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0

First off, your result should have some absolute value bars around it: $$y = \ln|x+1| +C$$ Other than that you are correct. Next step is to utilize the fact that $y(0)=0$. What this is saying is that at $x = 0$, $y = \ln|x+1| +C = 0$. $$y = \ln|0+1| +C$$ $$ = \ln{1} +C$$ $$ = 0+C = 0$$ Thus $C = 0$. Furthermore since we now know $y = \ln|x+1|$, ...


0

I'm assuming that you define $a(k)=a_{[k]}$, where $[k]$ is the floor function. You could similarly define it with a ceiling function and the answer should remain the same. First review the statement and proof of the integral test. The gamma function is strictly increasing in $k$, and lets assume $a_kx^k/k!$ is decreasing (it has to go to zero regardless), ...


1

Let $x=u^2$. Then $dx=2udu$ and then the integral becomes $$\int\limits_{0}^{\infty}e^{-u^2}u(2udu)=\int\limits_{0}^{\infty}2e^{-u^2}u^2du$$ Then you can use integration by parts to get $$u^2|_0^\infty \int\limits_{0}^{\infty}e^{-u^2}du-\int\limits_0^\infty 4e^{-u^2}udu$$ and you will plug in the values and use integration by parts again. I think you can ...


1

First do an integration by parts to get $\int_0^\infty e^{-x}\frac{\mathrm{d}x}{2\sqrt{x}}$, then try the substitution $u = \sqrt{x}$. It would be easier this way.


0

Edit: The function $s$ is defined from $\gamma_\pi^*$ which is the free energy (see page 15 http://arxiv.org/pdf/1308.5376v1.pdf). They are all functions of $t$ the time variable. so when one reads $$s (\mu(t+1)-\mu(t)) $$ this is meant as a scalar multiplications and not a function evaluation As one begins with a particular portfolio position one desires ...


0

I think you can use the Abel's summation formula to get the equality between the two functions as it allows us to transform a series to integral.


2

Another partial answer: $$ \int_{-\infty}^\infty \erf(x) \dfrac{d^n}{dx^n} \erf(x)\; dx = \cases{0 & $n$ odd\cr (-1)^{n/2} 2^{(n+1)/2}\Gamma(\frac{n-1}{2})/\pi & $n$ even }$$


1

Partial answer: $$\int _{-\infty }^{\infty }\! \left( {{\rm erf}\left(x\right)} \right) ^{m}{\frac {d}{dx}}{{\rm erf}\left(x\right)}{dx}= \cases{0&$m= {\it odd}$\cr 2\, \left(m+1 \right) ^{-1}&$m= even$\cr} $$ $$\int _{-\infty }^{\infty }\! \left( {{\rm erf}\left(x\right)} \right) ^{2\,n}{\frac {d^{2}}{d{x}^{2}}}{{\rm erf}\left(x\right)}{dx}=0 ...


5

Note that $\sqrt{x}+1 \ge \sqrt{x}$ implies that $\frac{1}{\sqrt{x}+1}\le\frac{1}{\sqrt{x}}$. Thus, we have for $x>0$ $$\left|\frac{1}{x(\sqrt{x}+1)}\right|\le\frac{1}{x^{3/2}}$$ Since $$\int_1^{\infty}x^{-3/2}dx=2$$ then by the comparison test, we conclude that $$\int_1^{\infty}\frac{dx}{x(x^{1/2}+1)}\,\,\text{converges}$$


1

Your integral is convergent since $\frac{1}{x(\sqrt{x}+1)}$ is equivalent to $\frac{1}{x^{3/2}}$ when $x$ is large and $3/2>1$ so by the Riemann criterion we have $\int_{1}^{+\infty} \frac{1}{x^{3/2}}$ is convergent so we get the convergence of the integral $\int_{1}^{+\infty}\frac{1}{x(\sqrt{x}+1)}.$


3

The statement after "We all know that" is completely false. $\sin(cx)$ for any $c$ is an odd function, so any integral on an interval symmetric around $0$ will give 0. The correct result is: $$\int_{-a}^a\sin(cx)dx=\int_{0}^a\sin(cx)dx+\int_{-a}^0\sin(cx)dx,$$ and now do a change of variables on the second integral $u=-x$ and use the fact that ...


