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0

Note $$\frac{1}{(i - 2it)^2 + i} = \frac{1}{i^2(1 - 2t)^2 + i} = \frac{1}{i - (1 - 2t)^2} = \frac{-i - (1 -2t)^2}{1 + (1 - 2t)^4}.$$ So $$\frac{-2i}{(i-2it)^2+i} = \frac{-2 + 2i(1 - 2t)^2}{1 + (1 - 2t)^4}.$$ Using this you can write $$\int_{\gamma} \frac{dz}{z^2 + i} = -\int_0^1 \frac{2}{1 + (1 - 2t)^4}\, dt + i\int_0^1 \frac{2(1 - 2t)^2}{1 + (1 - ...


0

How about this (as inspiration): Using parameterization $z(t) = it \quad (t: 1 \to -1)$. $$\int_\gamma f(z) \operatorname d z = \int_1^{-1} \frac{i}{i-t^2} \operatorname d t$$ $$= i \int_1^{-1} \frac{1}{(e^{i\frac{\pi}{4}} -t)(e^{i\frac{\pi}{4}} +t)} \operatorname d t $$ $$= i \frac{1}{2e^{i\frac{\pi}{4}}}\int_1^{-1} \left(\frac{1}{e^{i\frac{\pi}{4}} -t} ...


4

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 ...


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.


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

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} = ...


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.


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} ...


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]{} ...


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}}\\ ...


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 ...


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 $H(s)=\frac{2}{s^2+4}$.


0

Having been quite a bit of time I thought about this some more! If we cut our original radical at a finite time we have (as pointed out by Aryabhata) we have $$y = n!\frac{d^n}{dx^n} $$ Solutions to this include $$ y= C_1 e^{x\sqrt[n]{n!}_1} +C_2 e^{x\sqrt[n]{n!}_2}... C_n e^{x\sqrt[n]{n!}_n} $$ For all possible nth roots of unity. Now consider ...


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 ...


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.


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 ...


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

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 ...


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.)


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.


0

Hint: $I = \displaystyle \int \dfrac{d(\sqrt{x})}{(\sqrt{x})^2 + 1}$


0

They can't be the same, the first is unbounded while the second is bounded. At any rate, your second answer is correct. For completeness, here's a derivation. Let $u=\sqrt{x}$, then $du=\frac{dx}{2 \sqrt{x}}$ and $x=u^2$. (This was a good choice precisely because the factor of $2\sqrt{x}$ in the denominator gets absorbed into the $du$.) So your integral is ...


0

Remember that $E(I_A) = P(A)$ for any event $A$ and $E(I_AI_B)=P(A \text{ and } B)$. $F_X(x)=P(X\le x)$ and $F(x,y)=P(X \le x\text{ and } Y\le y)$.


1

Let $u=t^2$, then the integral becomes: $$\int_0^{\sqrt{b-a}} \frac{2dt}{a+t^2} $$ which is relatively easy to do, using that $\int \frac{dt}{1+t^2} = \arctan t + C$.


0

$$ \int_0^{b-a} \frac{1}{\sqrt{u} (a+u)} \,du $$ If $w=\sqrt{u}$ then $w^2=u$ so $2w\,dw=du$, and you have $$ \int_0^{b-a} \frac{1}{\sqrt{u} (a+u)} \,du = \int_0^{\sqrt{b-q}} \frac{1}{w(a+w^2)} (2w\,dw) = \int_0^{\sqrt{b-a}} \frac{2\,dw}{a+w^2} $$ and you get an arctangent.


1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} ...


0

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} ...


0

Hint: For $\beta>0$, we have: $$\int_{-\infty}^\infty \max(0,\alpha+\beta x)\phi(x)\,dx = \int_{-\alpha/\beta}^\infty (\alpha+\beta x)\phi(x)\,dx =\alpha N(\alpha/\beta) + \beta\phi(\alpha/\beta), $$ by using $$ \int \phi(x)dx = N(x) +C, \; \int x\phi(x)dx = -\phi(x) +C. $$ To use it in your calculation apply change of variable $x = ...


1

Generally, this is not true: the assumption $o(w^2h^2)$ as $wh\to 0$ does not yield the conclusion. For example, let $k(w)=c/(1+w^4)$ with $c>0$ chosen so that the integral of $k$ is $1$. The function $w^4h^4$ satisfies the assumption $o(w^2h^2)$ as $wh\to 0$, but $$\int k(w) w^4h^4\,dw = \infty \quad \text{ for all } h>0$$ So, you need some ...


