# Tag Info

8

To be be somewhat explicit. One may perform the change of variable, $q=e^{-x}$, $dq=-e^{-x}dx$, giving $${\large\int}_0^\infty e^{-x}\prod_{n=1}^\infty\left(1-e^{-24\!\;n\!\;x}\right)dx={\large\int}_0^1 \prod_{n=1}^\infty\left(1-q^{24n}\right)dq\tag1$$ then use the identity (the Euler pentagonal number theorem) $$... 7 First, to sketch such a graph, you want to consider the distance from the origin as the angle from the x-axis changes. Just like when sketching the graph of a function in rectangular coordinates it is good to evaluate at particular values of x and see the height of the function, when sketching a curve in polar coordinates, evaluate the function are a ... 6 An alternative: Consider$$ J(a)=\int_0^{\pi/2} \cos^a(\phi)d\phi $$differentiating w.r.t a gives us$$ \partial_a J(a)\big|_{a=1}=-I $$But on the other hand J(a) is just a Wallis integral and therefore$$ I=-\frac{\sqrt{\pi }}{2}\partial_a\left( \frac{\Gamma \left(\frac{a+1}{2}\right)}{\Gamma \left(\frac{a}{2}+1\right)}\right)\big|_{a=1} $$... 6 Sketching this in polar coordinates is pretty straightforward. Draw your axes and you know that the radial value is equal to the angle, so your curve would start at the origin when \theta is zero, it would be \frac{\pi}{2} at 90 degrees, etc. As for calculating the area, you are correct that you integrate it, but I'm not sure what it means physically ... 5 A Riemann sum for$$\int_a^b f(x)\,dx$$using n subintervals is$$\frac{b-a}n\sum_{k=1}^n f(x_k)\ ,$$where x_k is a point in the kth subinterval. Compare this with the given sum$$\frac1{30}\sum_{k=1}^{60} e^{k/30}\ .$$By looking at the upper limit of summation, n=60. Then from the fraction at the front, b-a=2, and since all your answer options ... 5$$\frac{\pi^2}{24}-\frac{2\pi}3+\frac1{36\sqrt{2}}\left[5\pi^2+12\left(4+\ln\left(\frac{1+\sqrt2}2\right)\right)\left(\pi-2\ln\left(1+\sqrt2\right)\right)-48\operatorname{Li}_2\left(\sqrt2-1\right)\right]$$5$$\int_{0}^{4}(\left \lceil x+1 \right \rceil)dx=\\ \int_{0}^{4}(\left \lceil x \right \rceil+1)dx=\\\int_{0}^{4}(\left \lceil x \right \rceil)dx+\int_{0}^{4}1dx=\\$$hence$$=\int_{0}^{4}(\left \lceil x \right \rceil)dx+\int_{0}^{4}1dx=\\ \int_{0}^{1}(\left \lceil x \right \rceil)dx+\int_{1}^{2}(\left \lceil x \right \rceil)dx+\int_{2}^{3}(\left \lceil x ...

4

Sorry, I first did not read that you only wanted a hint. I suggest to use L'Hospital rule,

4

Hint: $$\lim_{n \to \infty} \frac1{n} \sum_{k=1}^{\color {red} n} f(k/n) = \int_0^1 f(x) dx$$

4

I think you're falling into a trap many new students to of notation first, meaning second. You are writing things down and then asking what they mean. This is akin to writing a bunch of words and then asking "what does this sentence mean?". I'm not going to go through each of your 7 integrals to say which does or does not make sense, nor will I give you a ...

4

One may just apply the Jacobi triple product to the Dedekind eta function, then perform a termwise integration that leads to a multiple of $\zeta(2)$.

4

We have $$\int_{0}^{1}{\dfrac{1-x}{\log x}(x+x^{2}+x^{2^{2}}+x^{2^{3}}+\cdots)}\:dx=-\log 3. \tag1$$ Proof. One may recall that, using Frullani's integral, we have $$\int_{0}^{1}\frac{x^{a-1}-x^{b-1}}{\log x}\:dx=\log\frac ba \quad (a,b>0). \tag2$$ Considering a finite sum in the integrand, we get \begin{align} ... 4 If we substitute x=\tan y, we get: I = \int_{0}^{\pi/3}\frac{\arcsin(\sin(2y))}{\cos^2 y}\,dy = \int_{0}^{\pi/4}\frac{2t}{\cos^2 t}\,dt+\int_{\pi/4}^{\pi/3}\frac{\pi-2t}{\cos^2 t}\,dt $$hence:$$ I = \left(\frac{\pi}{2}-\log 2\right)+\left(\log 2+\frac{\pi}{\sqrt{3}}-\frac{\pi}{2}\right)=\color{red}{\frac{\pi}{\sqrt{3}}} $$as wanted. 3 This is not an answer. This is just something to offer some ideas. Actually, this more of a comment on steroids.$$\int_{0}^{\pi/2} \arctan(x)\cot(x) \text{d}x=I$$Now, what we do is rewrite the equation in terms of i. A nice way to do that is to do two things. We first substitute x=ia, then multiply and divide -i to the argument of the integral. (And ... 3 Using Mathematica I have arrived at the result$$I=\frac{1}{4} (-\text{Li}_3(-1-i)-\text{Li}_3(-1+i)+\text{Li}_3(1-i)+\text{Li}_3(1+i))+\frac{1}{4} \left(-\text{Li}_3\left(-\frac{1}{2}-\frac{i}{2}\right)-\text{Li}_3\left(-\frac{1}{2}+\frac{i}{2}\right)+\text{Li}_3\left(\frac{1}{2}-\frac{i}{2}\right)+\text{Li}_3\left(\frac{1}{2}+\frac{i ...