0

@RobertIsrael provided the way forward here. I thought that it would be instructive to show a simple "trick" that works well in other situations. To that end, we assume that $g(x)$ satisfies the Holder condition $$|g(x)-g(y)|\le C|x-y|^{\alpha} \tag 1$$ with Holder exponent $\alpha \ge 1/2$. Then we can write $$\begin{align} ...


2

If there is a blowup somewhere, then $f$ may be integrable when $f^2$ is not, because squaring "makes the blowup worse". For instance $f(x)=1/\sqrt{x}$ on $(0,1]$ is improperly integrable but its square is not. If there is cancellation between positive and negative pieces, then $f$ may be integrable when $f^+$ is not, because the cancellation was preventing ...


1

The Fourier transform maps the function s(t) in the time domain to a function S(f) in the frequency domain. We often call the function s(t) a 'waveform' and usually is some periodic function representing a wave (sound, radio, etc). What this means is that the function s(t) is broken down into a number (infinite possibly) of separate simple waves which have ...


2

For example, $\displaystyle\int_0^\infty \dfrac{\sin(x)}{\sqrt{x}}\; dx$ exists but $\displaystyle\int_0^\infty \left(\dfrac{\sin(x)}{\sqrt{x}}\right)^2\; dx$ and $\displaystyle\int_0^\infty \left(\dfrac{\sin(x)}{\sqrt{x}}\right)^+\; dx$ do not.


1

Observe that $I_0(a)=1-a$. For $n>0$ we can use integration by parts to find, $$I_n(a)=\int_a^1\ln^n(x)dx=-a\ln^n(a)-n\int_a^1\ln^{n-1}(x)dx=-a\ln^n(a)-n I_{n-1}(a).$$ By L'Hopital's rule, $$\lim_{a\to 0}-a\ln^n(a)=\lim_{a\to 0}\frac{-\ln^n(a)}{1/a}=\lim_{a\to 0}\frac{n\ln^{n-1}(a)}{1/a}=...=(-1)^nn!\lim_{a\to 0}\frac{1}{1/a}=0.$$ Thus, $$\lim_{a\to 0} ...


0

The $2 \pi$ is not essential; that just converts between angular frequency and ordinary frequency. There are definitions of the Fourier transform which don't use this factor at all. These are common in PDE theory. The gist of what's going on is very much like the situation in Fourier series: if you have a complex-valued square integrable function $f$ and a ...


0

Using $x=e^t$ and $s=-t$, we get $$ \begin{align} \int_a^1\log(x)^n\,\mathrm{d}x &=\int_{\log(a)}^0t^ne^t\,\mathrm{d}t\\ &=(-1)^n\int_0^{-\log(a)}u^ne^{-u}\,\mathrm{d}u \end{align} $$ Taking the limit as $a\to0$ gives $$ \begin{align} (-1)^n\int_0^\infty u^ne^{-u}\,\mathrm{d}u \end{align} $$ Now look up the Gamma Function.


0

Mathematica 8 says it is $$ \int_{-L}^{L}\exp\!\left(-\frac{\left(x-x_{0}\right)^{2}}{2\sigma^{2}}\right)\cos\!\left(x\right)\mathrm{d}x = \frac{1}{2}\sqrt{\frac{\pi}{2}}s\exp\!\left(-\frac{1}{2}s^{2}-ix_{0}\right) \cdot $$ $$ \cdot ...


1

Off to a good start. Draw the picture! The hard part is determining the integration limits. Really, you just need to understand what the lines of constant $r$ and $s$ mean. Basically, they are lines at 45-degree angles to the axes. So, really, imagine that if $r \in [0,1]$, then what are the limits of $s$? Well, in fact, $s$ ranges from $0$ at the ...


