Tag Info

New answers tagged

0

Hint: Swap y and z, and then switch to cylindrical. The numerical value should be about 0.62, you can check it afterwards.


0

There is no general rule in introducing an extra variable to apply the Feynman differentiation trick. In this case it looks reasonable to consider $\arctan(x)$ as $\arctan(ax)|_{a=1}$ because the derivative of the arctangent is a rather simple algebraic function. So, let: $$ I(a) = \int_{0}^{1}\frac{\arctan(ax)}{\sqrt{1-x^2}}\,dx. \tag{1}$$ We trivially have ...


1

First $$ \int_{S_{n}}^{S_{n+1}}\dfrac{dx}{x\ln(x)}\leq\int_{S_{n}}^{S_{n+1}}\dfrac{dx}{S_{n}\ln(S_{n})}=\dfrac{1}{S_{n}\ln(S_{n})}\int_{S_{n}}^{S_{n+1}}dx=\dfrac{a_{n+1}}{S_{n}\ln(S_{n})} $$ for $S_{n+1}\geq S_n$ and $x\ln(x)\downarrow$. So $$ ...


3

Hint. We have, as $x \to 0$, $$ \ln(1+x^4)=x^4+\mathcal{O}(x^5) $$ giving $$ \frac{\ln^{\alpha}(1+x^4)}{x^4}=\frac1{x^{4-4\alpha}}+\mathcal{O}\left(\frac1{x^{3-4\alpha}}\right) $$ and $$ \frac{\ln^{\alpha}(1+x^4)}{x^4}\cos \frac1x=\frac1{x^{4-4\alpha}}\cos \frac1x+\mathcal{O}\left(\frac1{x^{3-4\alpha}}\right). \tag1 $$ Thus $$ ...


0

Though answers were already given, I'd like to suggest a different approach: We see that the absolute value of $\cos(x) + \sin(x)$ is required. Also, we are dealing with sinusoidal functions, which can be easily dealt with the square function, so: $$ \begin{align} \int\limits_0^\pi \lvert \cos(x) + \sin(x) \rvert &= \\ &= \int\limits_0^\pi ...


0

In your simplified form, try substituting $u=\cos\left(\frac x2\right)$


0

Hint Use the double-angle identity $$\cos 2u = 2 \cos^2 u - 1,$$ (and set $u = \frac{x}{2}$) to express write of the trigonometric functions as functions the common argument $\frac{x}{2}$.


2

The standard way of proving this identity is to write $x^{-x} = e^{-x\log x}$ and then expand $e^{-x\log x} = \sum_{n=0}^{\infty} \frac{(-x\log x)^{n}}{n!}.$ Now the integral is $$\int_{0}^{1} \sum_{n=0}^{\infty} \frac{(-x\log x)^n}{n!} \mathrm{d}x = \sum_{n=0}^{\infty} \int_{0}^{1}\frac{(-x\log x)^{n}}{n!}\mathrm{d}x= \sum_{n=0}^{\infty} ...


2

Let $$ R_n=\dfrac{1}{n}\sum\limits_{i=0}^{n-1}\int_{i}^{i+1}f(\dfrac{i}{n})g(x)dx $$ First we prove that $$ \lim\limits_{n\to\infty}R_n=\int_{0}^{1}f(x)dx\int_{0}^{1}g(x)dx $$ Since $f(x)$ is uniform continuous on $[0,1]$, $\forall \epsilon>0,\space\exists N>0, \forall n>N$, let $\delta=\dfrac{1}{n},\space \forall x_1,x_2\in [0,1], ...


1

It seems your fundamental difficulty is with definitions of probability concepts, not so much with the process of integration. If $X \sim Unif(50, 150),$ then the density function $f_X(x)$ has three parts: (i) For $x < 50,\;f_X(x) = 0;\;$; (ii) for $50 \le x \le 150,\;f_X(x) = 1/100 = 0.01;$ and (iii) for $x > 150,\;f_X(x) = 0.$ If you want to ...


0

If you are going to use Parseval, then you need to evaluate $a_{n}=\frac{1}{\pi}\int_{-\pi }^{\pi }\sin ^{2}t\cos ntdt$. Since the integrand is even, it suffices to integrate from $0$ to $\pi $ and multiply by $2$. $b_{n}$ is a similar integral with $\sin nt$.But the integrand in this case is odd, so these terms vanish. So you are left with the integrals ...


