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0

we have $$g(c) = \int_{ca}^{cb} {\frac{{f(x)}}{x}dx}$$ now use the following substitution \eqalign{ & u = \frac{x}{c} \cr & du = \frac{1}{c}dx \cr} to get \eqalign{ & g(c) = \int_{ca}^{cb} {\frac{{f(x)}}{x}dx} = \int_a^b {\frac{{f(cu)}}{{cu}}cdu} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \int_a^b {\frac{{f(cu)}}{u}du} = ... 3 We suppose 0 < a < b. For \epsilon > 0, you can find \delta > 0 such that for \vert x \vert \le \delta you have \vert f(x)-f(0) \vert \le \epsilon (by continuity of f at 0). Then for \vert c \vert \max(a,b) \le \delta, we have [ca,cb] \subset [0,\delta]. Hence:\left\vert \int_{ca}^{cb} \frac{f(x)}{x} dx - \int_{ca}^{cb} ...

0

Hint: Note that $f$ is continuous at $0$, so that we can say that over the interval of interest, $f$ satisfies $$\frac{f(0) - \epsilon}{x} \leq \frac{f(x)}{x} \leq \frac{f(0) + \epsilon}{x}$$ when $c$ is sufficiently small.

2

Hint Substitute $$u := \frac{x}{c}, \quad du = \frac{dx}{c}.$$

-1

log(1+z) is analytic if we remove the ray (-inf,-1] and the given curve is in the resulting region. Thus from Cauchy formula f''(a)=2/2.pi.i.integral round C of f(z)/(z-a)^3 we get that the integral is (log(1+z))''.pi.i at z=1/2. It is i.pi.1/(9/4)

0

First of all, make an appropriate linear substitution $x=At+B$, so that the cubic polynomial $ax^3+bx^2+cx+d$ becomes $t^3+\alpha t+\beta$. So you are left with integrating $\exp\Big[i(t^3+\alpha t+\beta)\Big]$. But $e^{i\beta}$ is a constant, and can be therefore extracted outside the integral sign. Then, setting the limits of integration to ...

1

Notice, $$\int\frac{dx}{1+x^4}=\int\frac{dx}{x^2\left(\frac{1}{x^2}+x^2\right)}$$ $$=\int\frac{\frac{1}{x^2}}{\frac{1}{x^2}+x^2}dx=\frac{1}{2}\int\frac{\frac{2}{x^2}}{x^2+\frac{1}{x^2}}dx$$ $$=\frac{1}{2}\int\frac{\left(1+\frac{1}{x^2}\right)-\left(1-\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}}dx$$ ...

6

To find this decomposition, we have to factorize $x^4 + 1$. As $x^4 + 1$ does not have real roots, it does not have real linear factors. This gives the ansatz $$x^4 + 1 = (x^2 + ax + b)(x^2 + cx + d)$$ We have $$(x^2 + ax + b)(x^2 + cx + d) = x^4 + (a+c)x^3 + (b + d + ac)x^2 + (bc + ad)x + bd$$ Comparing the coefficients with $x^4+ 1$, we have ...

0

$$\int y\, dx = xy - \int x\, dy$$ If you can integrate a function then you can integrate its inverse function. You can prove this geometrically. There's a more common trick where you do $\int y \cdot 1\,dx = xy - \int x\frac{dy}{dx}\,dx$ which is apparently just as useful.

4

I am giving a hint. $$\int \frac {dx}{1+x^4}=\frac 12\left(\int\frac {1+x^2}{1+x^4}dx-\int\frac {x^2-1}{1+x^4}dx\right)\\=\frac 12\left(\int\frac {1+\frac 1{x^2}}{x^2+\frac 1{x^2}}dx-\int\frac {1-\frac 1{x^2}}{x^2+\frac 1{x^2}}dx\right)$$ Now, write $x^2+\frac 1{x^2}=\left(x-\frac 1x\right)^2+2$ and $x^2+\frac 1{x^2}=\left(x+\frac 1x\right)^2-2$, for the ...

0

When you change variables the first time putting $u$ in terms of $x,$ when $x=0$ you should have $u=\pi \cos 0=\pi,$ and for $x=1$ it is $u=\pi \cos (\pi)=-\pi.$ With this change it comes out $7.77939,$ which agrees with what maple found for the original integral without any initial substitution.

1

Let $u=x+\frac{a}{2}$ \begin{align} \int_0^a\sqrt{\frac{x+a}{x}}\,dx&=\int_{0}^{a}\frac{\sqrt{x^2+ax}}{x}\,dx\\ &=\int_0^a\frac{\sqrt{\left(x+\frac{a}{2}\right)^2-\frac{a^2}{4}}}{x}\,dx\\ \end{align} Now, let $x+\frac{a}{2}=\frac{a}{2}\sec t$, then \begin{align} ...

