# Tag Info

3

According to Theorem 5.13 in Rudin's "Principles of mathematical analysis" (slightly restated) If $f$ and $h$ are real and differentiable in $(0,1)$ and if $h'(x)\neq 0$ in a neighborhood of $1$, and $h(x)\to +\infty$ as $x\to 1^-$, and in addition $f'(x)/h'(x)\to A$ as $x\to 1^-$, then $f(x)/h(x)\to A$ as $x\to 1^-$. Applying this to $$f(x)=\... 0 One way to express this integral as a function of the Elliptic Integral K, is as follows. Do the change of variables x^2 = \frac{x_0+^2}{1-t^2}. This converts the integral into :$$ \frac{1}{\sqrt{x_+^2-x_-^2}}\int_0^1 \frac{dt}{\sqrt{(1-t^2)(1+\frac{x_-^2}{x_+^2-x_-^2}t^2)}} $$Which can be written formally as \frac{1}{\sqrt{x_+^2-x_-^2}}K(-\frac{x_-^2}{... 0 If$$f_h=\frac{1}{\sqrt{2 \pi \sigma ^2}}e^{-\frac{x^2}{2 \sigma ^2}}\qquad \text{and}\qquad f_\rho=\frac{f_h\left(\sqrt{\frac{x}{\rho }}\right)}{\sqrt{\rho x}}$$then$$f_\rho=\frac{e^{-\frac{x}{2 \rho \sigma ^2}}}{\sqrt{2 \pi \rho \sigma ^2} \sqrt{ x}}$$Let$$t^2=\frac{x}{2 \rho \sigma ^2}\implies x=2 \rho \sigma ^2 t^2\implies dx=4 \rho \sigma ^...

0

$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\... 1 This is false. Consider the function $$f(x)=\begin{cases} \frac{1}{\sqrt{-x}} & x\leq -1 \\ -\frac{1}{4}x^2+\frac{5}{4} & -1< x< 1 \\ \frac{1}{\sqrt{x}} & x\geq 1 \end{cases}$$ This function is differentiable for all$x\in\mathbb{R}$and converges to$0$at$\pm \infty$. The derivative is $$f'(x)=\begin{cases} \... 1 No, it's not true. Every continuous function f on \mathbb R such that f \to 0 at \pm \infty is uniformly continuous. Try e.g. f(x) = 1/(1+\sqrt{|x|}). 1 You should assume that b \not= 0 and c < L|b|. Then choose \epsilon > 0 small enough that (L - \epsilon)|b| > c, and select M > a with the property that s \ge M implies f'(s) > L-\epsilon. If a+t|b| > M then$$\int_a^{a+t|b|} f'(s) \, ds = \int_a^M f'(s) \, ds + \int_M^{a + t|b|}f'(s) \, ds > f(M) - f(a) + (L-\epsilon)... 0$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\...