5

We have: $$\sin^2(x) = \frac{1-\cos(2x)}{2}\tag{1}$$ hence Parseval's identity implies: $$ \int_{-\pi}^{\pi}\sin^4(x)\,dx = 2\pi\cdot\frac{1}{4}+\pi\cdot\frac{1}{4}=\color{red}{\frac{3\pi}{4}}.\tag{2}$$


0

A simpler way. Since $\cos(z) = \Re(e^{iz})$, $\begin{array}\\ I(n) &=\int_0^\pi e^{-t}\cos nt dt\\ &=\Re \int_0^\pi e^{-t}e^{int} dt\\ &=\Re \int_0^\pi e^{-t+int} dt\\ &=\Re \frac{e^{t(-1+in)}}{-1+in}\big|_0^\pi\\ \end{array} $. $\begin{array}\\ \frac{e^{t(-1+in)}}{-1+in} =\frac{e^{-t}+e^{int}}{-1+in}\\ ...


2

$$\begin{eqnarray*} I &=& \frac{1}{n}\int_{0}^{n\pi}e^{-t/n}\cos t\,dt =\frac{1}{n}\sum_{j=0}^{n-1}\int_{j\pi}^{(j+1)\pi}e^{-t/n}\cos t\,dt\\&=&\frac{1}{n}\int_{0}^{\pi}\cos t\,e^{-t/n}\sum_{j=0}^{n-1}(-1)^j e^{-j\pi/n}\,dt\tag{1}\end{eqnarray*}$$ Now: $$\sum_{j=0}^{n-1}(-1)^j ...


2

Since: $$\frac{d}{dx}\sqrt{1+x^4} = \frac{4x^3}{2\sqrt{1+x^4}}$$ we have: $$ \int_{0}^{1}\frac{x^3}{\sqrt{1+x^4}}\,dx = \frac{\sqrt{2}-1}{2}.$$


1

hint What about the change of variable $1+x^4=u$ such that $4x^3dx=du$ So the integral becomes $$\frac{1}{4}\int_1^2 u^{-\frac{1}{2}}du$$ Can you take it from there


6

The General Case: Consider instead the integral \begin{align} J(\eta) &=\Re\int^{2\pi}_0x^\eta\ {\rm Li}_2(e^{ix})^2\ {\rm d}x\\ \end{align} for $\eta\in\mathbb{N}_0$. Deforming the contour around $z=0$, we get \begin{align} J(\eta) &=\Re\oint_{|z|=1}\frac{{\rm Li}_2(z)^2\ln^{\eta}(z)}{i^{\eta+1}z}{\rm d}z\\ &=\Re\int^1_0\frac{\left((\ln(x)+2\pi ...


3

$I(1)$ can be evaluated in the following way also. One has to note that $$\int_0^{2\pi} x\sin(mx)\sin(nx)\,dx=\begin{cases} 0 & n \neq m \\ \pi^2 & n=m \end{cases}$$ To prove the above, write $$\int_0^{2\pi} x\sin(mx)\sin(nx)\,dx=\frac{1}{2}\int_0^{2\pi} x\cos((m-n)x) \,dx-\frac{1}{2}\int_0^{2\pi} x\cos((m+n)x)\,dx$$ and then use integration by ...


1

I get that the integral goes like $\frac{\pi(1-e^{-\pi})}{n^2}$ for even $n$. Here is the annoying long derivation. I'll assume that $n$ is even for now and write $n = 2m$. First, split the integral into $m$ parts. $\begin{array}\\ I(n) &=I(2m)\\ &=\int_0^\pi e^{-t}\cos n\, t\ dt\\ &=\int_0^\pi e^{-t}\cos (2mt) dt\\ ...


3

Here's one idea: $\cos(nt) \geq 1/2$ on $[0,\pi/3n]$, and is always $\geq -1$, so $$\int_0^\pi e^{-t} \cos(nt) \geq \int_0^{\pi/3n} e^{-t}/2 dt - \int_{\pi/3n}^\pi e^{-t} dt.$$ I don't know if this is tight enough for your purposes, however. If I needed this a little tighter I would repeat the same idea: cut off the region where $\cos(nt) \geq 1/2$ and ...


