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Let $$\mathcal{ I}=\int \ln(1 + e^x)\ \mathrm dx$$ By substituting $e^x=-t\iff e^x\,\mathrm dx=\dfrac{\mathrm dt}{t}$ \begin{align} \mathcal{ I} &=\int \frac{\ln(1 -t)}{t}\ \mathrm dt =-\int \frac{1}{t}\sum_{n=1}^{\infty}\frac{t^n}{n}\ \mathrm dt =-\int \sum_{n=1}^{\infty}\frac{t^{n-1}}{n}\ \mathrm dt\\ &=-\sum_{n=1}^{\infty}\int ... 5 The line you write is incorrect, 1+p+\ldots+p^{n-1} approaches to \frac1{1-p} assuming that |p|<1. Namely, if |p|<1, then \lim\limits_{n\to\infty}(1+\ldots+p^n)=\dfrac1{1-p}. This is true exactly because 1+\ldots+p^{n-1}=\frac{1-p^n}{1-p}, and when |p|<1, taking n to infinity yields p^n\to 0. 5 Since \arctan is increasing we get\frac{1}{ n^2}\int_{n\pi}^{2n\pi }\frac{t}{\arctan 2\pi n}dt \leq I \leq \frac{1}{n^2}\int_{n\pi}^{2n\pi}\frac{t}{\arctan n \pi}dt$$Now we can calculate both sides and we get$$ \frac{4\pi ^2 - \pi^2}{2\arctan 2\pi n} \leq I \leq \frac{4\pi ^2 - \pi^2}{2\arctan \pi n} $$So$$ \frac{3\pi ^2 }{2\arctan 2\pi n} ...

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It's easier to shift indices, so that the summand looks the same between the sums you are comparing. Your sum is $$\sum_{k=2}^{n+1} \frac{t^k}{k!}.$$ You can now "add and subtract" the first two terms of the sum for the exponential: $$\sum_{k=2}^{n+1} \frac{t^k}{k!} = \sum_{k=0}^{n+1} \frac{t^k}{k!} - \sum_{k=0}^1 \frac{t^k}{k!} \\ = \sum_{k=0}^{n+1} ... 4 Yes. Consider the change of variable, y=kx. As x \rightarrow \infty, y \rightarrow \infty 4 Let \varepsilon > 0, \exists N_0 > 1 \text{ such that if } x > N_0 \Rightarrow |f(x)-L| < \varepsilon. Now choose N_1 = \dfrac{N_0}{k} > 1 \Rightarrow \text{ if } x > N_1 \Rightarrow kx > k\left(\dfrac{N_0}{k}\right) = N_0 \Rightarrow |f(kx)-L| < \varepsilon 3 f^{100}(x) = 2\cdot 100!. Can you see how? Take a lesser power example: g(x) = 2x^3 \to g'(x) = 2\cdot 3x^2\to g''(x) = 2\cdot 3\cdot 2x \to g^{3}(x) = 2\cdot 3\cdot 2\cdot 1 = 2\cdot 3!. Generalize to 100. 3 You may recall the celebrated \Gamma function defined by$$ \Gamma(\alpha)=\int_0^\infty u^{\alpha-1} e^{-u}\:{\rm{d}}u, \quad \alpha>0 $$Observe that$$ \int_{-\infty}^\infty \frac{1}{x^{2n}+1} \:{\rm{d}}x=2\int_0^\infty \frac{1}{x^{2n}+1} \:{\rm{d}}x $$then you may write$$ \begin{align} \int_0^\infty \frac{1}{x^{2n}+1} ...

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In the body of the OP (but not in the title), we are differentiating $\exp(x\ln(1-1/x))$. We use the Chain Rule. Note that the derivative of $x\ln(1-1/x)$ is $$x(-1)(-1/x^2)\frac{1}{1-1/x}+\ln(1-1/x).$$ This simplifies to $$\frac{1}{x-1}+\ln(1-1/x).\tag{1}$$ For the derivative of our exponential, multiply by $(1-1/x)^x$.

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Or you can use $$|\sqrt{x} - 3| = \frac{|x-9|}{\sqrt{x}+3}\leqslant\frac{|x-9|}{3}.$$