3

We know that $|\sin(x)| \geq \frac12$ for $x \in [2\pi n + \pi/6, 2\pi n + 5\pi/6]$ (this holds for any $n \in \mathbb{N}$; to see why it's true just visualize the unit circle). Therefore we can estimate \int_\pi^\infty \frac{\sin^2x}{x} \,dx \geq \sum_{n = 1}^{\infty} \int_{2\pi n + \pi/6}^{2 \pi n + 5\pi/6} \frac{\sin^2x}{x} \,dx \geq \sum_{n=1}^{\... 1 \newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\... 2 \begin{align*} \int_{\pi}^{\infty} \frac{|\sin(x)|}{x} = \sum_{m=1}^{\infty}\int_{m\pi}^{(m+1)\pi}\frac{|\sin(x)|}{x}&\geq \sum_{m=1}^{\infty}\frac{1}{(m+1)\pi}\int_{m\pi}^{(m+1)\pi}|\sin(x)| \\&= \sum_{m=1}^{\infty}\frac{1}{(m+1)\pi}\int_{\pi}^{2\pi}|\sin(x)|\\ & =\sum_{m=1}^{\infty}\frac{2}{(m+1)\pi} \end{align*} The last quantity diverges. 3 @YvesDaoust is right. For n\ge1,\int_{n\pi}^{(n+1)\pi}\frac{|\sin x|}{x}dx=\int_0^\pi\frac{\sin x}{x+n\pi}dx\ge\frac{1}{(n+1)\pi}\int_0^\pi\sin xdx=\frac{2}{\pi}\frac{1}{n+1}.$$0 \newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\... 0 Well$$ \int_{\pi}^{\infty}\frac{\sin^2x}{x^2}dx\leq\int_{\pi}^{\infty}\frac{dx}{x^2} $$converges, but$$ \int_{\pi}^{\infty}\frac{|\sin x|}{x}\sqrt{1+\frac{\cos^2x}{x^2}-\frac{2\sin x}{x^3}+\frac{\sin^2x}{x^4}} \ dx\geq\int_{\pi}^{\infty}\frac{|\sin x|}{x}dx $$diverges. 0 \newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\... 1 Substitute$$x\to \frac{1-t}{t+1};\;dx=-\frac{2dt}{(t+1)^2};\;t\to \frac{1-x}{x+1}$$Integral becomes$$\int_1^{-1} \frac{(-2) e^{-\frac{2 (1-t)}{t+1}} \left(e^{-\frac{1-t}{t+1}}+1\right)}{(t+1)^2} \, dt=\int_{-1}^1 \frac{2 e^{\frac{2 (t-1)}{t+1}} \left(e^{\frac{t-1}{t+1}}+1\right)}{(t+1)^2} \, dt$$Notice that \underset{t\to -1}{\text{lim}}f(t)=0 so you ... 0$$\lim_{x\to \pm \infty}\text{erf}(x)=\lim_{x\to \pm \infty}\frac{2}{\sqrt \pi}\int_0^x e^{-t^2}dt = \frac{2}{\sqrt \pi} \int_0^{\pm\infty}e^{-t^2}$$using the fact e^{-t^2} is an even function:$$ \pm \frac{2}{\sqrt \pi} \int_0^\infty e^{-t^2}$$The integral is known as the gaussian integral(half of it) and is known to equal \frac{\sqrt \pi}{2} and thus,... 0 Hint Break the integral into$$\int_1^\infty \frac{x^2sin x}{e^x ln x}dx=\int_1^e \frac{x^2sin x}{e^x ln x}dx+\int_e^\infty \frac{x^2sin x}{e^x ln x}dx$$The second integral is convergent compared to x^2e^{-x} and for the first one, the function is \sim {1^2\times \sin 1\over e^1\ln x}. Now what are the natures of$$ \int_1^e{dx\over \ln x} $$and$$ \...

0

$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\... 3 You don't have to care about the exact value of the integral. Simply use asymptotics. To see how the integrand functions behaves asymptotically as$x \to \infty$, compute the sum of the two fractions. $$\frac{r}{x+1} - \frac{3x}{2x^2+r} = \frac{(2r-3)x^2 -3x + r^2-1}{(x+1)(2x^2+r)}$$ Now consider two cases: If$2r-3 = 0$, then asymptotically as$x \to \...

2

By using the comparison test, we know that if we integrate a rational function $\int_a^{\infty} p(x)/q(x)dx$ such that $q(x)\neq 0$ on the interval, then we converge if and only if $\deg(p)-\deg(q)<-1$. However, actually combining and simplifying the rational functions in this problem is tedious algebra, so one can ask if there is a better way. In fact, ...