2

I suppose your function is $f(x) = g(x)/\sqrt{x-\epsilon}$ where $g$ is bounded, and you want to integrate this on $(\epsilon, b)$. The change of variables $x = \epsilon + t^2$ gives you $$ \int_\epsilon^b \dfrac{g(x)}{\sqrt{x-\epsilon}}\; dx = 2 \int_0^{\sqrt{b-\epsilon}} g(\epsilon + t^2)\; dt$$ which gets rid of the singularity. Similarly to integrate ...


2

Let us assume that $f(z)$ has a Maclaurin series expansion that converges on the unit circle. Then $$ \text{Re} \int_{0}^{\infty} \frac{\sin^{3} x}{x} f(e^{2ix})\, dx = \int_{0}^{\infty} \frac{\sin^3 x}{x} \sum_{n=0}^{\infty} \frac{f^{n}(0)}{n!} \cos(2nx) \, dx .$$ Now assuming we can switch the order of summation and integration (which would be more ...


1

$\qquad$ Even the simpler-looking $\displaystyle\int e^{x^2}~dx$ and $\displaystyle\int x^x~dx$ do not possess a closed form expression, much less the integral you mentioned. See Liouville's theorem and the Risch algorithm for more information.


4

$$\begin{eqnarray*}\int_{0}^{1}x^{x^2}\,dx &=& \int_{0}^{1}\exp\left(x^2 \log x\right)\,dx = \sum_{n\geq 0}\frac{1}{n!}\int_{0}^{1}x^{2n}\log^n x\,dx\\&=&\sum_{n\geq 0}\frac{1}{n!}\cdot\frac{n!(-1)^n}{(2n+1)^{n+1}}=\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^{n+1}}\tag{1}\end{eqnarray*} $$ and the last series converges extremely fast. It is just a ...


6

The indefinite integral $$\int x^{x^2} \, \mathrm{d}x$$ has no closed form in terms of elementary functions. You can, however evaluate the definite integral, for some limits. For example, we have $$\int_0^1 x^{x^2} \, \mathrm{d}x \approx 0.896489$$ The derivation of this can be seen in Jack's answer!


-1

I don't say that this post is an answer, but take a look for the complication even in simple cases : CASE 1 : $P_{3}=x^{3}$ CASE 2 : $P_{3}=2x^{3}$ CASE 3 : $P_{3}=2x^{3}+1$


6

$f$ is continuous on $[0,2]$ and thus Riemann integrable.


0

Try to use method, which is used in the calculation of Green function of wave equation.


1

All solutions come in terms of some error function. I would like the answer in terms of a. Since those error functions had a in their argument, you already got what you wanted.


0

For line integral appears in homework or exercise (instead of real world), it is very common one don't need to parametrize anything to compute the integral. Let $\mathcal{E}$ be the ellipse $$x^2 + 4y^2 - 8y + 3 = 0 \iff x^2 + 4(y-1)^2 = 1.\tag{*1}$$ The integral we want to evaluate is $$\mathcal{I} \stackrel{def}{=} \int_\mathcal{E} xdx + xydy + x^2yz ...


1

First of all the ellipse's equation will be: $$x^2+4(y^2-2y+1)-1=0\Rightarrow x^2+4(y-1)^2=1\Rightarrow \vec r=(\sin t,\frac{\cos t}{2}+1)\Rightarrow \vec {dr}=(cost,-\frac{sin t}{2})$$ if we assume a clockwise orientation then $\vec F=(\sin t,\sin t(\frac{cos t}{2}+1),1)$ then $\vec F.\vec{dr}=\sin t\cos t-\frac{\sin^2 t}{2}(\frac{cos t}{2}+1)$ then ...