0

Hint: Let: $$ \int_a^x f(t)dt=F(x) \qquad \int_a^x g(t)dt=G(x) $$ From the fundamental theorem of calculus we have $$ \dfrac{d}{dx} F(x)=f(x) \qquad \dfrac{d}{dx} G(x)=g(x) $$ and, using the product rule for derivative: $$ \dfrac{d}{dx}\left[\int_a^x f(t)dt\int_a^x g(t)dt\right]=\dfrac{d}{dx}\left[F(x)G(x)\right]=G(x)f(x)+F(x)g(x) \ne f(x)g(x) $$


3

Just find two pairs of functions $f_1, g_1$ and $f_2,g_2$ such that $$\int f_1 = \int f_2, \quad\quad \int g_1 = \int g_2$$ but $$\int f_1 g_1 \neq \int f_2 g_2$$ Probably any two pairs you come up with will work.


5

I can confirm $I(1)$: Using the series expansion for Clausen's function we have $$ I(p)=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\int_0^{2 \pi}x^p\frac{\sin(n x)}{n^2}\frac{\sin(m x)}{m^2}\,\mathrm dx $$ now setting $p=1$. Furthermore we may use the orthogonality of sine $\int^{2 \pi}_{0}{\sin(n x)\sin(m x)}\,\mathrm dx=\pi\delta_{mn}$ and integrate by ...


1

Your integral with limit $\int_{r_e}^{r_o}$ will produce $\frac{q^2}{8\pi r_e}-\frac{q^2}{8\pi r_o}$ (in fact this follows from $\int_{r_e}^{r_o}=\int_{r_e}^\infty-\int_{r_o}^\infty$), so it does make sense to replace the infinite upper limit with a finite one (which would however feel arbitrary), but even then you cannot replace the lower limti with $0$.


2

$$I=\int_{0}^{a}J_0(b\sqrt{a^2-x^2})\cosh(cx)\,dx = a\int_{0}^{1}J_0(ab\sqrt{1-z^2})\cosh(acz)\,dx$$ The trick is now to expand both $J_0$ and $\cosh$ as Taylor series, then to exploit: $$ \int_{0}^{1}(1-z^2)^{n}z^{2m}\,dz = \frac{\Gamma(n+1)\,\Gamma\left(m+\frac{1}{2}\right)}{2\cdot\Gamma\left(m+n+\frac{3}{2}\right)}\tag{2}$$ hence, by assuming $b>c$ and ...


3

The question is: Which logarithm function are you using? If you are using the real-valued $\ln \colon (0,+\infty) \to \mathbb{R}$, your integrand is simply not defined on the interval $\bigl[0,\frac{\pi}{2}\bigr]$, and hence the integral neither. I put it into Alpha and it gave me -1/2 as answer. Unless you tell it not to, Wolfram Alpha happily uses a ...


0

Even integrals can be undefined. In your case there will be no curve drawn for$$f(x)=e^{- \frac12sin2x}$$ so the area is also undefined.


1

This is really a notational mistake. A volume element $ds$ is just a one form on $\partial D$ so that $|ds| =1$. Now the boundary is parametrized by $$ \gamma(\theta) = P + r e^{i\theta}\Rightarrow \gamma'(\theta) = ir e^{i\theta}\Rightarrow ||\gamma'(\theta)|| = r$$ Thus $ds = r d\theta$.


8

Divergent Integral Using the substitution $x=e^{-t}$, we get $$ \int_0^{\pi^2}\frac1{\log(x)}\,\mathrm{d}x =-\int_{-2\log(\pi)}^\infty\frac{e^{-t}}t\,\mathrm{d}t\tag{1} $$ This is not convergent near $t=0$, that is, near $x=1$. However, we can compute the Cauchy Principal Value. Principal Value First, we look at an interval symmetric about $t=0$ and ...


1

The Riemann-Lebesgue Lemma states that if $f$ is $\mathscr{L}^1$ integrable, then $$\lim_{\omega \to \infty}\int_{-\infty}^{\infty}f(t)e^{i\omega t}dt=0$$. Obviously, for real-valued $f$, we have $$\lim_{\omega \to \infty}\int_{-\infty}^{\infty}f(t)e^{i\omega t}dt=\lim_{\omega \to \infty}\int_{-\infty}^{\infty}f(t)\cos(\omega t)dt+i\lim_{\omega \to ...


0

If all you need is help taking the integral from -1 to 0, try looking at a plot of $\tan x$ from -1 to 1. The symmetry about the vertical axis should be helpful. It should be a matter of taking what you say you know how to do and adding a minus sign.