0

I have finally managed to answer the question :) It is trivial to see that the integral is divergent for $r = 0\$, so let $r \neq 0$. If $r \neq 0$, note that we can always choose $a$ large enough such that $$\int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x = \int^a_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x + \int^{\... 1 Try to bring the logarithms together! :)$$\lim_{a\to \infty} \biggl(r\ln\frac{a+1}{2}+\frac 3 4\ln|\frac{2+r}{2a^2+r}|\biggl)$$You maybe should take r < -2 and r > -2 because of the absolute value, but it gives the same result so I'll go with normal brackets.$$\lim_{a\to \infty} (\ln(\frac{a+1}{2})^r+ \ln(\frac{2+r}{2a^2+r})^\frac 3 4 )\...

0

$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\... 2 Note that when you calculate$F$you don't do definite integrals. So, similar to what you wrote in your question, $$F(x)=\begin{cases}−xe^{−x}−e^{−x}+C_1, x\ge 0\\ −xe^x+e^x+C_2, x<0\end{cases}$$ Here$C_1$and$C_2$are some constants. From your condition at$x\to\infty$, you get$C_1=-5$. To get$C_2$, you must put the condition that$F$is continuous ... 0$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\...

1

$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\... 0$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\...

3

$\newcommand{\S}{\operatorname{sinc}}$ A 'brute force' solution to a beautiful problem. I don't claim this answer is as insightful as the others and the question is somewhat old, but I feel it is relevant and unique enough to merit posting. As usual, we will let $\S(x) = \sin(x)/x$ with $\S(0)=1$. The big idea: use angle-addition, Taylor series, and ...

