According to Wolfram Alpha, $\int_{-\infty}^{\infty}\sin(x)dx$ does not converge.  This makes no sense to me, intuitively, which I'll justify with a plot:    As we see, the positive and negative areas 'cancel out', so, for any $\alpha \in \mathbb{R}$, $\int_{-\alpha}^{\alpha}\sin(x)dx=0$ (I'm just thinking geometrically- in no way is this supposed to be a rigourous justification).  So, why is is that $\int_{-\infty}^{\infty}\sin(x)dx$ diverges?  A natural and intuitive reason, more than a rigourous one, would be best.  Thanks
113518
I was determining whether   $$\int_{-\infty}^{\infty} \sin x \, \mathrm{dx}$$  was divergent or convergent. So, I did the following steps:  $$\begin{align} \int_{-\infty}^{\infty} \sin x \, \mathrm{dx} &amp;= \int_{0}^{\infty}\sin x \, \mathrm{dx}+\int_{-\infty}^0\sin x \, \mathrm{dx} \\ &amp;=\lim_{t\rightarrow\infty} \left(-\cos x |^{t}_{0}\right) + \lim_{a\rightarrow-\infty} \left(-\cos x |^{0}_{a}\right)\\ &amp;=\lim_{t\rightarrow\infty} -\cos (t) + \cos 0 + \lim_{a\rightarrow-\infty} -\cos 0 + \cos a\\ &amp;=\lim_{t\rightarrow\infty}1 - \cos t + \lim_{a\rightarrow-\infty} -1+\cos a \end{align}$$  Now, at this point, it would be reasonable to say that both the limits are undefined and therefore, the integral is divergent but then if I try something like the following   \begin{align} \quad\quad&amp;=\lim_{t\rightarrow\infty}1 -\cos t + \lim_{a\rightarrow\infty} -1+\cos a \\ \quad\quad&amp;=\lim_{t\rightarrow\infty}-1 -\cos t + \lim_{a\rightarrow\infty} \cos a+1 \\ \quad\quad&amp;=\lim_{b\rightarrow\infty}-1 +\cos b - \cos b+1 \\ \quad\quad&amp;= 0 \end{align}  So, as you can see, it was shown before that the integral is divergent but with some manipulation, we came at an answer of $0$ but is that valid? I assume, a similar technique can be applied to $\int_{-\infty}^{\infty} \frac{1}{x} \, \mathrm{dx}$.
108648
Suppose that $\int_{\mathbb R} |f(x)| dx &lt; \infty$.     Can we say that $\int^{\infty}_{n} |f(x)| dx \to 0$ as $n \to \infty$ If yes, how to justify?
42702
Of course, it is easy to see, that the integral (or the antiderivative) of $f(x) = 1/x$ is $\log(|x|)$ and of course for $\alpha\neq - 1$ the antiderivative of $f(x) = x^\alpha$ is $x^{\alpha+1}/(\alpha+1)$.  I was wondering if there is an intuitive (probably geometric) explanation why the case $\alpha=-1$ is so different and why the logarithm appears?  Some answers which I thought of but which are not convincing:   Taking the limit $\alpha=-1$ either from above or below lead to diverging functions. Some speciality of the case $\alpha=-1$ are that both asymptotes are non-integrable. However, the antidrivative is a local thing, and hence, shouldn't care about the behavior at infinity.
