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0

From my prospective, the concept of line (curve, path) integral is not much different from the concept of regular one dimensional (definite) Riemann integral. One way to interpret the Riemann integral is to perceive it as the area under the curve. Very often Riemann integral is introduced via Riemann sums, which plays well with its "area-under-the-curve" ...


2

Recall that $$g(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}G(\omega)e^{i\omega t}d\omega$$ Therefore, $$g(0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}G(\omega)e^{i\omega 0}d\omega=\frac{1}{2\pi}\int_{-\infty}^{\infty}G(\omega)d\omega$$ Now substitute $\omega \to -\omega$. The limits flip, but by absorbing the minus sign on the "differential", we have ...


1

Hint, for the first equality, note that $f(x)|_{a} = f(-x)|_{-a}$ and that you're integrating over a symmetric interval. For the second equality, look at the formula for the inverse Fourier transform. What happens to the exponential term in the integrand when $t=0$?


1

By decomposition, then by parts, $$\int\sqrt{1-x^2}dx=\int\frac{dx}{\sqrt{1-x^2}}-\int\frac{x^2\,dx}{\sqrt{1-x^2}}=\arcsin(x)+x\sqrt{1-x^2}-\int\sqrt{1-x^2}\,dx.$$ You can conclude.


-1

I assume that you mean to find a primitive of the function $x \mapsto x(x^{2}+4)^{5}$. But we have $$ \int x (x^{2}+4)^{5}\ dx = \frac{1}{2}\int (x^{2}+4)^{5} d(x^{2}+4) = \frac{(x^{2}+4)^{6}}{12} + constant. $$


2

Notice, $$\int x(x^2+4)^5dx$$ Let $x^2+4=t\implies 2xdx=dt$ $$\int t^5\frac{dt}{2}$$ $$=\frac{1}{2}\int t^5dt$$ $$=\frac{1}{2}\left(\frac{t^6}{6}\right)+c$$ $$=\frac{1}{2}\left(\frac{(x^2+4)^6}{6}\right)+c$$ $$=\frac{(x^2+4)^6}{12}+c$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\int x(x^2+4)^5dx=\frac{(x^2+4)^6}{12}+c}}$$


1

Defining $x=r\cos(\theta)$ we can find that $\text{d}x=-r\sin(\theta)\text{d}\theta$ and $\theta$ goes from $\pi$ to $0$ the integral becomes $$\int_{-r}^{r}\sqrt{r^2-x^2}\text{d}x=-r\int_{\pi}^{0}\sqrt{r^2-r^2\cos^2(\theta)}\sin(\theta)\text{d}\theta=-r^2\int_{\pi}^{0}\sin^2(\theta)\text{d}\theta$$ and you may continue.


2

By recalling $$\sin^2 x + \cos^2 x = 1,$$ we are tempted to make a substitution $x = \sin t$ in $(2)$. Then, the indefinite integral becomes $$\int \sqrt{1-\sin^2t} \cos t \,dt = \int \cos^2 t \, dt.$$ Using integration by parts you will get your desired result. What similar substitution can you do in order to solve $(1)$?


2

There is a standard error bound for TRAP that in principle could be used. If $|f''(x)|\le M$ on our interval, then the absolute value of the error, when we use a subdivision into $n$ equal parts, is $\le \frac{|b-a|^3}{12n^2}M$. However, this bound can be difficult to use, particularly in an automated setting, and often produces error estimates that are ...


0

Let us first carry out the integral given in (x,y). $$ \int_0^1 \left( \int_{1-x}^{4-x} \frac{1}{x+y} dy \right) dx + \int_1^4 \left( \int_{0}^{4-x} \frac{1}{x+y} dy \right) dx = 3. $$ This integral is bounded by the lines x=0, y=0, y=4-x and y=1-x. To be able to find the limits of the integral given in (u,v), we need to find what these lines become in the ...