0

First suppose $f$ is a continuous function on $[0,1)$ with compact support. Then $f$ is uniformly continuous, so given $\varepsilon > 0$, there corresponds a $\delta > 0$ such that for all $x$ and $y$, $|x - y| < \delta$ implies $|f(x) - f(y)| < \varepsilon$. Let $N$ be a positive integer such that $\frac{1}{N} < \delta$. If $n \ge N$, then ...


1

You're almost done. Note that $g'(x)$ dominates $\sqrt{x} g'(x)$ on this interval, and the integral of $g'(x)$ is less than or equal to $g(1)-g(0)$. This doesn't require absolute continuity for $g$. See Proposition 22 here.


4

let $-I = \int_0^1{\ln x \over 1 + x^2} \ dx.$ we will need $$a_k = \int_0^\infty xe^{-kx}\ dx = {1 \over k^2} \int_0^\infty xe^{-x} \ dx = {1 \over k^2}$$ $I = \int_0^1{\ln x \over 1 + x^2} \ dx,$ by a change of variable $ u = -\ln x, x = e^{-u}, $ the integral $\begin{eqnarray} I &=& \int_0^\infty{ ue^{-u} \over 1 + e^{-2u}} \ du \\ ...


3

\begin{align} \int_{0}^{1}\frac{\ln x}{1+x^2}\mathrm dx&=\int_0^{1}\sum_{n=1}^\infty(-1)^n\, x^{2n}\ln x\;\mathrm dx\tag1\\ &=\sum_{n=1}^\infty(-1)^n\int_0^{1} x^{2n}\ln x\;\mathrm dx\tag2\\ &=-\sum_{n=1}^\infty\frac{(-1)^n}{(2n+1)^2}\tag3\\ &=\bbox[5pt,border:3px #FF69B4 solid]{\color{red}{\large-\text{G}}} \end{align} Explanation : ...


5

Use integration by parts to get $$ \displaystyle\int_0^1 \frac {\ln x}{1 + x^2} \, \mathrm{d}x = \left[ \arctan(x) \ln(x) \right]_0^1 - \displaystyle\int_0^1 \frac {\arctan(x)}{x} \, \mathrm{d}x = - \displaystyle\int_0^1 \frac {\arctan(x)}{x} \, \mathrm{d}x. $$Now, we use the Taylor expansion of $\arctan(x)$ to get $$ -\int_0^1 ...


2

Continuing from Chris's sis answer, consider the partial sum: $$\frac{1}{2}\sum_{n=1}^m n(n+1)\left(\ln(n)-2\ln(n+1)+\ln(n+2)\right)+1-\frac{1}{n}$$ $$=\frac{m-H_m}{2}+\frac{1}{2}\sum_{n=1}^m\left( \ln\left(\frac{n}{n+1}\right)^{n(n+1)}+\ln\left(\frac{n+2}{n+1}\right)^{n(n+1)}\right)$$ $$=\frac{m-H_m}{2}+\frac{1}{2}\left(\ln\left(\dfrac{\displaystyle ...


4

Let us evaluate the general form of the integral \begin{align} \int\frac{\mathrm dx}{(x^2+a^2)\sqrt{x^2-b^2}}&=\int\frac{a\sec^2t}{(a^2\tan^2t+a^2)\sqrt{a^2\tan^2t-b^2}}\mathrm dt\tag1\\[7pt] &=\frac{1}{a}\int\frac{\cos t}{\sqrt{a^2\sin^2t-b^2\cos^2t}}\mathrm dt\tag2\\[7pt] &=\frac{1}{a}\int\frac{\cos ...


4

First substitute $x=e^y$ ("$dx=e^y dy$"): $$\int\frac{1-x^2}{1+3x^2+x^4}dx=\int\frac{e^y-e^{3y}}{1+3e^{2y}+e^{4y}}dy=\int\frac{e^{-y}-e^y}{e^{-2y}+3+e^{2y}}dy=\int\frac{-2\sinh y}{(2\cosh y)^2+1}dy$$ Now substitute $z=2\cosh y$ ("$dz = 2\sinh y\,dy$"): $$\int\frac{-2\sinh y}{(2\cosh y)^2+1}dy=\int\frac{-1}{z^2+1}dz=\text{arccot}\,z$$ So the answer should be ...


1

Using series representation you specified above, I got that your integral gets reduced to $$\sum _{n=1}^{\infty } \frac{(n+1) n^2 (\log (n)-2 \log (n+1)+\log (n+2))+n-1}{2 n}$$ Can you take it from here?


