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0

Leave out $1/x^2$, to be reinserted later: \begin{align} \int xe^x\,dx &=\sum_{k=0}^{\infty}\frac{x^{k+2}}{(k+2) k!}\\ &=\sum_{k=0}^{\infty}\frac{(k+1)x^{k+2}}{(k+2)!}\\ &=\sum_{k=2}^{\infty}\frac{(k-1)x^k}{k!}\\ &=\sum_{k=2}^{\infty}\frac{kx^k}{k!}-\sum_{k=2}^{\infty}\frac{x^k}{k!}\\ ...


6

We have $$\int_0^t e^{\lambda x}dx = \dfrac{e^{\lambda t}-1}{\lambda}$$ Differentiating with respect to $\lambda$, we obtain $$\int_0^t xe^{\lambda x}dx = \dfrac{\lambda t e^{\lambda t} - e^{\lambda t }+1}{\lambda^2}$$ Set $\lambda = 1$, to obtain $$\int_0^t xe^{x}dx = t e^{t} - e^{t }+1$$ EDIT To complete your approach, note that $$\sum_{k=0}^{\infty} ...


0

This is an alternative inspired by the technique used in this answer. I may or may not delete this answer depending on your answer to my comment in the question. Take the ansatz $\displaystyle \int \cos(x)e^{-x}\mathrm dx=(A\cos(x)+B\sin(x))e^{-x}$ with $A,B$ yet to be determined. Differentiating both sides yields, for all $x\in \mathbb R$, ...


0

Letting $u = e^{-x}$ and $dv = \cos x \, dx$ so that $du = -e^{-x} \, dx$ and $v = \sin x$, we obtain: \begin{align*} \int e^{-x}\cos x \, dx &= (e^{-x})(\sin x) - \int (\sin x)(-e^{-x} \, dx) \\ &= e^{-x}\sin x + \int e^{-x}\sin x \, dx \end{align*} Letting $u = e^{-x}$ and $dv = \sin x \, dx$ so that $du = -e^{-x} \, dx$ and $v = -\cos x$, we ...


0

Using $e^{-x}=dv$ and $cos(x)=u$: $$-e^{-x}cos(x)-\int e^{-x}sin(x)$$ $$\int e^{-x}sin(x)=-e^{-x}sin(x)-\int -e^{-x}cos(x)$$ Be careful with the signs, and you will get an expression like this: $$2\int e^{-x}cos(x)=-e^{-x}cos(x)+e^{-x}sin(x)$$ Divide by two, and that is your solution.


2

First, set $u = \cos x$ and $dv = e^{-x} dx$, so $du = - \sin x \,dx$ and $v = - e^{-x}$. We get $$\int \cos (x) e^{-x} \,dx = (\cos (x))(-e^{-x}) - \int (-\sin (x))(-e^{-x})\, dx$$. Now, set $u = - \sin x$ and $dv = -e^{-x}\,dx$ to get $du = - \cos x\, dx $ and $v = e^{-x}$. This gives us $$\int \cos (x) e^{-x} \,dx = (\cos(x))(-e^{-x}) - (-\sin(x))(e^{-x}) ...


1

$$\int\cos xe^{-x}dx=Re\int e^{ix}e^{-x}dx=Re\int e^{x(i-1)}dx$$


0

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0

The Riemann integral for vector functions with values in a Banach space $X$ is essentially the same as for scalar functions. $$ \int_{a}^{b} F(t)\,dt = \lim_{\|\mathcal{P}\|\rightarrow 0}\sum_{\mathcal{P}} F(t_{j}^{\star})\Delta_{j}t, $$ where $\mathcal{P}$ is a partition with partition points $$ a = t_{0} < t_{1} ...


1

Here is an elementary treatment: First note that $\displaystyle2+\cos x=\frac{3+\tan ^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$. Also note that for $\displaystyle f(x)=\frac{3+\tan ^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$, it holds that $\displaystyle f(x+\frac{\pi}{2})=f(x)$ for $0<x<\pi/2$. Therefore $$\begin{align}\int_0^{2\pi}\frac{1}{2+\cos ...


0

$$I=\int_0^1\frac{\log(u)-\log(1-u)}{u}\,dx = \int^1_0\frac{\log(u)}{u}\,du+\zeta(2)$$


1

A place to begin: I think the search for a friendlier closed form could benefit from breaking the series up into more manageable components. Start by splitting the sum into the sums of odd and even terms: $$\begin{align} S &=\sum_{p=1}^{\infty}\frac{\psi{\left(\frac{p+1}{2}\right)}}{\binom{2p}{p}}\\ ...