1

Pardon me, it may be better to do this directly. I assume you mainly have problems with the parametrization. After completing the square, you should get $$x^2 + \frac{(y - 1)^2}{1/4} = 1$$ The most natural parametrization is $x = \cos t$ and $y = 1 + (1/2)\sin t$. The bounds are $0 \leq t \leq 2\pi$. I assume you can handle the rest as it is just plug ...


1

let $S$ be the square $(0,0),(1,0),(1,1),(0,1)$ and set $P(x,y)=e^{xy}$. then by Green's theorem: $$ \int_0^1 \int_0^1 xe^{xy}dx dy =\int_S\frac{\partial P}{\partial y} dx dy = -\int_{\partial S} Pdx \\ = -\int_0^1 dx - \int_1^0 e^x dx \\ = e-2 $$


3

$$\int_0^1 \int_0^1 x e^{xy}\, dy \, dx$$ We anti-differentiate $x e^{xy}$, noting that $x$ is just a constant since we are doing this with respect to $y$. $$\int_0^1 \left| e^{xy} \right|_{y=0}^1 \, dx$$ $$\int_0^1 (e^x - 1) \, dx$$ This is just one variable, so we should be in more familiar territory. We anti-differentiate with respect to $x$ and ...


0

Hint. The integrand is positive. You may just write $$ \int_{0}^1 e^{xy}\, dy=\frac{e^x-1}x $$ then multiply by $x$ and integrate from $x=0^+$ to $1$.


1

If $y=3(2-x)$ then $x=\frac{6-y}{3}$ Next we would the bounds for the integration. $y=0$ is given. The other is whatever $y$ value makes $x=0$. When $x=0$, $y=6$ So$$V=\pi\int_0^6 ({\frac{6-y}{3}})^2\,dy$$


1

You can use $\displaystyle S=\int_1^2 2\pi R(y)\sqrt{1+(f^{\prime}(y))^2}dy$ where $R(y)=y$ is the distance from a typical point $(x,y)$ on the curve to the axis of rotation $y=0$, so this gives $\displaystyle S=\int_1^2 2\pi y\sqrt{1+y^2(y^2+2)}dy=2\pi\int_1^2y\sqrt{(y^2+1)^2}dy=2\pi\int_1^2y(y^2+1)dy$


1

By replacing $\sin x$ with $u$ we have: $$I=\int\frac{\sin x}{\tan x+\cos x}\,dx = \int \frac{\sin x\cos x}{1+\sin x-\sin^2 x}\,dx = \int \frac{u}{1+u-u^2}\,du\tag{1}$$ but the roots of $1+u-u^2$ occur at $u=\frac{1\pm\sqrt{5}}{2}$, so that: $$ \frac{u}{1+u-u^2} = ...


4

First complete the square: $$ u^2-u-1 = \left(u^2-u+\frac 1 4\right)-\frac 5 4 = \left( u - \frac 1 2 \right)^2 -\frac 5 4 = \left( u - \frac 1 2 - \frac{\sqrt 5} 2 \right)\left( u - \frac 1 2 + \frac{\sqrt 5} 2 \right) $$ Then use partial fractions. $$ \frac{-u}{u^2-u-1} = \frac A {u - \frac 1 2 - \frac{\sqrt 5} 2} + \frac B {u - \frac 1 2 + \frac{\sqrt 5} ...


1

This is not at all correct. Try e.g. $f(x) = x^2$. Then $(f(x)-f(y))/(x-y) = x + y$, and the requirement on $g$ is that $\int_{\mathbb R} g(y)\; dy = 0$ and $\int_{\mathbb R} y g(y)\; dy = 0$. Somewhat more generally, if $f$ is a polynomial of degree $d$ you need the moments $\int_{\mathbb R} y^k g(y)\; dy = 0$ for $0 \le k \le d-1$.