3

This is equivalent to American Mathematical Monthly Problem 11148 published in April 2005. Let $u=x-1$. Then rewrite the integral as $$\int_{0}^{\infty}{\frac{u^8-4u^6+9u^4-5u^2+1}{u^{12}-10u^{10}+37u^8-42u^6+26u^4-8u^2+1}}du.$$ The value of this integral is equal to the value of the original integral. The method to solve this equivalent integral can be ...

3

The reason is that the integral you evaluated doesn't give the area of the quadrilateral $R$. Indeed, called $A$ its area, you have $$\int\int_R dx dy=A$$ and $$\int\int_R x dx dy\neq \int\int_R dx dy.$$

2

You have, taking $u=-\frac{1}{t}$ (and so $u'=\frac{1}{t^2}$), $v=e^t$ and using the integration by parts formula $(\int u'v=[uv]-\int uv')$ that : $$F(x)=\int_1^x\frac{e^t}{t^2}dt=\Big[-\frac{e^t}{t}\Big]_1^x+\int_1^x\frac{e^t}{t}dt,$$ and so $$F(x)=e-\frac{e^x}{x}+G(x).$$

2

Hint: Use the fundamental theorem of calculus and L'Hospital rule.

2

I understand what you're asking now. If you want to integrate a function $f\colon [0,\infty) \to [0,\infty)$, then you can write $\int_{0}^{\pi/2} \frac12 r^2(\theta) d\theta$. You're asking if $\int_{0}^{\infty} \frac12 r^2(\theta) d\theta=\infty$. Well unless $r=0$ then yes of course. You're adding up a positive number infinitely many times. You are ...

2

Draw the Graph and find the area under it.

2

Hint: Draw it out. It should look like stairs rising towards the right. From $0 < x <= 1$, $y = 2$. Extrapolate from this to draw out the function. Then, find the area under the graph.

2

One may recall that for any Riemann integrable function over $[a,b]$, as $n \to \infty$, one has $$\sum_{k=0}^n\frac{(b-a)}nf\left(a+\frac{k(b-a)}n \right) \to \int_a^bf(x)dx$$ Then you may apply it to $\displaystyle f(x)=\sqrt{1+2x}$, $a=0$, $b=1$ giving the limit $$\int_0 ^1\sqrt{1+2x}\:dx.$$ By the change of variable $1+2x \to u$ you also get $$... 2 The integrand has a closed-form antiderivative in terms of elementary functions and polylogarithms. It can be found using Mathematica after expressing inverse trig functions through logarithms of complex arguments, and can be manually checked for correctness using differentiation. After subtracting its limits at \infty and 0 and simplification, we can ... 2 What you did is OK, you get$$ \mathcal{L}(\cos^2(\omega t))=\frac{1}{2}\left(\frac{1}{s} + \mathcal{L}(\cos(2\omega t))\right)=\frac{1}{2 s}+\frac{s}{2 \left(s^2+4 w^2\right)} $$where we have used$$\mathcal{L}(\cos\omega t) = \frac{s}{s^2 + \omega^2}.$$2 Explaining David G. Stork's result,$$ \begin{aligned} I &=\int_{\sqrt a}^\infty e^{-t^2+\beta t} \sin(\beta t) \, dt \\ &= \Im \, \int_{\sqrt a}^\infty e^{-t^2+\beta t + i\beta t} \, dt \\ &= \Im \left[ e^{(\beta + i\beta)^2/4} \int_{\sqrt a - \beta(1+i)/2}^\infty e^{-u^2} \, du \right] \\ &= \Im \left[ \frac{\sqrt\pi}{2} \, e^{i\beta^2/2} ...

2

Morera's theorem is probably the simplest approach (for questions such as these -- Did's hint gives a quicker solution in this particular case). First show that $f$ is continuous. (I'll leave that to you.) Then, if $\gamma$ is any closed curve in $\mathbb{C}$, \begin{align} \int_{\gamma} f(z)\,dz &= \int_{\gamma} \left( \int_0^1 \frac{e^{tz}}{1+t^2}\,dt ...

2

You are interested in the integral $$f(x) = \int_1^x y\sin\!\left(\frac{2\pi (y-1) x}{y}\right) dy,$$ and this is very close to the integral $$g(x) = \int_0^x y\sin\!\left(\frac{2\pi (y-1) x}{y}\right) dy,$$ which is (basically) expressible in closed form. To see this, first make the substitution $y = ux$ to get $$g(x) = x^2 \int_0^1 u ... 1 Let$$ \begin{align} I(\alpha,k) &= \int^\infty _{-\infty} \frac{e^{-i \alpha x}}{x^2 + k^2}\,\mathrm{d}x \tag{1} \\ &= \int^{\infty} _{-\infty} \frac{e^{i \alpha x}}{x^2 + k^2}\,\mathrm{d}x \tag{2} \\ &= \frac12\int^{\infty} _{-\infty} \frac{e^{i \alpha x} + e^{-i \alpha x}}{x^2 + k^2}\,\mathrm{d}x \tag{3} \\ &= \int^{\infty} _{-\infty} ...

1

The arbitrary constant has to be added whenever you are dealing with indefinite integrals. In other words, if you are told that $F$ is the antiderivative of a function $f$, meaning that $$F(x) = \int f(x) dx \qquad\mbox{ (integral without limits)},$$ then any other funtion $G(x)=F(x)+C$ will be an antiderivative of $f$ too. In fact, $$F'(x) = G'(x) = f(x).$$ ...

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