2

First of all, you forgot the $dx$'s. Here's a hint. You can integrate $\tan x$ by writing it as $\sin x / \cos x$ and making the substitution $u=\cos x$. For $\sin^2 x$, use the trig identity $$ \sin^2 x = \frac{1-\cos{2x}}{2} $$ Don't forget to change the upper and lower limits of your integral when you do the $u$ sub.


3

This type of integral is called a logarithmic integral function $\mathrm{li}(x)$. A logarithmic integral function is used in physics and number theory. In the case of having $0$ and $\pi^2$ as your bounds, the function would be $\mathrm{li}(\pi^2) - \mathrm{li}(0)$ which simplifies to $\mathrm{li}(\pi^2)$. Calculating it would be tedious and you would ...


1

Do not waste your time on random integrals like this, since chances are there is no closed form or nice expression for this. If you really need to compute this integral for some purpose, the right way out is to find an appropriate numerical quadrature capable of handling the singularity at the origin.


0

You may obtain $$ \int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\:dx=\frac12 \log^2 (1+\sqrt{2}) . \tag1 $$ Proof. First observe that $$\begin{align} \int_0^\pi \frac{x\cos x}{1+\sin^2 x}\:dx &= \int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\:dx + \int_{\pi/2}^{\pi} \frac{x\cos x}{1+\sin^2 x}\:dx \\ & = \int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\:dx - ...


2

Sub $x=y^2$ and the integral becomes $$4 \int_0^{\infty} dy \frac{\log{y}}{1+y^4} $$ So consider $$\oint_C dz \frac{\log^2{z}}{1+z^4} $$ where $C$ is a keyhole contour of outer radius $R$ and inner radius $\epsilon$ about the positive real axis. In the limit as $R \to \infty$ and $\epsilon \to 0$, the contour integral is equal to $$-i 4 \pi ...


0

Hint: Let $u=\sqrt{x}$ to avoid the square root and then $x=u^2$ and the integral becomes $$ I=4\int_0^\infty\frac{\log u}{1+u^4}du. $$ Then use the contour $\Gamma=\{z| r\le|z|\le R\}\setminus [r,R]$ to work on the function $$ f(z)=\frac{\log^2 z}{1+z^4} $$ and you will get the anwser.


3

Riemann-Lebesgue lemma states If $f$ integrable on $[a,b]$, then $$\lim_{p\to +\infty } \int\limits_a^bf(x)e^{ipx}\,dx=0.$$ Using Euler formula ($e^{i\varphi} = \cos\varphi + i\sin\varphi$) one can show that also $$\lim_{p\to +\infty } \int\limits_a^bf(x)\sin px\,dx=0 \quad\text{ and }\quad \lim_{p\to +\infty } \int\limits_a^bf(x)\cos ...


1

In general, if $f(x)$ is integrable in $[a,b]$ then $$\lim_{n\to \infty}\int_a^bf(x)\cos(nx)\,dx=0$$Here, $f(x)=\frac{1}{x^2+4}$ is continuous in $[0,2\pi]$ and so integrable.


0

Actually I was working in @Lucian's hint when I realized there is no need to "forget to derive" the exponential part. Let me do it step by step We derive $I(a)$ w.r.t. $a$ $$I'(a)=\int_{-\infty}^\infty dt\,\left(2 i a e^{-a^2-i b t-t^2} \text{erfi}(a+i t)-\frac{2 i e^{-a^2+(a+i t)^2-i b t-t^2}}{\sqrt{\pi }}\right)\\ =-2a\,I(a)-i \sqrt{2} e^{-\frac{1}{8} ...


2

Since setting $\sin x=u$ gives you $$x=\arcsin u,\ \ \ \mathrm{d}u=\cos x \ \mathrm{d}x,$$ you have $$\begin{align}\int_{0}^{\pi/2}\frac{x\cos x}{1+\sin^2x}\ \mathrm{d}x&=\int_{0}^{1}\frac{\arcsin u}{1+u^2}\ \mathrm{d}u\\&=\int_{0}^{1}(\arctan u)'\arcsin u\ \ \mathrm{d}u\\&=[\arctan u\arcsin u]_{0}^{1}-\int_{0}^{1}\frac{\arctan u}{\sqrt{1-u^2}}\ ...


0

It's a change of variables. If $y=\sin(x)$, $x=\arcsin(y)$ and $dy=\cos(x)dx$, so $x\cos(x)dx=\arcsin(y)dy$. Then you also have to change the bounds of the integral. Knowing that : the derivative of $x\mapsto \arcsin(x)$ is $x\mapsto\frac{1}{1-\sqrt{x^2}}$ the derivative of $x\mapsto \arctan(x)$ is $x\mapsto\frac{1}{1+x^2}$ the second part is simple if ...