2

If $I(n, m) =\int_0^1 x^n \operatorname{li}(x^m)\,dx $, setting $y = x^m$, $x = y^{1/m}$ so $dx = \frac1{m}y^{1/m-1} dy $. Therefore $\begin{array}\\ I(n, m) &=\int_0^1 y^{n/m} \operatorname{li}(y)\,\frac1{m}y^{1/m-1}dy\\ &=\frac1{m}\int_0^1 y^{n/m+1/m-1} \operatorname{li}(y)\,dy\\ &=\frac1{m}\int_0^1 y^{(n+1-m)/m} \operatorname{li}(y)\,dy\\ ...


1

The integral converges iff $a = b$. Let's rewrite the expression using big-O notation. So $$\sqrt{x+a}-\sqrt{x} = x^{1/2}(\sqrt{1+a/x}-1) = x^{1/2}(1+a/2x+O(x^{-2})),$$ Therefore $$\sqrt{\sqrt{x+a}-\sqrt{x}} = x^{1/4}(a/4x+O(x^{-2}))$$ and similarly $$\sqrt{\sqrt{x}-\sqrt{x-b}} = x^{1/4}(b/4x+O(x^{-2})).$$ Hence $$ \int\limits^{\infty}_{b} \left ...


6

$$\int_{0}^{1}x^n\text{li}(x^m)\,dx = \frac{1}{m}\int_{0}^{1}z^{(n+1)/m-1}\text{li}(z)\,dz $$ but integration by parts gives: $$\int_{0}^{1}z^{\alpha-1}\text{li}(z)\,dz = \left.\frac{z^\alpha-1}{\alpha}\text{li}(z)\right|_{0}^{1} - \int_{0}^{1}\frac{z^{\alpha}-1}{\alpha\log z}\,dz $$ and the last integral can be computed through the substitution $z=e^{-t}$ ...


0

Let's deal with the first term in the difference. We use $(1+h)^{1/2}= 1 + h/2 +O(h^2)$ throughout. $$(x+a)^{1/2}-x^{1/2} = x^{1/2}[(1+a/x)^{1/2} -1] = x^{1/2}[1+a/(2x) + O(1/x^2) -1] = x^{1/2}[a/(2x) + O(1/x^2)] = (a/2)(1/x^{1/2})[1 + O(1/x)].$$ Now take the square root of that to get $$(a/2)^{1/2}(1/x^{1/4})[1 + O(1/x)].$$ Similarly, the second term in ...


1

I think you really want $$\begin{align}\int_{-\infty}^{\infty} d\tau \left (1-\frac{|\tau|}{2 T}\right ) \theta(2 T-|\tau|) e^{i \omega \tau} &= \int_{-2 T}^{2 T} d\tau \left (1-\frac{|\tau|}{2 T}\right ) e^{i \omega \tau}\\ &= \frac{2 \sin{2 \omega T}}{\omega} - \frac{1}{T} \int_0^{2 T} d\tau \, \tau \, \cos{\omega \tau}\\ &= \frac{2 \sin{2 ...


0

The key to showing that this limit converges to $1$ is to show that the function $A^{1/x}$ decreases quickly towards $1$ as $x$ increases beyond $1$, so that the integral isn't significantly larger than $\int_1^A 1\,dx \sim A$. To this end we will focus on showing that the difference (which is clearly positive) is small enough: $$\int_1^A (A^{1/x} - 1)\,dx ...


1

We have, $$y'+2xy=e^{-x^2}$$ Compare above ODE with Leibniz equation $\frac{dy}{dx}+P(x)y=Q(x)$, we get, $P(x)=2x $ & $Q(x)=e^{-x^2}$ Now, Integration factor $$I.F.=e^{\int P(x)dx}=e^{\int 2xdx}=e^{x^2}$$ Hence, the general solution is given as $$y(I.F.)=\int Q(x)(I.F.) dx+c$$ $$\implies y(e^{x^2})=\int e^{-x^2}(e^{x^2}) dx+c$$ $$\implies ...


0

$${ y }^{ \prime }+2xy={ e }^{ -{ x }^{ 2 } }\\ \left( y{ e }^{ { x }^{ 2 } } \right) ^{ \prime }=1\\ y{ e }^{ { x }^{ 2 } }=x+C\\ y={ e }^{ -{ x }^{ 2 } }\left( x+C \right) $$


2

When the DE is in this form: $y'+p(x)y=q(x)$ the integrating factor is $m(x)=e^{\int{p(x)dx}}$.