0

You can rewrite your integral as $\displaystyle\int_1^4\frac{\sqrt{200y-28\sqrt{y}+1}}{2\sqrt{y}}dy$, so $u=\sqrt{y}$, $du=\frac{1}{2\sqrt{y}}dy$ gives $\displaystyle\int_1^2\sqrt{200u^2-28u+1}du=$ $10\sqrt{2}\int_1^2\sqrt{\left(u-\frac{7}{100}\right)^2+\left(\frac{1}{100}\right)^2}du.$ Now let $u-\frac{7}{100}=\frac{1}{100}\tan\theta$ to get ...


0

This is for Differentiation-under-Integral-sign lovers Consider following integral $$I(\alpha )=\int_0^\infty \frac{\sin (\alpha x)}{e^{x}} \,\mathrm dx=\int_0^\infty e^{-x}\sin (\alpha x) \,\mathrm dx$$ We have $I(0)=0$ $$I'(\alpha )=\int_0^\infty xe^{-x}\cos (\alpha x) \,\mathrm dx=\int_0^\infty xe^{x}\cos (\alpha x) \,\mathrm dx$$ Now consider ...


3

We have $$\eqalign{I_m(a,b)= \int_0^\pi\frac{\cos(mx)}{a^2-2ab\cos x+b^2}dx &=\left\{\matrix{\frac{\pi}{a^2-b^2}&\hbox{if}&m=0\\\cr \frac{\pi}{a^2-b^2}\left(\frac{b}{a}\right)^m&\hbox{if}&m\ne0 }\right.}$$ Proof can be seen here. Hence \begin{align} \int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\, dx&=\frac{1}{2}\int_0^{\pi} ...


4

Such integral is related with the topological degree of a closed curve. We have: $$ I = \frac{1}{2}\int_{0}^{2\pi}\frac{\sin^2 x}{\|a-be^{ix}\|^2}\,dx = -\frac{i}{2}\oint_{\|z\|=1}\frac{\left(\frac{z-z^{-1}}{2}\right)^2}{(a-bz)(az-b)}\,dz$$ hence we just have to consider the residues of the last integrand function in $z=0$ and $z=\frac{b}{a}$ (since ...


2

Let $x_n$, $n=1,2,...$ be such that $\lambda+\tan(x_n\lambda)=0$, with $x_{n+1}>x_n>0$. Let $x_0=0$. Now note that $\int_0^{\infty}\sin(\lambda x)e^{-x}dx=\sum_{n=0}^{\infty}\int_{x_n}^{x_{n+1}}\sin(\lambda x)e^{-x}dx$. Using integration-by-part twice we obtain $\int_0^{x_{1}}\sin(\lambda x)e^{-x}dx=\frac{\lambda-\lambda\cos(\lambda x_1)-\sin(\lambda ...


0

Ok, I think I got it now... Take a slight variation of the famous example: $$F:(0,1]\to\ell(0,1]:t\mapsto\chi_t$$ This is measurable as: $$\forall\varphi\in\ell(0,1]:\quad\|\varphi-\chi_s\|<\varepsilon<1\implies\varphi(s)>1-\varepsilon>0$$ so that small balls are measurable: ...


2

Here I prove $\displaystyle\int_0^{2\pi}\frac{\sin^2(2n+1)x}{\sin^2x}dx=2\pi(2n+1).$ Since $\displaystyle\sum_{j=1}^n \cos 2jx = -\frac{1}{2}+\frac{\sin(2n+1)x}{2\sin(x)}$, we have that $\displaystyle1+\sum_{j=1}^n \cos 2jx = \frac{\sin(2n+1)x}{\sin(x)}$. Now take both sides to the power of $2$ and integrate term-by-term. ...


3

$a>0$ \begin{align*} I&=\int_0^{+\infty}e^{-ax}\sin bx\mathrm{d}x=\int_0^{+\infty}e^{-ax}\mathrm{d}\left(-\frac{\cos bx}{b}\right)\\ &=e^{-ax}\left(-\frac{\cos bx}{b}\right)\mid_0^{+\infty}-\frac{a}{b}\int_0^{+\infty}e^{-ax}\cos bx\mathrm{d}x\\ &=\frac{1}{b}-\frac{a}{b}\int_0^{+\infty}e^{-ax}\mathrm{d}\left(\frac{\sin bx}{b}\right)\\ ...


0

Using the first suggestion, i.e. $y=x^2$ to change the variable to $x$ we have $\int_1^4\sqrt{50+\frac{1}{4y}-\frac{7}{\sqrt{y}}}dy=\int_1^2\sqrt{\frac{1+(7-100x)^2}{50}}dx$. Now choose $u=7-100x$ so that you have a "standard" integral involving $\sqrt{1+u^2}$: use $\arcsinh$ to do a new change of variable. You will then obtain ...


