Show without evuluating the integral that $\int_{-\infty}^{0} \frac{x^2}{(x^2+1)^2(x^2 + 2x + 2)} dx$ converges. There is a similar question posted  whereby nobody actually showed that the above integral is convergent without evaluating the integral. Is it right that I first let $u = -x$, then $$\int_{-\infty}^{0} \frac{x^2}{(x^2+1)^2(x^2 + 2x + 2)} dx = -\int_{0}^{\infty} \frac{u^2}{(u^2 + 1)^2(u^2 -2u + 2)} du$$ Afterwards I can apply the Comparison test on the Right hand side? The only comparison test I know of is the one stated in Stewart's book,  Suppose $f$ and $g$ are continuous functions with $f(x) \geq g(x) \geq 0$ for $x\geq a$. If $\int_{a}^{\infty} f(x) dx$ is convergent, then $\int_{a}^{\infty} g(x) dx$ is convergent. If $\int_{a}^{\infty} g(x) dx$  is divergent, then $\int_{a}^{\infty} f(x) dx$ is divergent.  Or what is the better approach here?
122186
Show that $$\int_{-\infty}^{\infty} \dfrac{x^2}{(x^2+1)^2(x^2+2x+2)} \ dx$$ is convergent.   $$\int_{0}^{\infty} \dfrac{x^2}{(x^2+1)^2(x^2+2x+2)} \ dx$$ is convergent by comparison test with $\int_{0}^{\infty} \dfrac{1}{x^4} dx$.  How do i show that $$\int_{-\infty}^{0} \dfrac{x^2}{(x^2+1)^2(x^2+2x+2)} \ dx$$ is convergent?
101866
In a course, my teacher told us that the following integral is convergent and used the comparison test to prove it; my question is how to find the antiderivative in closed form? It seems to exist; if, however, it doesn't exist, can someone prove it?       $$\int\sqrt{\dfrac1{1+x^3}}\mathrm dx$$
57620
Let $f: [1, \infty) \to \mathbb{R}$ be a continuous function such that the improper integral $$\int_1^\infty f(x) \ dx$$ exists. Show or disprove that $\lim \limits _{x \to \infty} f(x) =0$.  Our professor said that there are counterexamples but after frying my brain out I still could not find any. It does sound logical.
76859
I have been asked to prove that $$\int_0^\infty \frac{x\,\log x}{(1+x^2)^2}\,dx = 0$$  I tried proceeding as follows: \begin{align} \int_0^\infty \frac{x\,\log x}{(1+x^2)^2}\,dx &amp;= \int_0^1 \frac{x\,\log x}{(1+x^2)^2}\,dx + \int_1^\infty \frac{x\,\log x}{(1+x^2)^2}\,dx \\ &amp; = \int_0^1 \frac{x\,\log x}{(1+x^2)^2}\,dx + \int_{u=1}^0 \frac{u\,\log u}{(1+u^2)^2}\,du\quad (\text{where }u=1/x) \\ &amp;= \int_0^1 \frac{x\,\log x}{(1+x^2)^2}\,dx - \int_0^1 \frac{u\,\log u}{(1+u^2)^2}\,du \\ &amp;=^? \int_0^1 0 \,dx \\ &amp;= 0 \end{align}  But I think that I should first show that the improper integral $\int_0^1 {x\,\log x}/{(1+x^2)^2}\,dx$ is convergent.  Do I need to show it? If yes, how to do so?  (I have been taught basic methods such as comparison, limit comparison and Dirichlet test.)
76881
$$\int^\infty_1\frac{1}{x^{1+1/x}}dx$$  I'm preparing for exams. I would also like to know what are some commonly used methods to show an improper integral diverges?
2553
Is this convergent or divergent? if convergent, find the sum. If divergent explain why.  $$\sum _{n=1}^{\infty }\:\frac{2n^2}{5n^2+2n+1}$$  I want to use the divergent test which is $\mathrm{If\:}\lim _{n\to \infty }a_n\ne 0\mathrm{\:then\:}\sum a_n\mathrm{\:diverges}$   $\frac{2n^2}{5n^2+2n+1}$  I would assume to take out the largest $n$ in both the top and bottom   $\lim _{n\to \infty }\left(\frac{n^2}{n^2}\cdot \frac{2}{5+\frac{2}{n}+\frac{1}{n^2}}\right)\:$  having the $n^2$ cancel  $\lim _{n\to \infty }\frac{2}{5+\frac{2}{n}+\frac{1}{n^2}}$ after taking the limit  $\frac{2}{5+0+0}$  so $\frac{2}{5}$$\ne $ $\:0\:$ so it's divergent?
72000
For what values $s$ does $\int_{0}^{\infty} \frac{\sin x}{x^s}dx$ converge?  Solution says $0&lt;s&lt;2$.  Any ideas?
25859
I'm really struggling on this question.  I've been thinking about it for awhile now but this is one of those where I don't really have much intuition on what to do.  Problem 3. Prove that if the limit $\lim_{x\to +\infty}f(x) =: L$ exists (finite or infinity) and the improper integral. $$\int_a^{+\infty}f(x) dx$$ is convergent, then $L = 0$.  Here is some facts that I know that I think I should perhaps use. Facts/Knowledge I know: The above integral can be rewritten as:  $$\int_a^{\infty}f(x)dx = \lim_{A\to \infty}\int_a^{A}f(x)dx$$  We know that limit as x goest to infinity of $f(x)=L$.  I need to somehow show that limit L is equal to 0. If I take the integral I get $\lim_{A \to \infty} F(A)-f(a)$ and I know this does not equal infinity as the integral converges.  I'm not sure if this is on the right track or what to do next. I was thinking maybe perhaps instead I could use the Cauchy Criterion to come up with a proof.  Note: I'm not used to doing proofs with Improper integrals.  Theorem 1 (Cauchy Criterion). The improper integral (1) converges if and only if for every $\epsilon &gt; 0$ there is an $M\geqslant a$ so that for all $A, B \geqslant M$ we have $$\Bigg|\int_A^B f(x) dx\Bigg| &lt; \epsilon$$
107304
The problem is to show  $$\int_0^\infty \frac{e^{-x}-e^{-xt}}{x}dx = \ln(t),$$  for $t \gt 0$.  I'm pretty stuck. I thought about integration by parts and couldn't get anywhere with the integrand in its current form. I tried a substitution $u=e^{-x}$ and came to a new integral (hopefully after no mistakes)  $$ \int_0^1 \frac{u^{t-1}-1}{\log(u)}du, $$  but this doesn't seem to help either. I hope I could have a hint in the right direction... I really want to solve most of it by myself.  Thanks a lot!
