right|<\varepsilon,\forall t\in [c,d]$ and at the same time $\left|\int_b^g f(x,y_0)dx\right|<\varepsilon.$

$$\begin{aligned}\left|\int_b^g(f(x,y)-f(x,y_0))dx\right|&=\left|\int_b^g\int_{y_0}^y\partial_2f(x,t)dtdx\right|\\&=\left|\int_{y_0}^y\int_b^g\partial_2f(x,t)dxdt\right|\\&\le\int_{y_0}^y\left|\int_b^g\partial_2f(x,t)dx\right|dt\\&\le (d-c)\varepsilon,\end{aligned}$$ whenever $a_0<b<g.$

$$\begin{aligned}\implies\left|\int_b^gf(x,y)dx\right|&\le\left|\int_b^g(f(x,y)-f(x,y_0))dx\right|+\left|\int_b^g f(x,y_0)dx\right|\\&\le(d-c)\varepsilon+\varepsilon\\&=(1+d-c)\varepsilon,\forall y\in [c,d]\end{aligned}$$ it follows that $\int_a^\infty f(x,y)dx$ exists and converges uniformly on $[c,d].$

Now, let's define a function $\color{purple}{G(y):=\int_a^\infty\partial_2f(x,y)dx}.$ Then $G$ is continuous (a result proven in the script priorly). It remains to notice that $$\int_c^ydt\int_a^\infty\partial_2f(x,t)dx=\int_a^\infty dx\int_c^y\partial_2f(x,t)dt.$$

In all the results, we used the fact the set was unbounded. In old materials, I've come across the following task:

Prove that the  function $F(a)=\int_0^{1/a}\frac{e^{ax}-1}xdx$ is constant.

I might be wrong, but I tried to use the substitution $x=\frac1t$ in order to get to an unbounded interval and then use the results above. This got a bit messy, so I was wondering if it makes sense to even consider the $\boldsymbol{\text{ result 2}}$ and write something as $F'(a)=\int_0^{1/a}\partial_2 f(x,a)dx.$ I'm primarily interested in justification for that.

I apologize for the length of my post and thank you for reading in advance!

  • 1
    $\begingroup$ If I understand you correctly, you want to differentiate w.r.t $a$. Notice that $\int_0^{a+h}f(x,a+h)dx -\int_0^{a}f(x,a) = \left( \int_0^a f(x,a+h)dx - \int_0^{a}f(x,a) \right) + \int_a^{a+h}f(x,a+h)$. Next consider that if $f$ is continuous in a neighbourhood of $(a,a)$ then $(1/h)\int_a^{a+h}f(x,a+h)dx \approx f(a,a)$ using uniform continuity. So under mild conditions, you only really need to be worried about the term in the parentheses. $\endgroup$ Apr 12, 2022 at 0:36
  • $\begingroup$ @MarkoKarbevski, thank you for the comment. It makes sense. (: $\endgroup$
    – Invisible
    Apr 12, 2022 at 5:12

1 Answer 1


The answer to the question you have is actually a lot easier than you've made it out to be. With only theorems related to change of variables on bounded intervals we have

$$\int_0^\frac{1}{a} \frac{e^{ax}-1}{x}\:dx \xrightarrow{u = ax} \int_0^1 \frac{e^u-1}{u}\:du$$

which has no $a$ dependence at all.


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