72610
Does a function's limit when x goes to infinity must be zero for its integral to converge?  I had proved in my homework that if the function is non-negative, it's not necessarily true.  Now I read a solution that uses this as a fact - For an improper integral to converge, function's limit must be zero.  I hadn't found it in my books and notebook, and I can't find a reliable place says it is.  any help would be appreciate,  thanks
99793
I need to show that $$ \int\limits_1^{\infty}x^\alpha\sin(x^\beta)dx $$ converges if and only if $\alpha + 1 &lt; \beta$. I used substitution and integration to get $$ \lim\limits_{t\to\infty}\int\limits_1^tx^\alpha\sin(x^\beta)dx \overset{u = x^\beta}{\underset{du = \beta x^{\beta-1}}{=} }\lim\limits_{t\to\infty}\left(\frac{1}{\beta}\int\limits_1^{t^\beta}u^{\frac{\alpha + 1 - \beta}{\beta}}\sin{u}du\right) $$ If $\alpha + 1 -\beta &lt; 0$, then we have a product if a function with bounded anti-derivative and a differentiable decreasing function which tend to 0, and from Dirichlet's criterion, the integral converges.   For the converse, I used integration by parts $$  = \frac{1}{\beta}\lim\limits_{t\to\infty}\left(\left.-u^{\frac{\alpha + 1-\beta}{\beta}}\cos{u}\right|_1^{t^\beta} + \frac{\alpha+1-\beta}{\beta}\int\limits_1^{t^\beta}u^{\frac{\alpha + 1-2\beta}{\beta}}\cos{u}du\right) $$ For $\alpha + 1 -\beta\geq 0$, I get that the first term indeed diverges, but this is not enough to show total divergence. How can one show it?
98630
Can anybody tell me why $\sin(nx)$ converges weakly in $L^2(-\pi,\pi)$. I can't see how $\sin(nx)$ can converge? Explanation with any other example will be nice as well.
122758
I tried looking it up but many websites just state it without proof and without intuition.  I'm hoping to learn it a little bit better so that I don't forget how to compute the Jacobian when working with surface integrals where the divergence theorem is not applicable.  If you have a good online reference instead, please feel free to provide it :-)  Thanks,
75170
Let $\{x_n\}_{n=1}^\infty$ be a squence of nonnegative real numbers such that $\displaystyle \lim_{n \to \infty} x_n =x \in \Bbb R$. If $\alpha \in (0, \infty)$, is it true that $\displaystyle \lim_{n \to \infty} x_n^\alpha =x^\alpha$?  I read the proofs for $\alpha$ integer and for $\alpha=1/2$, but I wonder what happens with the other cases.
81277
I have this statement and I need to say if it is true or false:  Let $\{a_n\}$ be a real sequence.  $$\lim_{n\to +\infty} a_n = 0 \quad \implies \quad \sum_{n=1}^{\infty}(-1)^na_n \quad \text{converges}$$  I know, from the Leibniz criterion that:     If $a_n \to 0$, $a_n$ is decreasing and positive then $\sum_{n=1}^{\infty}(-1)^na_n$ converges   From this fact I believe that the statement is false but I couldn't come up with an infinitesimal sequence that isn't decreasing and for that reason is divergent. I tried some function with $sin(\frac{1}{n})$ without any luck. Any help would be very appreciated, thank you!
64675
$$\int_{0}^{\infty }\sin(x^{2})dx$$  I'm confused with this integral because the square is on the x, not the whole function. How can I integrate it? Thank you.  I have not done complex analysis (only real analysis as I am a high school student) so how can I evaluate it using elementary functions (without complex analysis)?
44167
I have this integral:  $$ \int_0^1{\frac{\sin x}{x}dx} $$ And I should prove that it is convergent. I have understand that if the resulting area is finite, then this integral is convergent, right?  So, I do this:  $$ \int_0^1{\frac{\sin x}{x}dx} = \int_0^1{\sin x\cdot \frac{1}{x}dx} =\\= \begin{bmatrix} -\cos x\cdot \ln x\end{bmatrix}_0^1 = (-\cos1\cdot \ln1)-(\cos0\cdot \ln0) = 0-(-\infty) = \infty $$  But, I get an infinit value, which I think is wrong. What am I doing wrong?  Ok, so apparently I messed some things up. If I do this again and use the integration by parts method I get this:  $$ \int_0^1{\frac{\sin x}{x}dx} = \int_0^1{\sin x\cdot \frac{1}{x}dx} = \sin x \ln x- \int_0^1{(\cos x\ln x) dx} = \sin x \ln x -\begin{bmatrix} \sin\frac{1}{x} \end{bmatrix}_0^1 \Rightarrow \lim_{t \rightarrow 0} {\int_t^1\frac{\sin x}{x} dx} = \lim_{t \rightarrow 0}{(\sin x \ln x-\frac{sin1}{1}-\frac{\sin t}{t})} $$  Is that more correct? And how do I proceed?
