Integral Definition and 1000 Threads

## \int_0 ^ {2 \pi} \frac {dx} {3 + cos (x)} ## las únicas formas que probé fueron, multiplicar por ## \frac{3-cos (x)}{3-cos (x)} ## pero no me gusta esto porque obtengo una expresión muy complicada. También recurrí a la sustitución ## t = tan (\frac {x} {2}) ## que me gusta bastante, pero...

Sorry - I wish I had some way of writing equations in this forum so the "relevant equations" section is easier to read. The answer to the first part is (a) so the rest follows from using the electric field given in B. If anyone is interested this question comes from Griffith's 3rd edition...

Problem: Evaluate $$ \int_{0}^\infty \frac{e^{3x} - e^x}{x(e^x + 1)(e^{3x} + 1)}\ dx $$ Attempt. I substituted $y=e^x$, thus $dx = dy/y$, which turns the above integral to $$ \int_{1}^\infty \frac{y^2 - 1}{(\log y)(y+1)(y^3+1)}\ dy = \int_{1}^\infty \frac{y-1}{(\log y) (y^3+1)} \ dy $$ I am...

i) I approximate the solenoid as a cylinder with height L and radius R. I am not sure how I am supposed to place the solenoid in the coordinate system but I think it must be like this, right? The surface occupied by the cylinder can be described by all vectors ##\vec x =(x,y,z)## so that...

Naturally there are vector equivalents of the Kirchhoff Integral. Taken from Jackson (10.113) ##\vec{E} \left( \vec{r} \right) = \frac{ie^{ikr}}{r} a^2 E_0 \cos \alpha \left( \vec{k} \times \vec{\epsilon}_2 \right) \frac{J_1 \left( \sin \theta \right)}{\sin \theta}##Where I just let ##\alpha =...

Hi, I was just working on a homework problem where the first part is about proving some formula related to Stokes' Theorem. If we have a vector \vec a = U \vec b , where \vec b is a constant vector, then we can get from Stokes' theorem to the following: \iint_S U \vec{dS} = \iiint_V \nabla...

I am trying to evaluate an integral with unknown variables ##a, b, c## in Mathematica, but I am not sure why it takes so long for it to give an output, so I just decided to cancel the running. The integral is given by, ##\int_0^1 dy \frac{ y^2 (1 - b^3 y^3)^{1/2} }{ (1 - a^4 c^2 y^4)^{1/2} }##

Dear Everyone, I am wondering how to use the integral formula for a holomorphic function at all points except a point that does not exist in function's analyticity. For instance, Let f be defined as $$f(z)=\frac{z}{e^z-i}$$. F is holomorphic everywhere except for $$z_n=i\pi/2+2ni\pi$$ for all...

I know the value of this integral is equal to 0, but I would like to see if there is any tricks to spot this answer using symmetries or even odd propreties? Thanks in advance

Dear Everyone, I am wondering how to use the integral formula for a holomorphic function at all points except a point that does not exist in function's analyticity. For instance, Let $f$ be defined as $$f(z)=\frac{z}{e^z-i}$$. $f$ is holomorphic everywhere except for $z_n=i\pi/2+2ni\pi$ for...

calculate the double integral over the region of integration is x^2 + y^2 ≤ 4; x^2 + (y/4)^2 ≥ 1 the integrals have been made over two regions my problem is that when I go to the polar coordinate for the ellipsis and use the jacobian i got 2 instead of 8 ( the following is the professor...

It is from David Tong's note for QFT. The equation states ##\left . \int d^4 p \cdot \delta \left ( p^2_0 -{\vec p}^2 -m^2 \right ) \right | _{p_0>0} =\left . \int \frac {d^3 p} {2 p_0} \right |_{p_0=E_{\vec p}}## where ##p## is a 4-vector ##p=\left ( p_0, \vec p \right )##. In my...

I already have the solution in which the region of integration has been divided into two regions but I was wondering if I can only use one region considering the polar coordinate system) the disk equation for me is r=2cos(θ) and the theta goes from 0 to (pi/4) 0<r<2cos(θ) and the 0 <θ<pi/4...

I'm studying the proof of this theorem (Zorich, Mathematical Analysis II, 1st ed., pag.136): which as the main idea uses the fact that a diffeomorphism between two open sets can always be locally decomposed in a composition of elementary ones. As a remark, an elementary diffeomorphism...

Hello (A continued best wishes to all, in these challenging times and a repeated 'thank you' for this site.) OK, I have read that Newton figured out that differentiation and integration are opposites of each other. (This is not the core of my question, so if that is wrong, please let it go.)...