0

For first order differential equations which can be solved using the integrating factor method, you take the function in front of the $y$ and integrate it, the raise $e$ to it as an exponent: So your integrating factor is $$e^{\int 2x dx} =e^{x^2} $$.


1

Integration factor in this case is $e^{x^2}$, multiplied both sides by it: $$e^{x^2}y'+2xye^{x^2}=1$$ $$(e^{x^2}y)'=1$$ So: $$e^{x^2}y=x+C$$ Finally: $$y(x)=xe^{-x^2}+Ce^{-x^2}$$


0

The function in question here is the anti-derivative of $f(x)$ starting at $a$. $$F(x) = \int_a^x f(x) dx$$ This function is zero at $F(a)$ and equal to $\int_a^b f(x) dx$ at $F(b)$. At an intermediate value between $a$ and $b$ it will have differing values, but there is a point in between ($k$) where everything balances out: $$\int_a^b f(x) dx = ...


5

The main ingredient here is the integral representation $$\operatorname{Li}_n(z)=\frac{(-1)^{n-1}}{(n-2)!}\int_0^1 \frac{\ln\left(1-zx\right)\ln^{n-2}x\,dx}{x},\tag{$\spadesuit$}$$ valid for $|z|<1,n\in\mathbb{N}_{\ge 2}$. The derivation goes as follows: Rewrite the initial integral as \begin{align*} ...


1

The IVT does not say "there exists a value $L$ between $f(a)$ and $f(b)$ such that $f(c)=L$". It says that if $L$ is any value between $f(a)$ and $f(b)$ then there exists $c\in(a,b)$ such that $f(c)=L$. Note the "any". Consider the interval $[m,M]$. Let $L=\frac{1}{b-a}\int_a^b f(t)\,dt$. Then $L$ is a value between $f(m)$ and $f(M)$ (one of the ...


1

HINT....consider the graph of $f(x)$ which is an isosceles triangle with base from $x=1$ to $x=5$ and height $2$


0

If you whant to find the area of the circle using integrals you can use polar coordinates: for $r=1$ for the area of $x^2+y^2=1$ using double integral $0\leq r\leq 1$ $x:=r\cos\theta,y:=r\sin\theta$ The Jacobian=r $$4\int_{0}^{\pi /2}\int_{r=0}^{r=1}rdrd\theta=4\int_{0}^{\pi /2}\frac{1}{2}d \theta=2\cdot \frac{\pi}{2}=\pi$$


4

We have $$ K(\sqrt{k}) = \int_{0}^{1}\frac{dt}{\sqrt{1-t^2}\sqrt{1-k t^2}}\tag{1} $$ hence: $$ K(\sqrt{k})^2 = \iint_{(0,1)^2}\frac{dt\,ds}{\sqrt{1-t^2}\sqrt{1-s^2}\sqrt{(1-kt^2)(1-ks^2)}}\tag{2}$$ and since: $$\begin{eqnarray*} ...


0

A slight variation on the top-voted solution begins on line $3$: \begin{align} I &= \int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx \\ &= \int_{0}^{\infty} \frac{dx}{1 + e^{bx}} - \int_{0}^{\infty} \frac{dx}{1 + e^{ax}} \\ &= \int_{0}^{\infty} \frac{e^{-bx} dx}{(1 + e^{bx})e^{-bx}} - \int_{0}^{\infty} \frac{e^{-ax}dx}{(1 + ...


0

You problem is essentially equivalent to computing the Laplace transform of $\frac{1}{x^2+(1+i)}$, but the Laplace transform of $\frac{1}{x^2+a^2}$ depends on $\text{Si}(as)$ and $\text{Ci}(as)$, the sine integral and cosine integral, so there is no residue method that is able to provide a nice-looking answer.


