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6

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} ...


4

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} ...


3

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 $(1-1/x)^x$.


3

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


3

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 of $x$ ...


1

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.


1

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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