My teacher told me that not only do we have to use the erf function to approximate error, but that it is proved impossible to integrate in real analysis (at least not Riemann-integrable). Is there a name for this proof, and can I have it? I am not a mathematician, but such a simple function has been around and it irks me terribly and need a detailed proof for closure.
8216
I was wondering what the antiderivative of $e^{-x^2}$ was, and when I wolfram alpha'd it I got $$\displaystyle \int e^{-x^2} \textrm{d}x = \dfrac{1}{2} \sqrt{\pi} \space  \text{erf} (x) + C$$  So, I of course didn't know what this $\text{erf}$ was and I looked it up on wikipedia, where it was defined as:   $$ \text{erf}(x) = \dfrac{2}{\sqrt{\pi}} \displaystyle \int_0^x e^{-t^2} \textrm{d}t $$   To my mathematically illiterate mind, this is a bit too circular to understand. Why can't we express $\int e^{-x^2} \textrm{d}x$ as a 'normal function'? Also, what is the use of the error function?
25200
I am working my way through , where the author states that     "there is no closed form expression for the normal cumulative density   function"   but does not provide any further details as to why this is the case, simply saying that the alternative is to write it in terms of the error function.  Is there some way to intuit why  the Normal Distribution can not be expressed as a closed form function?
58107
$u$ substitution doesnt work. I don't see any connection with the Weierstrass substitution either. integration by parts results in a infinite integral series.
120929
We know that if $f\in L^1(\mathbb{R})$, then $\|f(\cdot+1/n)-f(\cdot)\|_{L^1}\to 0$ as $n\to \infty$, which implies that there exists a subsequence $f_{n_k}=f(x+\frac{1}{n_k})$ such that $f_{n_k}\to f$ a.e.   My problem is that  Is there any $f\in L^1(\mathbb{R})$  function such that the almost convergence result  $$ f(x)=\lim_{n\to\infty}f(x+1/n)\quad \text{a.e.} $$  is NOT valid?   I intend to believe that there exits such function. Otherwise, there would be a beautiful statement such that $f(x)=\lim_{n\to\infty}f(x+1/n)$ a.e. for any $f\in L^1(\mathbb{R})$.   Thank you for the long commenting below @David C. Ullrich. This question has been reduce to find a compact set $K$ such that for any $x\in K$, there are infinite $n$ such that $x+\frac{1}{n}\not\in K$.
10579
Consider the following problem from Axler's &quot;Measure, integration and real analysis&quot; (p72):  Honestly, I'm not even sure where to begin. This problem appears in the section that proves Lusin's and Egorov's theorem. The abstract Lebesgue integral is not yet introduced so I can not use that. Otherwise I could just show that the sum has finite integral and could conclude. I think we can write $f= \lim_k f_k$ as a pointwise limit of the partial sums but this probably does not quite help. Further attempt: I tried to show $$\{x: f(x) &lt; 1\}$$ contains a set of infinite measure but was unsuccesful. A hint to get started is appreciated!
99105
I'm wondering, for no particular reason: are there differentiable vector-valued functions $\vec{f}(\vec{x})$ in three dimensions, other than the constant function $\vec{f}(\vec{x}) = \vec{C}$, that have zero divergence and zero curl? If not, how would I prove that one doesn't exist?  I had a vague memory of learning some reason that such a function doesn't exist, but there's a pretty good chance my mind is just making things up to trick me ;-) But I thought about it for a little while and couldn't think of another divergence-free curl-free function off the top of my head, so I'm curious whether I was thinking of real math or not.
32074
In an exercise session of an analysis course (which covers Riemann integration and differentiation in one dimension rigorously) the following question came up:  Suppose $f$ is strictly positive and integrable (on some compact interval $[a,b]$ on the real line). Can we show that $\int_a^b f &gt; 0$?  The proof by using measure theory and Lebesgue integration is easy, but also beyond the students at this point. So can one do without machinery such as sets of zero measure?    Tools in use: Mean value theorems, fundamental theorem of calculus, Riemann's condition for integrability (upper and lower integral within every epsilon of each other implies integrability), Riemann-Darboux integral, usual integration and differentiation techniques, as well as elementary real analysis in the epsilon-delta style.
12405
In the past, I've come across statements  along the lines of &quot;function $f(x)$ has no closed form integral&quot;, which I assume means that there is no combination of the operations:  addition/subtraction multiplication/division raising to powers and roots trigonometric functions exponential functions logarithmic functions  which when differentiated gives the function $f(x)$. I've heard this said about the function $f(x) = x^x$, for example. What sort of techniques are used to prove statements like this? What is this branch of mathematics called?  Merged with &quot;&quot; by : Sometimes we are told that some functions like $\dfrac{\sin(x)}{x}$ don't have an indefinite integral, or that it can't be expressed in term of other simple functions. I wonder how we can prove that kind of assertion?
109804
Give me an example of two Riemann-integrable functions $f,g:[0,1]\to[0,1]$ such that $g\circ f$ isn't integrable!  I already know the following example: $$f(x)=\begin{cases} 0,     &amp;     \text{if $x$ is irrational} \\ 1,     &amp;     \text{if $x=0$}\\ \frac1q, &amp;     \text{if $x$ is rational and $x=\frac pq$ such that $q\in\Bbb N$ and $(p,q)=1$}  \\ \end{cases}$$ $$g(x) = \begin{cases} 1,  &amp; \text{if $x$ is of the form $\frac 1q$such that $q\in \Bbb N$} \\ 0, &amp; \text{otherwise}  \\ \end{cases}$$ now observe that $g\circ f$ is a famous example of non-integrable function!