9

We have $$\int_{0}^{\infty}\textrm{Li}_{2}\left(e^{-\pi x}\right)\arctan\left(x\right)dx=\sum_{k\geq1}\frac{1}{k^{2}}\int_{0}^{\infty}e^{-\pi kx}\arctan\left(x\right)dx=\frac{1}{\pi}\sum_{k\geq1}\frac{1}{k^{3}}\int_{0}^{\infty}\frac{e^{-\pi kx}}{1+x^{2}}dx $$ and this is the Laplace transform of $\frac{1}{1+x^{2}} $ at $s=\pi k $. This can be calculated ...


0

Hint: as your contour, use an infinite quarter circle that takes up the first quadrant.


0

Since $g$ is differentiable on $-\pi/4 <x<\pi/4$ your integral changes to Riemann integral simply by the following theorem: $$\int_a^bf(x)dg(x)=\int_a^bf(x)g'(x)dx$$ so you will have $\int_{- \pi/4}^{\pi/4} f(x)dg(x)= \int_{- \pi/4}^{\pi/4}\frac{\sin^4x}{\cos^2x}d(1+\sin x)=\int_{- \pi/4}^{\pi/4}\frac{\sin^4x}{\cos^2x}\cos x dx$


4

Using integrating factor gives (I replace the integration variable $u$ with $\tau$) $$ \left[u(t)\exp\int_T^t b(\tau)\,d\tau\right]'=-a(t)\exp\int_T^t b(\tau)\,d\tau\ \Rightarrow \ u(t)=\frac{\int_t^T a(s)\exp\int_T^s b(\tau)\,d\tau\,ds+c}{\exp\int_T^t b(\tau)\,d\tau}. $$ Two exponential functions can be glued to one $$ \frac{\exp\int_T^s ...


9

Since, as it was pointed out, $\cos(2x) = 1-2\sin^2 x$, you can see that $$\sin^2(x) + c = -\frac12 \cos(2x) + (\frac12 + c)$$ This means that both integrals describe the same set of solutions. Remember that anytime you are integrating a function $f$, you get a set of functions for which $F'=f$. The set always contains functions that differ by only a ...


9

Hint: recall that $$ \cos (2x) = 1-2\sin^2 x $$


1

$$\int{\frac{rd r}{ (r-a-b r^3) }} $$ Solve $r-a-b r^3=0$ for $r$. You find the roots, real or complex, $r_1 , r_2 , r_3$ Hense : $$r-a-b r^3=-b(r-r_1)(r-r_2)(r-r_3)$$ $$\int{\frac{rd r}{ (r-a-b r^3) }} = -\frac{1}{b} \int{\frac{rd r}{ (r-r_1)(r-r_2)(r-r_3) }} $$ $$\frac{1}{ (r-r_1)(r-r_2)(r-r_3) } = \frac{r_1}{ (r-r_1)(r_1-r_2)(r_1-r_3) } + ...


3

Hint Considering $$\begin{equation} I=\int{\frac{d r}{1-\frac{a}{r}-b r^2}}=-\int{\frac{r}{b r^3-r +a}}\,dr \end{equation}$$ Let $x,y,z$ be the roots of $b r^3-r +a=0$ so $$b r^3-r -a=b(r-x)(r-y)(r-z)$$ and $$I=-\frac 1 b \int \frac{r}{(r-x)(r-y)(r-z)}\,dr$$ Now, partial fraction decomposition followed by integration of each term would lead to a sum of ...


1

Guessing First I would make a quick guess based on the chart of function space. It groups together Sobolev spaces $W^{s,p}$ with the same value of $\frac n p -s$, because these are related by the embedding theorem. While the inclusion provided by this theorem is strict, the sharpness of the theorem still makes "the spaces with equal $\frac n p -s$ are ...


2

We have the integral of interest $I(x)$ given by $$I(x)=\frac{1}{\pi}\int_{-\infty}^{0} \int_{-\infty}^{\infty}\frac{df(y)}{dy}\frac{x-y}{(x-y)^2+z^2}\,dy\,dz$$ Integrating the inner integral by parts and exploiting that $f$ is a test function gives $$\begin{align} I(x)&=-\frac{1}{\pi}\int_{-\infty}^{0} ...