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Here is an approach. $\displaystyle \mathcal{L}(\sin 2t) = \frac{2}{s^2 + 2^2} = \frac{2}{s^2 + 4}$, using the table. $\displaystyle \mathcal{L} (e^{-t} \sin 2t) = \frac{2}{(s + 1)^2 + 4}$, using frequency shifting. $\displaystyle \mathcal{L}( t e^{-t} \sin 2t) = -\frac{d}{ds}\!\left(\frac{2}{(s + 1)^2 + 4}\right) = \frac{4 ... 3 We can write the integral as $$I=\int\frac{\sqrt{1+x^4}}{1-x^4}\textrm{d}x=\int\frac{\left(1+x^4\right)}{\left(1-x^4\right)\sqrt{1+x^4}}\textrm{d}x.$$ Let$t=\frac{1+x^4}{1-x^4}$so that$x^4=\frac{t+1}{t-1}$and$\textrm{d}x=\pm\left(\frac{t+1}{t-1}\right)^{\frac{1}{4}}\cdot\frac{-1}{2(t-1)(t+1)}\textrm{d}t. The integral above becomes $$\pm \int ... 2 Hint 1: Let's suppose that \delta < 1. Then, |x+1| < 1, so 0 < x < 2. This means that |x+2| < 4, so |x+1||x+2| < 4|x+1|. Can you add an additional restriction on \delta so 4|x+1| < \epsilon? Hint 2: Alternatively, take \delta < 1 again. Then \delta(\delta+1) < 2\delta. What additional restriction can you put ... 2 If you're familiar with the Squeeze Theorem, then you know that \lim_{x}h(x)=\lim_{x}g(x) (exists, is finite) implies \lim_{x}f(x) exists and is equal to the former. Your condition is weaker. So for example h(x)=g(x)=f(x)=x satisfies the problem assumption but clearly the limit is infinite. Or take g(x)=h(x)=f(x)=(-1)^{[x]} where the limit doesn't ... 2 Letting u = x - 3 we have that du = dx and 2u + 5 = 2x -1.$$\begin{align}\int \frac{2x- 1}{x^2-6x + 13}dx &= \int \frac{2x- 1}{(x-3)^2 + 4}dx\\&=\int \frac{2u + 5}{u^2 + 4}du\\&=\int \frac{2u}{u^2 + 4}du + \int \frac{5}{u^2 + 4}du\\&=\ln |u^2 + 4| + \frac{5}{2}\arctan\Big(\frac{u}{2}\Big) \end{align}$$Because \frac{1}{u^2 + 4} = ... 2 The only problem is at 0. The absolute value is irrelevant (just multiply by (-1)^{n+1} the final result when no absolute value is used), so we can consider$$ \int_a^1(\log x)^n\,dx $$with a>0. An integration by parts gives$$ \Bigl[x(\log x)^n\Bigr]_a^1-\int_a^1 x\cdot n(\log x)^{n-1}\frac{1}{x}\,dx $$Can you go on from here? 2 Since the integrand is an even function, we can rewrite$$I_n=2\int_0^\infty\frac{\mathrm dx}{1+x^{2n}}$$and we have a well-known result for the latter integral, namely$$\int_0^\infty\frac{\mathrm dx}{1+x^{2n}}=\frac{\pi}{2n}\csc\left(\!\frac{\pi}{2n}\!\right)$$Hence$$I_n=\frac{\pi}{n}\csc\left(\!\frac{\pi}{2n}\!\right)$$and as n\to\infty, we get ... 2 e^{t}=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+ \cdots, then$$e^t-1-t=\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+\cdots=\sum_{2}^{\infty}\frac{t^k}{k!}$$2 The question is trivial if A is finite, so let's assume A is infinite. Since \operatorname{sup}(A) is a limit point of A, then in particular, we can find an infinite number of points within any given \varepsilon-neighborhood of it. Further, all such points will be less than or equal to \operatorname{sup}(A) by definition of the supremum. So ... 2 Schaum's Outline in Complex Variables LINK Starts from the beginning. Contains a whole chapter on integrals by contours. If you can do all the problems at the end of that chapter, you will be ready to tackle problems here! 2 I liked the free book: "A first Course in Complex Analysis" by Matthias Beck, Gerald Marchesi, Dennis Pixton, which consists of plenty of exercises and hints/solutions. Available at: http://math.sfsu.edu/beck/complex.html 2 Multiplying \cos x you take the risk of multiplying by negative numbers and changing the inequality, since -1\leq \cos x \leq 1. On the other hand, you may notice that |\cos x| \leq 1 then from what you did$$|x^2\cos x - 0| = |x^2||\cos x| \leq x^2 < \delta^2 = \epsilon$$2 See when you differentiate 100 times all the terms except the first go to 0. All you have to do is see what the first looks like 2 Since \pi\csc(\pi z) has residue (-1)^n at z=n for n\in\mathbb{Z}, we will use the contours$$ \gamma_\infty=\lim\limits_{R\to\infty}Re^{2\pi i[0,1]}\qquad\text{and}\qquad\gamma_0=\lim\limits_{R\to0}Re^{2\pi i[0,1]} $$To sum over all n\in\mathbb{Z} except n=0, we use the difference of the contours, which circles the non-zero integers once ... 2 By definition we have \lim\limits_{x\to\infty}f(x)=L means$$\forall \epsilon>0,\; \exists A>0,\; x>A\implies |f(x)-L|<\epsilon$$so for the selected \epsilon>0, and for B=kA we have (kx>B\iff x>\frac Bk)\implies |f(kx)-L|<\epsilon hence we get$$\lim\limits_{x\to\infty}f(kx)=L$$2 \text{Line #11 is incorrect}. \text{Its} \infty. \text{We have:} \left|\dfrac{a_{n+1}}{a_n}\right| = \left|\dfrac{(n+2)!}{8(n+1)!}\right|=\dfrac{n+2}{8}\to \infty 2 we have f(x)=\frac{1}{x-2}+\frac{1}{x+2} then we get$$f'(x)=- \left( x-2 \right) ^{-2}- \left( x+2 \right) ^{-2}f''(x)=2\, \left( x-2 \right) ^{-3}+2\, \left( x+2 \right) ^{-3}f'''(x)=-6\, \left( x-2 \right) ^{-4}-6\, \left( x+2 \right) ^{-4}$$can you proceed? for your control the answer is$$f(x)^{(20)}=2432902008176640000\, \left( x-2 \right) ... 1 The key is knowing the definitions of numbers. What is a critical point? A simple google search will answer this. The answer is, a critical point is any point where the derivative is zero. What does it mean for a function to be increasing? decreasing? A function is increasing means its derivative is positive. So you have to find all the values ofx\$ ...

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I'd recommend the second book in the "Princeton Lectures in Analysis" series. It's titled Complex Analysis, written by Elias Stein and Rami Shakarchi.

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setting $$\sqrt{5+12x-9x^2}=xt+\sqrt{5}$$ we get $$x=\frac{12-2t\sqrt{5}}{t^2+9}$$ and we get $$dx=\frac{2(-9\sqrt{5}-12t+\sqrt{5}t^2)}{(9+t^2)^2}dt$$ and our integral is rational.

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