# Question on an improper integral of the form $\int_0^af(x,a)dx$

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. $$a_0 and at the same time $$\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

\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!

• 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. Apr 12, 2022 at 0:36
• @MarkoKarbevski, thank you for the comment. It makes sense. (: Apr 12, 2022 at 5:12

$$\int_0^\frac{1}{a} \frac{e^{ax}-1}{x}\:dx \xrightarrow{u = ax} \int_0^1 \frac{e^u-1}{u}\:du$$
which has no $$a$$ dependence at all.