1

Clearly (A) and (B) are wrong. Noting $$\lim_{a\to\infty} f(a,b) = \lim_{a\to\infty}\int_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx=\int_0^\infty \frac{1}{1+e^{bx}}dx<\infty,$$ we have to choose (E).


1

We know that $\sin \theta=x/a$. From the identity $(\sin \theta)^2+(\cos\theta)^2=1$, we find $\cos\theta=\sqrt{1-(x/a)^2}$ (assuming $\theta$ is in the first quarter). From this you can find $\tan\theta=\sin \theta/\cos \theta$, which agrees with what the book says.


3

Solution without trigonometric functions. Given $$ \int \frac{dx}{ \sqrt{ a^2 - x^2 }^3 }. $$ Write this as $$ \int \frac{dx}{ \sqrt{ a^2 - x^2 }^3 } = \frac{1}{a^2} \int \frac{a^2 - x^2 + x^2}{ \sqrt{ a^2 - x^2 }^3 } dx $$ or $$ \int \frac{dx}{ \sqrt{ a^2 - x^2 }^3 } = \frac{1}{a^2} \int \frac{dx}{ \sqrt{ a^2 - x^2 } } + \color{blue}{ ...


1

Ultimately you want to re-express your answer $\frac{\tan\theta}{a^2}$ in terms of $\sin\theta$ and then convert back using the substitution you already did. Observe that $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and that $\cos\theta = \sqrt{1 - \sin^2\theta}$. Then $$ \tan\theta \;\; =\;\; \frac{\sin\theta}{\cos\theta} \;\; =\;\; ...


1

You can form a right angled triangle from your substitution $x = a\sin\theta$. This'll give the hypotenuse as $a$, opposite side as $x$ and by Pythagoras' theorem, your adjacent side will have length $\sqrt{a^2-x^2}$. $$ \implies \tan \theta \ = \ \dfrac{\text{opposite}}{\text{adjacent}} \ = \ \dfrac{x}{\sqrt{a^2-x^2}}$$ $$ \therefore \ \int ...


0

Just a note. We can decompose the integrand as the following. $$ \frac{x y}{(x+1) (y+1) \log (x y)} = \frac{x}{(x+1)\log (x y)} - \frac{x}{(x+1) (y+1) \log (x y)} $$ Then we notice that $$ \int _0^1\int _0^1 \frac{x}{(x+1)\log (x y)} \ dx \ dy = \int_0^1 \frac{\operatorname{li}(x)}{x+1} \ dx $$ where $\operatorname{li}$ is the logarithmic integral. This ...


14

The integral being considered is, and is evaluated as, the following. \begin{align} I &= \int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx \\ &= \int_{0}^{\infty} \frac{dx}{1 + e^{bx}} - \int_{0}^{\infty} \frac{dx}{1 + e^{ax}} \\ &= \left( \frac{1}{b} - \frac{1}{a} \right) \, \int_{1}^{\infty} \frac{dt}{t(1+t)} \mbox{ where $t = ...


30

One indirect approach: Write $$f(a,b) = \int_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx$$ Then changing variables by $x = ku$, for some positive $k$, $$f(a,b) = k \int_0^\infty \frac{e^{kau}-e^{kbu}}{(1+e^{kau})(1+e^{kbu})}du = k\, f(ka,kb) $$ The only$^*$ answer which obeys the relation $$f(a,b) = k \,f(ka,kb)$$ is Option E. ...


7

Tale $a = 1$ and let $b\to 0^+.$ In the limit you get $$\int_0^\infty\frac{e^x-1}{(1+e^x)2}\,dx.$$ That integral equals $\infty$ because the integrand has a positive limit at $\infty.$ The only answer that fits this phenomenon is E.


0

I think asking if the integral is finite is the wrong question, because it suggests that the integral makes sense as a number in $[-\infty,\infty],$ which it doesn't. Certainly if we put absolute values on, the integral equals $\infty.$ You can see this just by noting $$\int_0^1 \frac{1}{|y-z|}\,dz = \infty$$ for each fixed $y.$ Without the absolute ...



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