1

No. Consider the sequence of functions given by $$f_n(x)=\begin{cases} \frac1n&\text{if}\ 0<x<n^2\\ 0&\text{otherwise.} \end{cases}$$ Then for any $x>0$, $|f_n(x)|\leq\frac1n$ so $f_n\to f:=0$ uniformly. However, $$\int_0^\infty f_n(x)\ dx=n\to\infty$$ whereas $$\int_0^\infty f(x)\ dx=0.$$ This counterexample still has $f$ integrable but you ...


1

We have$$\int_0^1\frac{\ln(x)}{x^2-1}\,\mathrm dx=\int_0^1\frac{\ln(1-x)}{(1-x)^2-1}\,\mathrm dx=\int_0^1\frac{\ln(1-x)}{x(x-2)}\,\mathrm dx$$ We will generalize by introducing parameter $\alpha$ such that $$I(\alpha )=\int_0^1\frac{\ln(1-\alpha x )}{x(x-2)}\,\mathrm dx$$ And we have $I(0)=0$ Then $$I'(\alpha )=-\int_0^1\frac{1}{(1-\alpha x)(x-2)}\,\mathrm ...


0

Doesn't any non-negative continuous function on $[a,b]$ satisfy the conditions given in the last two bullets? Also, using $\varepsilon$ for something a priori big is touche. I also do not think that you can construct such a counting function using integral (w.r.t. Lebesgue measure): currently you are trying to approximate the length of $\{f = 0\}$, not the ...


3

Consider $$ \int_1^2 \frac{\log (x-1)}{x(x-1)}dx = \int_0^1 \frac{\log u}{(1+u)u}du \sim \int_0^1 \frac{\log u}{u}du, $$ which is divergent, since $$\frac{d}{du} \frac12 \log^2 u = \frac{\log u}{u}.$$ Hence the whole integral is divergent. By the way, the integral "at infinity" converges.


0

Hints: Let $t=\log(x-1)$. Then $x=1+e^t$


1

Considering the problem of the antiderivtive , change variable $x=a+iy$ $$\int(x-a)^ke^{-i\omega x}dx=i^{k+1}e^{-ia\omega}\int y^k e^{\omega y}dy=-y^{k+1}i^{k+1}e^{-ia\omega} E_{-k}(-y \omega )$$ where appears the elliptic integral.


5

$$ \int \sqrt{1 + {1 \over t^2} + {2 \over t}} dt=\int {1\over t}\sqrt{t^2+2t+1} \,\,\,dt$$


4

Hints: $$ 1 + {1 \over t^2} + {2 \over t}=\frac{(t+1)^2}{t^2}$$


5

You may just integrate by parts $$\frac{1}{2\pi}\int_{-\pi}^{\pi}(x-a)^k e^{-i\omega x}dx =\left.-(x-a)^k\frac{e^{-i \omega x}}{2\pi i\omega }\right|_{-\pi}^{\pi} +\frac{k}{2\pi i\omega}\int_{-\pi}^{\pi}(x-a)^{k-1} e^{-i\omega x}dx$$ thus, with an obvious notation, you have $$ I_k(a,\omega) =\frac{(-1)^k }{2\pi i\omega }\left((a+\pi)^k e^{i \pi \omega }- ...


0

Take $\ \ \ f(x) = \int_1^x \sqrt{t^3-1} \ dt $ Since integrand is continous, you can write $\ \ f'(x) = \sqrt{x^3-1} $ Now apply the arc length formula.


3

Hint: You do not need to evaluate the integral. Recall that in the arc length formula, the function's derivative appears, then apply the fundamental theorem of calculus.


0

Think about the definition of the derivative: $$\frac{\partial g}{\partial r}(r,\theta) = \lim_{\Delta r\to0}\frac{g(r+\Delta r,\theta)-g(r,\theta)}{\Delta r} = \lim_{\Delta r\to0}\frac{\theta - \theta}{\Delta r} = \lim_{\Delta r\to0}\frac{0}{\Delta r}= 0$$ The same is true any time $g(r,\theta)$ only depends on $\theta$.


2

First, note that $f(x)=1/p(x)$ for $p(x)=(x^2+1)(x^2+2x+2)^2(x^2-2x+2)^2$, so that $p$ has roots (and $f$ has poles) at $\pm i$ of order $1$ and at $\pm i\pm 1$ each of order $2$. The key point to note is that the integral in question is part of a path that would include the upper half plane (UHP), by including for example the semicircular path ...


0

Gauges are directed: $$\delta\wedge\delta'(\omega):=\delta(\omega)\cap\delta'(\omega)$$ and every gauge admits a partition: $$\Omega=\bigcup_{n=1}^N\delta(a_n)=:\bigcup_{n=1}^NA_n:\quad\mathcal{P}_\delta:=\{A_1',\ldots,A_N'\}\quad(A_n':=A_n\setminus\cup_{m=1}^{n-1}A_m)$$ (Note that compactness was necessary!) That is gauges induce a net of partial sums: ...