So we have ##x=\beta(1/2 mv^2-\mu)##, i.e ##\sqrt{2(x/\beta+\mu)/m}=v##. ##dv= \sqrt{2/m}dx/\sqrt{2(x/\beta+\mu)/m}##. So should I get in the second integral ##(x+\beta \mu)^{1/2}##, since we have: $$v^2 dv = (2(x/\beta+\mu)/m)\sqrt{2/m} dx/\sqrt{2(x/\beta+\mu)/m}$$ So shouldn't it be a power...

if ## \gamma (t):= i+3e^{2it } , t \in \left[0,4\pi \right] , then \int_0^{4\pi} \frac {dz} {z} ## in order to solve such integral i substitute z with ##\gamma(t)## and i multiply by ##\gamma'(t)## that is: ##\int_0^{4 \pi} \frac {6e^{2it}}{i+3e^{2it}}dt=\left.log(i+3e^{2it}) \right|_0^{4...

Prove that if $[a,\,b]\subset \left(0,\,\dfrac{\pi}{2}\right)$, $\displaystyle \int_a^b \sin x\,dx>\sqrt{b^2+1}-\sqrt{1^2+1}$.

Evaluate $\displaystyle\int\limits_0^{\infty} \dfrac{x^2+2}{x^6+1} \, dx$.

Several weeks ago I had considered the question as to how one can start from the Schroedinger Equation, and after several transformations, derive F=ma as a limiting case. At some point in my investigations of this derivation, it occurred to me that this is simply too much work. While in...

I am not clearly understand what the question requests for, is it okay to continue doing like this ? Kindly advise, thanks

I got stuck here, how to integrate e^(y^2), I searched but it's something like error function

I am trying to solve it using cylindrical coordinates, but I am not sure whether the my description of region E is correct, whether is the value of r is 2 to 4, or have to evaluate the volume 2 times ( r from 0 to 4 minus r from 0 to 2), and whether is okay to take z from r^2/2 to 8

How to solve the following integral (in Maple notation): Int(y**k*exp(-u[0]*exp(-y)/a[0]-u[1]*exp(y)/a[1]),y=-infinity..infinity) with 0<a[0], 0<u[0], 0<a[1], 0<u[1]?

I have the following integral (in Maple notation): Int(exp(c[0]*ln(y)/a[0]+c[1]*M*ln(M-y)/a[1]), y = 0 .. M); with (in Maple notation): 0<a[0], 0<a[1], 0<c[0], 0<c[1], 0<y, y<M, 0<M. What is the solution of this integral? I suspect that the solution has something to do with a beta distribution.

[Ref. 'Core Principles of Special and General Relativity by Luscombe] Let ##\gamma:\mathbb{R}\supset I\to M## be a curve that we'll parameterize using ##t##, i.e. ##\gamma(t)\in M##. It's stated that: Immediately after there's an example: if ##X=x\partial_x+y\partial_y##, then ##dx/dt=x## and...

I've tried with this work in attachment. i&m not sure of my answer is correct.

If $$F(x)=\int_{a}^{b}f(x)dx$$ implies $$F'(x)=\int_{a}^{b}f'(x)dx$$?

Because the limit of the integral is multi-variable, which is not explained at the ML Boas's example, I tried to start from the basic. First, I use: $$\frac {dF}{dx}=f(x) \Rightarrow \int_a^b f(t) dt = F(b) - F(a)$$. In my case now: $$\int_{u(x)}^{v(x,y)} f(t) dt = F(v(x,y)) - F(u(x))$$ So...

This is my method, could you help me to continue?

Integrals are defined with the help of upper and lower sums, and more number of points in a partition of a given interval (on which we are integrating) ensure a lower upper sum and a higher lower sum. Keeping in mind these two things, I find the following definition easy to digest A function...

Not homework, just trying to understand a statement in the book. On page 158 in Fisher, the following statement is made: In these applications of the Residue Theorem, we often need to estimate the magnitude of the line integral of e^{iz} over the semicircle = Re^{i\theta}, \; 0 \le \theta \le...

I have a question like this; I selected lambda as 4 (I actually don't know what it must be) and try to make clear to myself like these limits (1,2) and (2,4) is x and y locations I think :) If I find an answer for part one of the integral following, I would apply this on another: My...