0

\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\... 2 Note $$I=\int\limits_0^{\infty}\frac{\ln (1+x^{4})}{\sqrt{x}(1+x)}dx\overset{x\to x^2}= 2\int\limits_0^{\infty}\frac{\ln (1+x^{8})}{1+x^{2}}dx$$ $$1+x^8=(1+e^{i \frac\pi4}x^2)(1+e^{-i \frac\pi4}x^2)(1+e^{i \frac{3\pi}4}x^2)(1+e^{-i \frac{3\pi}4}x^2)$$ Then, apply the known integral result $$J(a)=\int_0^\infty \frac{\ln(1+ax^2)}{1+x^2}dx= \pi\ln(1+a^{\frac12})... 1 Put t=\sqrt{x} or t^2=x, which yields 2t\,dt=dx and x^8=t^{16}. Then we have$$ \int_0^{\infty}x^8 e^{-\sqrt{x}}\,dx = 2\int_0^{\infty} t^{17}e^{-t}\,dt $$This is the well-known Gamma function. It converges as long as the power of t is non-negative. You can figure out the exact value yourself by doing integration by parts once or twice until you ... 1 One more solution: using \sum_{n=-\infty}^{\infty}\left(x+2\pi n\right)^{-2}=\frac{1}{4}\csc^{2}\frac{x}{2} and$$\sum_{n\ge0}C_{n}x^{n}=\frac{1-\sqrt{1-4x}}{2x}\implies\sum_{n\ge0}C_{2n}x^{2n}=\frac{\sqrt{1+4x}-\sqrt{1-4x}}{4x}$$(by taking the even part) in terms of the Catalan numbers C_{n}=\frac{\left(2n\right)!}{n!\left(n+1\right)!},$$\begin{align}I\... 2 Using the principal branch of the logarithm, we have \begin{align} I &= \int_{0}^{\infty} \frac{\ln(1+t^{4})}{\sqrt{t}(1+t)} \, \mathrm dt = 2\int_{0}^{\infty} \frac{\ln(1+x^8)}{1+x^2} \, \mathrm dx = \int_{-\infty}^{\infty} \frac{\ln(1+x^8)}{1+x^2} \, \mathrm dx \\ &= \sum_{n=0}^{7} \int_{-\infty}^{\infty} \frac{\ln \left(1-xe^{i \pi(2n+1)/8}\... 0 Good evening, I found a simpler faster way to deal with this integral : 1/ The numerator being a pair function of Sin(x) one can easily show that the integral is equal to : F(a)=Integral from 0 to Pi/2 Arctan(a.(Sin(x)^2))) / (Sin(x)^2) dx 2/ Then we apply Feynman’trick to F(a) and get : F’(a)=Integral from 0 to Pi/2 dx / (1+ a^2. (Sin(x))^4 3/ Then we use t ... 1 Note \begin{align} &\int_0^\infty \frac{1-x(2-\sqrt x)}{1-x^3}dx=\int_0^\infty \frac{1+x^{\frac32}-2x}{1-x^3}dx\\= &\int_0^\infty \frac{1+x^{\frac32}}{1-x^3}dx -\int_0^\infty \underset{x\to x^{\frac12} }{\frac{2x}{1-x^3}}dx= \int_0^\infty \frac{dx}{1-x^{\frac32}} -\int_0^\infty \frac{dx}{1-x^{\frac32}}=0 \end{align} 1 For the second one you can use direct comparison as follows: Note that for x>0 you havee^{-\sqrt x} = \frac 1{e^{\sqrt x}} = \frac 1{\sum_{n=0}^{\infty}\frac{x^{\frac n2}}{n!}} \stackrel{n=20}{<} \frac 1{\frac{x^{10}}{20!}}$$Hence,$$\int_{x=1}^\infty x^8 e^{-\sqrt x} dx< 20! \int_{x=1}^\infty \frac 1{x^2} dx < \infty$$2 Let g(x)=\sqrt{x}-10\ln(x). Then g is increasing for x>400. g(e^9)=e^{4.5}-90>0. So g is positive and increasing for x>e^9>400. So, for x>e^9 we have that \begin{eqnarray} \sqrt{x}-10\ln(x)>0\\ \ln(x^{10})<\sqrt{x}\\ x^{10}<e^{\sqrt{x}}\\ x^8e^{-\sqrt{x}}<\frac{1}{x^2} \end{eqnarray} You should be able to finish from ... 3 Note that \int\limits^{\infty}_0\frac{\tan^{-1}t}{(1+t)^{n+1}}=\frac1nI_n, where$$I_n=\int\limits^{\infty}_0\frac{dt}{(1+t^2)(1+t)^{n}}$$The integrand can be decomposed iteratively as$$A_n(t)= \frac{A_{n-1}}{1+t}=\frac{1}{(1+t^2)(1+t)^{n}} =\frac{a_n-b_n t}{1+t^2}+ \sum_{k=1}^{n}\frac{b_{n-k+1}}{(1+t)^k}\tag1 $$where the coefficients satisfy the ... 1 I was able to get some kind of recurrence relation for$$I_n=\int\limits_0^{\infty}\frac{dt}{(1+t^2)(1+t)^n},$$but I am not satisfied with it since you can't really do anything with it. It's still kind of an answer but I'll accept a better one. First substitute t\mapsto\frac1t so that$$I_n=\int\limits_0^{\infty}\frac{t^ndt}{(1+t^2)(1+t)^n}.