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5

Here's a little help: at first, it may seem to integral is not treatable using Cauchy, since the problem lies at $1/\sin z$ which has a zero at $z=0$. Now consider $f(z)= \dfrac{z}{\sin z}$. This is now analytic and zero free in $|z|\leqslant 1/2$, and the same holds for $g(z)=f(z)(z-1)^{-1}$. Your integral now has the form ...


5

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 ...


4

Denote the first integral by $I$ and the second by $J$. Then, $$\begin{aligned} J=&\int_0^{\pi/4} x\left(\frac{\pi}{2}-\arctan\sqrt{\frac{\cos 2x}{2\sin^2 x}}\right)\,dx \\ =&\frac{\pi^3}{64}-I \,\,\,\,\,\,\,(1) \end{aligned}$$ $I$ can be simiplified to: $$I=\int_0^{\pi/4} x\arccos(\sqrt{2}\sin x)\,dx=\left(\frac{x^2\arccos(\sqrt{2}\sin ...


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

$$$$ Let us start to calculate it. \begin{eqnarray} I&=&\int_{-1}^{1}\dfrac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}\\ &=&\frac{1}{2\sqrt{ab}}\int_{-1}^{1}\dfrac{dx}{\sqrt{\frac{a^2+1}{2a}-x}\sqrt{\frac{b^2+1}{2b}-x}}\\ &=&\frac{1}{2\sqrt{ab}}\int_{-1}^{1}\dfrac{dx}{\sqrt{m-x}\sqrt{n-x}}\\ ...


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 ...


2

As in differential geometry, we are defining the "distance" between two points to be the inf of the "lengths" of all paths joining the points. (For example, with the usual notion of lengths of paths in $\Bbb R^2-\{0\}$, what is the distance from $(-1,0)$ to $(1,0)$? Note that you cannot use the usual line segment joining them.)


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 ...


1

Noting that $f_n\colon x\in\mathbb{R}\mapsto \frac{1}{1+x^{2n}}$ converges pointwise towards $f = \chi_{(-1,1)} + \frac{1}{2}\chi_{\{-1,1\}}$, and that $f_n$ (non-negative) is dominated by $f$ (which is Lebesgue integrable) for all $x,n$, you can use the dominated convergence theorem to argue that $$\int_{\mathbb{R}} f_n \xrightarrow[n\to\infty]{} ...


1

$\quad$ All integrals of the form $~\displaystyle\int_0^\infty\dfrac{x^{k-1}}{(x^n+a^n)^m}~dx~$ can be evaluated by substituting $x=at$ and $u=\dfrac1{t^n+1}$ , then recognizing the expression of the beta function in the new integral, and lastly employing Euler's reflection formula for the $\Gamma$ function to simplify the result.


1

$$3I=\int\frac{3xdx}{\sqrt{3^2-(3x-2)^2}}$$ Setting $3x-2=3\sin\theta\implies dx=\cos\theta\ d\theta$ and $3x=2+3\sin\theta$ $$3I=\int\frac{\cos\theta(2+3\sin\theta)}{3\text{sign}(\cos\theta)\cos\theta}d\theta$$ $$\implies9I=\text{sign}(\cos\theta)\int(2+3\sin\theta)d\theta$$ $$=\text{sign}(\cos\theta)(2\theta-3\cos\theta+K)$$ Now, ...


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.


1

The pathlines are curves in space, and parametric functions of $t$. So if you eliminate t, you get an expression for $\frac{dy}{dx}$. $$ \frac{dy}{dx} = \frac{y}{x} $$ On integrating this, you have $y=cx$, so the pathlines are straight lines through the origin.


1

For $n$ even and $a>0$: $\int_{a}^{1} |lnx|^n dx=\int_{a}^{1}(x)^{'}(lnx)^ndx=xlnx_{a}^{1}-\int_{a}^{1}xn(lnx)^{n-1}\frac {1}{x}dx=xlnx_{a}^{1}-\int_{a}^{1}n(lnx)^{n-1}dx$. By induction $\int_{a}^{1} (lnx)^{n}dx$ converges iff $\int_{a}^{1} lnx dx$ converges. Same tactic for $n$ odd.


1

The definition given is $\bar{d}(p_1,p_2) = \inf\{\bar{L}(\gamma) : \gamma(0) = p_1, \gamma(1) = p_2\}$. The condition $\gamma(0) = p_1$, $\gamma(1) = p_2$ means the curve $\gamma(t)$ starts at point $p_1$ and ends at point $p_2$. Since $\bar{L}(\gamma)$ is the "length" of the curve $\gamma$, $\{\bar{L}(\gamma) : \gamma(0) = p_1, \gamma(1) = p_2\}$ is ...


1

Let $f(t)=te^{−t}\sin(2t)=tg(t)$ and $g(t)=e^{−t}\sin(2t)=e^{−t}h(t)$ with $h(t)=\sin(2t)$. So $F(s)=-G'(s)$ and $G(s)=H(s+1)$ with $H(s)=\frac{2}{s^2+4}$.



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