124228
When evaluating $f(x)=\left(\dfrac{e^x-1-x}{x^2}\right)$, we observe large relative error as $x \rightarrow 0$.   Find a method to evaluate $f(x)$ to full machine precision for all $|x|&lt;1$.
82533
In the book of Analysis on Manifolds by Munkress, at page 121, it is asked that     Let $A$ be a bounded open set in $\mathbb{R}^n $; let $f: \mathbb{R}^n &gt; \to \mathbb{R} $ be a bounded continuous function. Give an example   where $\int_{\bar A} f$ exists but $\int_A f$ does not.   I couldn't find any such example. I mean I do noticed that $A$ should not be rectifiable, and the only way I can think of that $f$ is not integrable on $A$ is that the set $E$ where the limit  $$\lim_{x \to x_0} f(x) = 0$$ fails for $x_0 \in \partial A$ has a non-zero measure, but then this set $E$ also contained in $\bar A$, so in such a case $\int_{\bar A} f$ also wouldn't exists.  In short, I'm looking for such an example.  Note: I have read  answer, but I have no idea what the answerer is talking about, so I wanted to ask a new question including my own thoughts.
78051
Today in my calculus class, we encountered the function $e^{-x^2}$, and I was told that it was not integrable.  I was very surprised. Is there really no way to find the integral of $e^{-x^2}$? Graphing $e^{-x^2}$, it appears as though it should be.   A  on Gaussian Functions states that   $$\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$$  This is from -infinity to infinity. If the function can be integrated within these bounds, I'm unsure why it can't be integrated with respect to $(a, b)$.  Is there really no way to find the integral of $e^{-x^2}$, or are the methods to finding it found in branches higher than second semester calculus?
56427
If a function is a combination of other functions whose derivatives are known via composition, addition, etc., the derivative can be calculated using the chain rule and the like. But even the product of integrals can't be expressed in general in terms of the integral of the products, and forget about composition! Why is this?
91512
I'd like to prove the following version of the Riemann-Lebesgue lemma:     Let $f: [0,1] \to \mathbb R$ be continuous. Then $$\int_0^1 f(x)\sin(nx) \, dx \xrightarrow{n \to \infty} 0$$   It's quite easy to show the lemma for the case $f\in C^1$. So I was hoping that one can approximate the $f\in C^0$ with polynomials (by using the Weierstrass approximation theorem) and then apply the already shown case for $f\in C^1$. Yet it didn't work out for me. Is this a feasible way?
113366
I am having trouble with the following problem:      $f:\mathbb{R}\to \mathbb{R}$ is a measurable function such that for all $a$:   $$\int_{[0,a]}f\,dm=0.$$   Prove that $f=0$ for $m$ almost every $x$ (here $m$ is the Lebesgue measure).    I have no problem proving this for $f$ non-negative, or under the assumption that $f$ is integrable. But the question only assumes that $f$ is measurable and no more.  My idea was the usual thing; we look at the set of points where $f$ is positive and negative and assume one of these has measure greater than zero. Then I wanted to estimate one of these by an open set, look at the integral on the open set and show that it had to be greater than zero, a contradiction. But a key part of this attack is the assumption of the absolute continuity of the integral, which only holds in the case where $f$ is integrable.  Alternatively, if it were integrable one could simply estimate $f$ by a continuous function, where the result is quite obvious.  Ultimately we are going to show that $f$ is integrable, but it is not clear to me how to show this before showing it is zero a.e. So there must be a simpler way. Does anyone have suggestions?
94405
Can somebody give rigorous proof that the circle cannot be embedded into the real line? Obviously, if so it must be necessarily continuous and I guess this suffices in this case since there is no continuous injection by intuition...
30293
$\newcommand{\erf}{\operatorname{erf}}$ This may be a very na&iuml;ve question, but here goes.  The  $\erf$ is defined by $$\erf(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2}dt.$$ Of course, it is closely related to the normal cdf $$\Phi(x) = P(N &lt; x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-t^2/2}dt$$ (where $N \sim N(0,1)$ is a standard normal) by the expression $\erf(x) = 2\Phi(x \sqrt{2})-1$.  My question is:     Why is it natural or useful to define $\erf$ normalized in this way?   I may be biased: as a probabilist, I think much more naturally in terms of $\Phi$.  However, anytime I want to compute something, I find that my calculator or math library only provides $\erf$, and I have to go check a textbook or Wikipedia to remember where all the $1$s and $2$s go.  Being charitable, I have to assume that $\erf$ was invented for some reason other than to cause me annoyance, so I would like to know what it is.  If nothing else, it might help me remember the definition.  Wikipedia says:     The standard normal cdf is used more often in probability and statistics, and the error function is used more often in other branches of mathematics.   So perhaps a practitioner of one of these mysterious "other branches of mathematics" would care to enlighten me.   The most reasonable expression I've found is that $$P(|N| &lt; x) = \erf(x/\sqrt{2}).$$ This at least gets rid of all but one of the apparently spurious constants, but still has a peculiar $\sqrt{2}$ floating around.