32312
Consider the $\displaystyle \int_0^\infty $sin$ (x^p) dx$. Does it converge when $p&lt;0$? Does it converge when $p&gt;1$?    My Work:  Let $x^p=y$, then $\displaystyle \int_0^\infty $sin$ \displaystyle (x^p) dx=\frac{1}{p}\sum_{n=1}^\infty \int_{(n-1)\pi}^{n\pi} \frac{\text{sin}  \; y}{y^r} dy$ , where $r=\frac{p-1}{p}&gt;1 $when $p&lt;0$ and $0&lt;r=\frac{p-1}{p}\leq 1 $when $p&gt;1$. I know that $\displaystyle \int_0^\infty $sin$ (\frac{1}{x}) dx$ diverges which is a special case of first case. But failed to show it generally. Can anyone please give me a hint to preceed?
37377
Recently I found a claim saying that  $$  \int_0^\infty \left( \frac{\sin ax}{x}\right)\left( \frac{\sin bx}{x}\right) \mathrm{d}x= \pi \min(a,b)/2  $$ from what I can see this seems to be true. I already know that $\int_{0}^\infty \operatorname{sinc}xy\,\mathrm{d}y = \pi/2$, and so independant of $y$. My suspicion is that this is closely related to the integral above.   Can someone give me some suggestions for evaluating the integral above? Also are there any generalizations for the integral? Eg $$ \int_{0}^{\infty} \left( \prod_{k=1}^N \frac{\sin (a_k \cdot x)}{x} \right) \,\mathrm{d}x $$ Where $a_k, \cdots, a_N$ are arbitrary positive constants. It seems related to the , but there are some subtle differences.
88402
I am trying to evaluate $$\int_{-\infty}^{\infty} \frac{\sin(x)^2}{x^2} dx $$ Would a contour work? I have tried using a contour but had no success. Thanks.  Edit: About 5 minutes after posting this question I suddenly realised how to solve it. Therefore, sorry about that. But thanks for all the answers anyways.
88183
This question is realy hard for me and I'd like to get some advises:  $$f(x) =   \begin{cases}      x^{\alpha}(1-\cos(1-x))^{\beta} &amp;  x\in (0,1)\\[6px]     \dfrac{1}{x^{\alpha}+x^{\beta}}  &amp;   x\ge 1   \end{cases}$$  I need to find the values of $\alpha$ and $\beta$ that make $\int_{0}^{\infty}f(x)\,dx$ converge.  Now I think I need to split the integral to $\int_{0}^{1}f(x)\,dx$ and  $\int_{1}^{\infty}f(x)\,dx$ and check the converge for each integral. But I'm stuck here.
6800
Question : The integral   $$ \int_{0}^{ \infty} \int_{0}^{ \infty} \frac { e^{-(x+y)}}{x+y} \mathop{dx}\mathop{dy} $$  is   (a) infinite   (b) finite but can not be evaluated in closed form   (c) 1  (d) 2 .  I tried substituting $u=x+y$ and $v=y$ that led me no where . I'm not even sure about convergence of integral .Any help would be greatly appreciated .
56816
$$ \sum_{n=1}^{\infty} \frac{(\sin n+2)^n}{n3^n}$$  Does it converge or diverge?   Can we have a rigorous proof that is not probabilistic?   For reference, this question is supposedly a mix of real analysis and calculus.