2

We split up $\partial D$ into $C_1, C_2, C_3$, where: \begin{align*} C_1:\qquad (f, g) &= (2(1 - t), 4(1 - t)) &\text{where } t \in [0, 1] \\ C_2:\qquad (f, g) &= (2t, 0) &\text{where } t \in [0, 1] \\ C_3:\qquad (f, g) &= (2, 4t) &\text{where } t \in [0, 1] \\ \end{align*} Note that: \begin{align*} \int_{C_1} f \, dx + g \, dy &= ...


1

I believe you mean that $$F(x)=\int_{0}^{x}f(t)dt.$$ Am I correct? If so, here's a hint: Fundamental Theorem of Calculus.


1

You can usually reduce to known form; specifically, if $f \sim g$, then $\int f$ exists if and only if $\int g$ does. For example, since you know that $$\int_1^\infty \frac{1}{x^\alpha}$$ converges if and only if $\alpha > 1$, then $\int_{1}^\infty \frac{x}{x^4 + 19}$ converges too because it is asymptotic to $\frac{1}{x^3}$ which converges. There ...


3

$||f||_{\infty}$ acts as an essential bound on the absolute value of $f$, exceeded, if at all, on a set of measure zero. so $g = \frac{f}{||f||_{\infty}}$ may be treated as a function whose absolute value does not exceed unity, and the proposition to be proved now becomes: $$ ||g||_p^p \le ||g||_2^2 $$ (interpreting the RHS as $\infty$ if $f$ is not square ...


0

Hint: let $M$ be the bound of $|f|$. You get $$\left|\int_a^b F(x) dx\right|= \left|\int_a^b \int_a^x f(y)dy dx \right|\leq \int_a^b\int_a^x |f(y)|dy\leq \int_a^b M(b-a) dx=M(b-a)^2$$


4

Assuming you can't use complex exponentials, I would use integration by parts. You will have to do it twice. For the first application of integration by parts, take $u = \sin x$ and $dv = e^{-x} dx$. Note that, after both applications of integration by parts, if this is done correctly, you should get an equation of the form $I = \text{ some stuff } - cI$, ...


2

There are probably other ways to do this, but using complex exponentials is just too nice for me to pass up: $$\int_0^\infty \dfrac{\sin (\lambda x)}{e^x}dx=\frac1{2i}\int_0^\infty e^{-x+i\lambda x}-e^{-x-i\lambda x} ~dx$$ Can you proceed from here?


1

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1

You're almost there. Let $\displaystyle h(x) = g(x) \int_a^b f(t) \ dt$. As $g$ is continuous, $h$ is also continuous. Without loss of generality, let $x_1 < x_2$. By what you've shown above, $\int_a^b f(x)g(x) \ dx$ is a number between $h(x_1)$ and $h(x_2)$. As $h$ is continuous, by the IVP there must be a value $x_0 \in (x_1, x_2)$ such that ...


3

The ML inequality is (essentially) a real inequality. It holds for all (sufficiently regular, e.g. piecewise differentiable) curves and [again, sufficiently regular so that the integral is defined] functions [or vector fields] in any $\mathbb{R}^n$, $\mathbb{C}^n$ or more generally, Riemannian manifold. Its proof uses a) the inequality for real intervals ...


1

Given only the information stated, the only reason we can assume that 800-p isn't negative is that we are taking its logarithm. This makes sense in terms of the model; 800 is functioning as a population ceiling, as the rate of increase slows down as $p$ approaches 800.


1

I think the answer is likely to be due to the fact that $800-p$ cannot be negative. I don't quite know what 800 represents but it seems likely that $800-p$ cant be a negative number otherwise it doesn't work out. Hope this makes sense. Would have added this as a comment but I cant presently.


0

We can use the following Taylor expansion of $\ln(1+x)$ to evaluate \begin{eqnarray} \ln(1+x)&=&\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}x^n. \end{eqnarray} Then \begin{eqnarray} \int_0^{\pi/2}\frac{1+\sin\phi}{\sin\phi}d\phi&=&\int_0^{\pi/2}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\sin^{n-1}\phi d\phi\\ ...


1

A possible contour is the semi circle in the upper half plane with a semi circle around the origin. Let $\int_{\Gamma}$ be large semi circle and $\int_{\gamma}$ be the small semi circle. Let $R$ be the radius of the $\Gamma$ and $\epsilon$ the radius of $\gamma$. Now as $R\to\infty$, $\int_{\Gamma}\to 0$ and similarly as $\epsilon\to 0$, $\int_{\gamma}\to 0$ ...



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