1

let me try a change of variable $$x = {1 + t^2 \over 1 - t^2},\ dx = {-4t \ dt \over (1-t^2)^2},\ x^2+1 = {2(1+t^4) \over (1-t^2)^2}, \ x^2-1 = {4t^2 \over (1-t^2)^2}$$ $$\int {1 \over (x^2+1) \sqrt{x^2-1}} dx = \int {t^2 - 1 \over t^4 + 1} \ dt = \int {1 \over t^2 + 1} dt - 2 \int {1 \over t^4 + 1} dt = \tan^{-1}t - 2\int{dt \over t^4+1}$$ i got to go ...


2

For the record, for ease of access, here is my solution. Notice that the degree of the denominator is double the degree of the numerator, so there is a possibility that the answer is of the form $ \arctan \left( f(x) \right) $, where $f(x)$ is a rational function. So we try it out. If this is the case, we have $$ \frac {f'(x)}{1 + \left[ f(x) \right]^2} ...


4

HINT: Divide the numerator & the denominator by $x^2$ and as $\int\left(1/x^2-1\right)dx=-x-\dfrac1x$ set $x+\dfrac1x=u$ and $\dfrac1{x^2}+3+x^2=\left(x+\dfrac1x\right)^2-2+3$


2

By symmetry, $$ I = \int_{[0,1]^2}|x-y|\,d\mu = 2 \int_{0}^{1} \int_{y}^{1} (x-y)\, dx\,dy = \int_{0}^{1}(1-y)^2\,dy=\frac{1}{3}.$$


5

$$u=\frac{\sqrt{x^2-1}}x$$ $$\frac{du}{dx}=-\frac{\sqrt{x^2-1}}{x^2}+\frac{2x}{x\cdot2\sqrt{x^2-1}}$$ $$=\frac{-(x^2-1)+x^2}{x^2\sqrt{x^2-1}}$$ $$\implies\int\frac{dx}{(1+x^2)\sqrt{x^2-1}}=\int\frac{x^2}{1+x^2} \frac{dx}{x^2\sqrt{x^2-1}}$$ Now $u^2=\dfrac{x^2-1}{x^2}\implies\dfrac1{x^2}=1-u^2$ ...


10

Notice for $|t| < 1$, $\frac{1}{\sqrt{1 - 2xt + t^2}}$ is the generating function for the Legendre polynomials: $$\frac{1}{\sqrt{1 - 2xt + t^2}} = \sum_{n=0}^\infty P_n(x) t^n\tag{*1}$$ It is known that for $x \in [-1,1]$, $|P_n(x)| \le P_n(1) = 1$. This means as long as $t$ is fixed and $|t| < 1$, the absolute values of the $n^{th}$ term is bounded ...


1

Yes, your result is correct. I would also recommend solving it using the Heaviside Unit Step Function. We can write your piecewise function (see my response Using laplace transforms to solve a piecewise defined function initial value problem) as: $$f(t) = 0 - 0 u(t-2) + t u(t-2) = t u(t-2)$$ The Laplace Transform (see item 27) is: $$\mathscr{L} (t~ ...


1

This is not an answer to the question, but it is too long for a comment. Here is a piece of information that I know about the integral that might help you to evaluate it. The integral above is known as one of the three of the Watson/van Peype Triple Integrals: \begin{align} I_1&=\frac{1}{\pi^3}\int_0^\pi\int_0^\pi\int_0^\pi\frac{\mathrm dx\;\mathrm ...


1

For $m=1$ $$\sum _{k=1}^n \left(\frac{1}{(a-2 i k)^1}+\frac{1}{(a+2 i k)^1}\right)=\frac{1}{2} i \left(\psi ^{(0)}\left(1+\frac{i a}{2}+n\right)-\psi ^{(0)}\left(1-\frac{i a}{2}+n\right)-\psi ^{(0)}\left(1+\frac{i a}{2}\right)+\psi ^{(0)}\left(1-\frac{i a}{2}\right)\right)$$ For $m=2$ $$\sum _{k=1}^n \left(\frac{1}{(a-2 i k)^2}+\frac{1}{(a+2 i k)^2}\right)$$ ...


2

$$\int_3^4 e^{-ut}\,dt=\left.-\frac1ue^{-ut}\right|_3^4=-\frac1u(e^{-4u}-e^{-3u})=\frac{e^{3u}-e^{4u}}u$$ and now substitute the above in the right hand of your equality


0

We have: $$I = \int_{0}^{1} \frac{x^2(x - 1)}{\log(x)}$$ Consider: $$I(b) = \int_{0}^{1} \frac{x^b(x - 1)}{\log(x)}$$ Recognize that $I(0) = \int_{0}^{1} (x-1)/\log(x) dx = \log(2)$, this is a particular solution for later. Differentiate partially with respect to $b$ $$I'(b) = \int_{0}^{1} \frac{\log(x)\cdot x^b(x - 1)}{\log(x)} dx$$ $$I'(b) = ...



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