1

I would start by trying to make the substitution $r+ir^2 = s$ to get $$ r\,dr = \left(-\frac{i}{2} \pm \frac{i}{2\sqrt{1+4is}}\right) ds, $$ where either $+$ or $-$ is chosen to pick the right branch of the square root. In either case we have $$ r\,dr \sim -\frac{i}{2}\,ds $$ for large $r$ and $s$, so to first order we should have $$ f(x) \approx ...


1

For any $b \in (0,1),$ we have $$\int_0^b \frac{dx}{\sqrt x}+ \int_b^1 nx^n\,dx \le I_n.$$ Let $n\to \infty$ to see $$\int_0^b \frac{dx}{\sqrt x}+ 1 \le \liminf I_n.$$ Now let $b\to 1$ to see $$\int_0^1 \frac{dx}{\sqrt x}+ 1 = 3 \le \liminf I_n.$$ For an estimate from above, note that for $u,v\ge 0, \sqrt {u+v} \le \sqrt u + \sqrt v.$ Thus $$I_n \le ...


0

A picture can help. $W$ is the tetrahedron with the grey plane as its base. The three sets of boundaries: $$\int_{y=0}^{2} \int_{x=0}^{y/2} \int_{z=0}^{2-y} {x\;dy\;dx\;dz} \\ \int_{x=0}^{1} \int_{z=0}^{2-2x} \int_{y=2x}^{2-z} {x\;dx\;dz\;dy} \\ \int_{z=0}^{2} \int_{y=0}^{2-z} \int_{x=0}^{y/2} {x\;dz\;dy\;dx}.$$ The reasoning for the first one, for ...


4

First of all, note that $$\frac{d}{dx}(e^{g(x)}f(x)) = e^{g(x)} \{g'(x)f(x) + f'(x)\}$$ Consequently, $$\int e^{g(x)} \{g'(x)f(x) + f'(x)\} dx = e^{g(x)}f(x) $$ Our aim is to reduce the given integral to the above form. We already have $g(x) = \sec(x)$ Let $$I = \int_{0}^{\frac{\pi}{4}} e^{\sec x} {\frac{\sin (x+\frac{\pi}{4})}{(1-\sin x)(\cos x)}} dx$$ ...


5

Let $I$ be given by $$I=\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4} \arccos\left(\frac{2x}{1+x^2}\right)dx$$ Upon substituting $x\to -x$ we find that $$\begin{align} I&=\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4} \arccos\left(\frac{-2x}{1+x^2}\right)dx\\\\ &=\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4} ...


4

Split the interval of integral into two subintervals $$[0,1]=[0,1-\epsilon]+[1-\epsilon,1] \ \ \ \ \ \ \ \epsilon \rightarrow 0^{+}$$ For very large $n$ in the first interval we have $\frac{1}{x} \gg n^2 x^{2n}$ and in the second $n^2 x^{2n} \gg \frac{1}{x}$. The integral becomes $$I=\int_{0}^{1-\epsilon} \frac{1}{\sqrt{x}} \text{d} x+\int_{1-\epsilon}^{1} n ...


4

Given the two identities: $$ -\log(\sin(x))=\sum_{k=1}^\infty\frac{\cos(2kx)}{k}+\log(2)\tag{1} $$ and $$ -\log(\cos(x))=\sum_{k=1}^\infty(-1)^k\frac{\cos(2kx)}{k}+\log(2)\tag{2} $$ proved here, and the orthogonality relation: $$ \int_{0}^{\pi/2}\cos(2k_1 x)\cos(2k_2 x)\,dx = \frac{\pi}{4}\,\delta(k_1,k_2)\tag{3} $$ it follows that $\int_{0}^{\pi/2}\log(\sin ...


2

Here is an alternate method you could use: Multiply $\displaystyle\int\frac{\sin x\cos x}{\sin x+\cos x}dx$ on the top and bottom by $\cos x-\sin x$ to get $\hspace{.6 in}\displaystyle\int\frac{\cos^2x\sin x}{2\cos^2 x-1}dx-\int\frac{\sin^2x\cos x}{1-2\sin^2 x}dx$. Now substitute $u=\cos x$ in the first integral and $u=\sin x$ in the second integral to ...



Top 50 recent answers are included