0

$$ \begin{align} \int_{\pi/4}^{\pi/2}\frac{\sqrt{\tan(x)}}{\sin(x)}\,\mathrm{d}x &=\int_{\pi/4}^{\pi/2}\frac{\sqrt{\tan(x)}}{\sin(x)}\frac{\mathrm{d}\tan(x)}{\sec^2(x)}\tag{1}\\ &=\int_1^\infty\frac{\sqrt{u}}{\frac{u}{\sqrt{1+u^2}}}\frac{\mathrm{d}u}{1+u^2}\tag{2}\\ &=\frac12\int_0^\infty\frac{\mathrm{d}u}{\sqrt{u(1+u^2)}}\tag{3}\\ ...


4

First of all, it is obvious that the minimum is when $1>t>0$, (since $e^{-x^2}<1$ on the interval of integration). \begin{align*}f(t)=\int\limits_0^1 x |e^{-x^2} - t|\ \mathrm{d}x &= \int\limits_0^{\alpha} x(e^{-x^2}-t)\ \mathrm{d}x + \int\limits_{\alpha}^1 x(t-e^{-x^2})\ \mathrm{d}x .\end{align*} Where: $\alpha =\sqrt{-\ln t}$. Now you have ...


1

It's true if $\xi$ is continuous at $0$ and $1$. Alternatively, it's true for arbitrary $\xi$ if we restrict to $f$ with $f(0)=0=f(1)$. I'm assuming here that $\xi$ is defined on $\mathbb R$, so that the convolution is defined. If $\xi$ is originally defined only on $[0,1]$ you can extend it. There's a simple proof if you know about Schwarz functions and ...


2

Technicality of the sort often passed over in silence: If a sequence converges in $L^2$ norm to one thing and pointwise to another the two limits are the same. Hence for $x>K$ we have simply $$Hf(x)=\frac1\pi\int_{x-K}^{x+K}f(x-t)\frac{dt}t.$$Hence$$Hf(x)=\frac1{\pi ...


0

Put the second parabola into standard form: $x - 100 = -(y-10)^2$ which has a vertex at $(100, 10)$ and opens to the left. By coincidence, this is also a point on both parabolas. The line $x = 102$ does not come into play (or else there is no region bounded by all three boundaries together). Because of the form of the two parabolas, it is easier to ...


1

Recall that $L^1(\Omega,\Bbb C):=\{f:\Omega\to\Bbb C\;\;\mbox{meas. s.t.}\;\;\int_{\Omega}|f|dm<+\infty\}$, where $dm$ is the Lebesgue measure (and $\Omega$ a measurable subset of $\Bbb R^n$), thus when integrating à la Lebesgue the answer to your question is yes. If otherwise we integrate à la Riemann the answer is no: just consider $f(x)=\frac{\sin ...


1

In the case of the Riemann integral: no. For instance, let $$ f(x) = \begin{cases} \frac{1}{1+x^2} & x \in A \\ \frac{-1}{1+x^2} & x \notin A \\ \end{cases} $$ where $A$ is the rationals. Then the integral of the absolute value exists, but the integral of $f$ does not.


4

Think about the analogous question for sums. The alternating harmonic sum: $$1 - 1/2 + 1/3 - 1/4 +...$$ converges (any alternating series with terms decreasing, in absolute value, to 0 converges). But the corresponding positive series diverges.


2

Between $x=n\pi$ and $x=n\pi-1/n$, we have $|\sin x|<1/n$, so $f(x)>1/(1+\pi^2)$. How does that affect the total integral.


3

$$\Gamma(y) = \int\limits_0^\infty y^{t-1}e^{-y} \, \mathrm{d}y=\int\limits_0^1 y^{t-1}e^{-y} \, \mathrm{d}y+\int\limits_1^\infty y^{t-1}e^{-y} \, \mathrm{d}y$$ We have that for $0 < y \leq 1 \implies y^{t-1}e^{-y}\le y^{t-1}$ and that $$\int\limits_0^1y^{t-1} \, \mathrm{d}y=\left.\frac{y^t}t\right|_0^1=\frac1t$$ Hence the integral converges for the ...



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