The vector field F which is given by $$\mathbf{F} = \dfrac{(x, y)} {\sqrt {1-x^2-y^2}}$$ And the line integral $$ \int_{C} F \cdot dr $$C is the path of $$\dfrac{\ (\cos (t), \sin (t))}{ 1+ e^t}$$ , and $$0 ≤ t < \infty $$ How do I calculate this? Anyone got a tip/hint? many thanks

I was trying the evaluate the integral $$\int_{25\pi/4}^{53\pi/4}\frac{1}{(1+2^{\sin x})(1+2^{\cos x})} dx$$ from I have since manipulated this integral into $$\int_{\pi/4}^{5\pi/4}\frac{7}{(1+2^{\sin x})(1+2^{\cos x})} dx$$ Any help on how to proceed further would be appreciated. The value...

Hi forgive my very ignorant question. How frequently does 'Path Integral', curved shot and normal shot happen out of say 100 shots with a photon?

I want to find the analytical solution to the integral given below. \int_{-\infty}^{\infty} \frac{ sinc^2(\frac{k_yb}{2})}{\sqrt{k^2 - k_x^2 - k_y^2}}dk_y In other words, \int_{-\infty}^{\infty} \frac{ \sin^2(\frac{k_yb}{2})}{(\frac{k_yb}{2})^2\sqrt{k^2 - k_x^2 - k_y^2}}dk_y Can this be...

$\dfrac{dy}{dx}=\dfrac{4y-3x}{2x-y}$ OK I assume u subst so we can separate $$\dfrac{dy}{dx}= \dfrac{y/x-3}{2-y/x} $$

The integral looks like Y_{mut, mn} = -j^{m+n}nm \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{2 ab (k^2 - k_x^2) \sin^2(\frac{k_yb}{2}) \cos^2(\frac{k_xa}{2})}{\omega \mu k_z (\frac{k_yb}{2})^2 [(n\pi)^2 - (k_xa)^2][(m\pi)^2 - (k_xa)^2]} dk_x dk_y Here, k_z = -1j \sqrt{(-(k_0^2 -...

I am not sure about finding the limit of the integral when it comes to finding the CDF using the distribution function technique. I know that support of y is 0 ≤y<4, and it is not a one-to-one transformation. Now, I am confused with part b), finding the limits when calculating the cdf of Y...

Hello, everyone. I know that it is feasible to exchange the order of one variation sign and one integral sign. But there gives a proof of this in one book. I wonder about a step in it. As below marked in the red rectangle: How can ##\delta y## and ##\delta y^\prime## be moved into the integral...

I tried integration by parts with both ##u = x^2, dv = J_0 dx## and ##u = J_0, du = -J_1 dx, dv = x^2 dx.## But neither gets me in a very good place at all. With the first, I begin to get integrals within integrals, and with the second my powers of ##x## in the integral would keep growing...

One of the maths groups I'm apart of on Facebook posts (usually) daily maths challenges. Typically they act as small brain teaser for when I wake up and I can solve them without much trouble. However, today's was more challenging: (Note: blue indicates a variable and red indicates a constant)...

I(k_x, k_y) = \int_{0}^{R} \int_{0}^{2\pi} J_{m-1}(\alpha \rho) \sin((m + 1) \phi) e^{j\rho(k_x \cos\phi + k_y \sin\phi)} \rho d\rho d\phi Is there any way to do it? J is the Bessel function of the first kind. I thought of partially doing only the phi integral as \int_{0}^{2\pi} \sin((m + 1)...

Hi all, I'm finding it difficult to start this line integral problem. I have watched a lot of videos regarding line integrals but none have 3 line segments in 3D. If someone can please point me in the right direction, it would help a lot. I've put down the following in my workings: C1...

Summary:: I'm solving an exercise. I have the following center of gravity problem: Having the function Y(x)=96,4*x(100-x) cm, where X is the horizontal axis and Y is the vertical axis, ranged between the interval (0, 93,7) cm. Determine: a) Area bounded by this function, axis X and the line...

What I've done so far: From the problem we know that the curve c is a half-circle with radius 1 with its center at (x,y) = (0, 1). We can rewrite x = r cos t and y = 1 + r sin t, where r = 1 and 0<t<pi. z stays the same, so z=z. We can then write l(t) = [x(t), y(t), z ] and solve for dl/dt...

I know how to find integral solutions of linear equations like x+y=C or x+y+z=C where C is a constant. But I don't have any idea how to solve these type of questions.I am only able to predict that both x and y will be greater than 243554.Please help.

the integral is: and according to mathematica, it should evaluate to be: . So it looks like some sort of Gaussian integral, but I'm not sure how to get there. I tried turning the cos function into an exponential as well: however, I don't think this helps the issue much.