113518
I was determining whether   $$\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}$$  Now, at this point, it would be reasonable to say that both the limits are undefined and therefore, the integral is divergent but then if I try something like the following   \begin{align} \quad\quad&amp;=\lim_{t\rightarrow\infty}1 -\cos t + \lim_{a\rightarrow\infty} -1+\cos a \\ \quad\quad&amp;=\lim_{t\rightarrow\infty}-1 -\cos t + \lim_{a\rightarrow\infty} \cos a+1 \\ \quad\quad&amp;=\lim_{b\rightarrow\infty}-1 +\cos b - \cos b+1 \\ \quad\quad&amp;= 0 \end{align}  So, as you can see, it was shown before that the integral is divergent but with some manipulation, we came at an answer of $0$ but is that valid? I assume, a similar technique can be applied to $\int_{-\infty}^{\infty} \frac{1}{x} \, \mathrm{dx}$.
117760
Consider the sequence of functions  $$ f_n(x) = \frac{x^2}{n} $$ for $x \in \mathbb{R}$  I have shown that $f_n(x)$ converges pointwise towards $0$ for $n \rightarrow \infty$ but I am not sure whether my attempt for showing that it does not converge uniformly towards $0$ for $n \rightarrow \infty$ is correct. Do you mind verifying?   By negation we have that $f_n(x)$ does not converge towards $0$ for $n \rightarrow \infty$ if   $$\exists \epsilon &gt; 0 \forall n \in \mathbb{N} \exists x \in \mathbb{R} \exists n \in \mathbb{N}: n \geq N \Rightarrow |\frac{x^2}{n}| \geq \epsilon$$  Let $\epsilon = 1$. Then for all $n \in \mathbb{N}$ we can find a $x \in \mathbb{R}$ such that for $n \geq N$ that $|x^2/n| \geq 1$. Can I then just pick $x = n^{1/2}$ so we have that $$ |x^2/n| = |(n^{1/2})^2/n| = |n/n| = 1 \geq 1 $$ which means that $f_n(x)$ does not converge uniformly towards $0$ for $n \rightarrow \infty$. Is this ok?
108084
Determine whether the following series is convergent or divergent  $$\sum_{n=0}^∞ \frac{(\ln(n)+5)}{n^2}$$  I tried using comparison test and p-test: $$\frac{\ln(n)+5}{n^2}$$&lt;=$$\frac{1}{n^2}$$  And since 1/n^2 is convergent by p-test, the series is convergent. But i am not sure if my solution is correct especially the part on determining the upper bound of $$\frac{\ln(n)+5}{n^2}$$
88631
Problem: Show that $\int_{0}^{\infty}|\frac{ \sin x}{x}|=\infty $ .  Here is what i have tried :  I found out that for all $x&gt;0$ , $x-\frac{x^3}{6} &lt; \sin x &lt;x $  . So $1-\frac{x^2}{6}&lt;\frac{ \sin x}{x}&lt;1$ . So applying integration to both side from $0$ to $M$ we have $M-\frac{M^3}{18}&lt;\int_{0}^{M}|\frac{ \sin x}{x}|&lt; M$ . Now we send $M \rightarrow \infty$ and observe that left-hand side goes to infinity . Hence proved .  Please point out if there is anything wrong with my approach . I will also be delighted if could provide alternative solutions . Thank you .
106185
$\sin x^2$ does not converge as $x \to \infty$, yet its integral from $0$ to $\infty$ does.  I'm trying to understand why and would like some help in working towards a formal proof.
76201
I need help solving this integral: for which values of the parameter $p$ is this improper integral convergent?  $$\int_{0}^{\pi/2}\tan^{p}(x) \; dx$$  Thanks a lot!
36066
So I need to know the convergence of this sum. The integral test does not seem to work (WolframAlpha gives a confusing answer which implies it's not the correct method). I also tried the limit comparison test for various harmonic series, but they are all inconclusive. I'm also not getting very far with the ratio test. $$ \sum_{n=2}^\infty \dfrac{\sqrt{n+1}}{n(n-1)} $$
21683
Question :    For each $n \in \mathbb N$, Define : $\gamma_n=1+\frac{1}{2}+\dots+\frac{1}{n}-\int_1^n \frac{1}{x}dx$    Prove that $\{\gamma_n\}$ is convergent.      Note 1 :  I know that $\sum_1^\infty \frac{1}{k}$ diverges. I can guess that this sequence is convergent to $0$ because of the similar terms inside the integration and outside of it.  The problem is that the integration is continuous. I mean that $x$ is not just integers.  What can we do with the values left?! ( For example, $\frac{1}{2.5}$ ) Also, Notice that this problem should be solved with some elementary tools.  ( It's from a part of my book which is even before any discussion about the Fundamental theorem of calculus. )   Note 2 : I'm not good at integration and i'm learning it. Any good details can help me.    Thanks in advance.