$$Notice that ... 3 This is not an answer but it is too long for comments. For the computation of$$I_n=\int\limits^{\infty}_0\frac{dt}{(1+t^2)(1+t)^{n+1}}$$it is amazing that a CAS gives a solution in terms of a generalized hypergeometric function which works very fine ... except when n is an integer ! What I think is that writing$$(1+t^2)(1+t)^{n+1}=(t+i)(t-i)(1+t)^{n+1}$$... 1 If by convolution you mean the standard$$(f * g)(t) = \int f(x)g(t-x)dx$$and by area$$A_f = \int f(x)dx$$then the question is whether the following is true$$\int\left[\int f(x)g(t-x) dx\right]dt = \left[\int f(x)dx\right]\cdot \left[\int g(x)dx\right]$$This is indeed true. By a simple change of variables$$u = xv = t - x$$with Jacobian$$\det \... 0 \begin{align}J&=\int_0^\infty \frac{\ln(1+x^2)\arctan x}{x(1+x^2)}dx\\ &=\int_0^1 \frac{\ln(1+x^2)\arctan x}{x(1+x^2)}dx+\int_1^\infty \frac{\ln(1+y^2)\arctan y}{y(1+y^2)}dy\\ &\overset{x=\frac{1}{y}}=\int_0^1 \frac{\ln(1+x^2)\arctan x}{x(1+x^2)}dx+\int_0^1 \frac{x\ln\left(\frac{1+x^2}{x^2}\right)\arctan\left( \frac{1}{x}\right)}{1+x^2}dx\\ &=... 0 Counterexample Setf_n(x)=n$for$x\in [0,\frac{1}{2n}]$and$f(x)=-2n^2x+2n$for$x\in [\frac{1}{2n}, \frac{1}{n}]$and$f$is zero for$x\geq \frac{1}{n}$. Then every$f_n$is continuous (in$C([0,\infty))$) and the sequence of the integrals goes to$0$but clearly$f_n$is not uniformly bounded 0 Let$f_n(x)=n^{3}(x-n)$for$n \leq x \leq n+\frac 1 {n^{2}}$,$f_n(x)=n^{3}(n+\frac 2 {n^{2}}-x)$for$ n+\frac1 {n^{2}} \leq x \leq n+\frac 2 {n^{2}}$and$0$elsewhere. Verify that this is a counter-example. [$f(n+\frac 1 {n^{2}}) =n \to \infty$] 0 To flesh out Kavi's hint, assuming that$C_2(x + \rho)$is positive (if it's negative, the integral doesn't converge) we can make the substitution$t = C_2(x + \rho)y$gives$dy = \frac{1}{C_2(x + \rho)} dt$and the integral becomes $$\int_0^\infty \left( \frac{t}{C_2(x + \rho)}\right)^{2m + 1} e^{-t} \left(\frac{1}{C_2(x + \rho)}dt\right) = \left(\frac{1}{... 1 Since you received good answers for the limit, let me go beyond it. Let x=n t$$I=\int_0^\infty\frac{\sin \left(\frac{x}{n}\right)}{x^2+x}dx=\int_0^\infty\frac{\sin (t)}{n t^2+t} dt=\int_0^\infty\frac{\sin (t)}{t} dt-\frac 1n\int_0^\infty\frac{\sin (t)}{t+\frac1n} dt\sin(t)=\sin \left(t+\frac{1}{n}-\frac{1}{n}\right)=\sin \left(t+\frac{1}{n}\right)\... 1 The problem here is: $$L=\lim_{x\to 0}\lim_{n\to\infty}\frac{\sin(x/n)}{x+x^2}$$ if we look at: $$\lim_{n\to\infty}\frac{\sin(x/n)}{x+x^2}=\lim_{n\to\infty}\frac{-x\cos(x/n)}{x(x+1)n^2}=\lim_{n\to\infty}\frac{-\cos(x/n)}{(x+1)n^2}=-\frac{1}{x+1}$$ and so: $$L=\lim_{x\to 0}\frac{-1}{x+1}=-1$$ so the integrand converges to a finite value for$x=0$now if we ... 1 So maybe this is another example of tricking the integral to allow the limit inside itself. We can do this in a lot of ways, but I'm very fond of leveraging dominated convergence theorem. Observe, for any$n,$we see: $$\int_0^{\infty} \frac{\sin(x/n)}{x + x^2} dx = \underbrace{\int_0^{\epsilon} \frac{\sin(x/n)}{x + x^2} dx}_{I} + \underbrace{\int_{\epsilon}^... 1 For x>1 we have$$\left|\frac{\sin x}{e^x-e^{-x}}\right|\leq \frac{1}{e^{x/2}}$$because for x>1$$e^x-e^{-x}>e^{x/2}$$thus$$\int_1^{\infty} \frac{\sin x}{e^x-e^{-x}} \,dx\le \int_1^{\infty} \frac{dx}{e^{x/2}}$$as the last integral converges, the given integral converges at$+\infty$. It converges at$x=0\$ as proved above by Mr.Claude ...

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