0

Use partial fraction method. Add and subtract 2 in the numerator and then the integration boggs down tona simple integration.

0

Hint 1 : Put , $x=\sqrt 2\sec \theta$. Then , $\,dx=\sqrt 2 \sec\theta \tan \theta$. Hint 2: $\frac{x^2}{x^2-2}=1+\frac{2}{x^2-2}$

2

What you need to care about is the behaviour of the integrand for $t$ close to zero. Therefore, find the first nonzero term in the Taylor expansion: $$\frac{t \log{(1+t)}}{t^4+4} = t(t+o(t))\left(\frac{1}{4}+o(t^2)\right) = \frac{t^2}{4}+o(t^2) \text{ as } t \to 0.$$ Integrating this preserves the order term: $$\int_0^x \frac{t \log{(1+t)}}{t^4+4} \, dt = ... 2 Using L'Hospital, given limit is$$\lim_{x\to 0}\frac{x\ln(1+x)}{3x^2(4+x^4)}=\frac{1}{12}$$2 Let me see if I can give a solution. The orthogonality relation for the Gegenbauer polynomials is$$ \int_{-1}^{1} P_m(x) P_n(x) dx = \frac{2}{2n+1} \delta_{m,n}. $$The forward shift operator for the Jacobi polynomials, e.g. http://homepage.tudelft.nl/11r49/askey/ch1/par8/par8.html, is given by$$ \frac{d}{dx} P_m^{(\alpha,\beta)}(x) = ...

2

Let us denote $$\mathcal{I}_{l,n}:=\int_{-1}^1P_l^{(4)}(x)P_n(x)\,dx.$$ If $l\leq n+3$, then $P_l^{(4)}(x)$ is a polynomial of degree at most $n-1$ and $\mathcal{I}_{l,n}$ vanishes because of orthogonality of Legendre polynomials. It also vanishes when $n$ and $l$ have different parity. Hence in the following it will be assumed that $l\geq n+4$ and ...

2

Hint Let $$G(x)=\int_0^x e^{-u^2}\mathrm dx=\int_0^x g(u)\mathrm du.$$ You have that $$f(x)=G^2(x).$$ Therefore $f'(x)=2G(x)G'(x)$ and by definition, $G'(x)=g(x)$. I let you conclude.

1

Hint: Since $f$ is integrable on $[a,r]$ for each $r$, and $f$ is bounded, let $M$ be such a bound. Then, take $r$ sufficiently close to $b$ so that, even if $f$ behaves as badly as $M$ at $b$, this behaviour is negligible and hence $f$ is integrable on $[a,b]$ by this partition. It's the common trick of either (i) controlling $f$ where you know it's ...

1

First of all, the expression is $$\int_{\partial \Omega} \phi F \cdot \vec n d\sigma.$$ To use divergence theorem, you better need to calculate $$\text{div} (\phi F) = \nabla \phi \cdot F + \phi \text{div} F.$$ Now there are two terms on the right hand side. Note that the second term is zero as $\text{div} F = 0$. The first term is zero too, as $F = ... 1 Given$\displaystyle \int\sqrt{\sin x}\;dx$Let$\displaystyle \sin x = t^2\Leftrightarrow \cos xdx = 2tdt\Leftrightarrow dx = \frac{2t}{\sqrt{1-t^4}}dt$So integral convert into$\displaystyle \int t.\frac{2t}{\left(1-t^4\right)^{\frac{1}{2}}}dt$So Integral is$\displaystyle 2\int\;t^2.\left(1-t^4\right)^{-\frac{1}{2}}dt$Now Using$\displaystyle ...

3

Your integral may be expressed in terms of an incomplete elliptic integral, a Legendre integral, one may prove that $$\int_0^x \sqrt{\sin t}\ \text dt=\sqrt{\frac{2}{\pi }} \Gamma\left(\frac{3}{4}\right)^2-2 \text{EllipticE}\left[\frac{1}{4} (\pi -2 x),2\right],\quad 0\leq x \leq \pi,$$ where $\text{EllipticE}\left[\phi,m\right]$ is a special ...

3

From the equation $x^2+y^2=r^2$, you may express your area as the following integral $$A=\int_0^r\sqrt{r^2-x^2}\:dx.$$ Then substitute $x=r\sin \theta$, $\theta=\arcsin (x/r)$, to get \begin{align} A&=\int_0^{\pi/2}\sqrt{r^2-r^2\sin^2 \theta}\:r\cos \theta \:d\theta\\ &=r^2\int_0^{\pi/2}\sqrt{1-\sin^2 \theta}\:\cos\theta \:d\theta\\ ... 0 For the homogenous equationT'-k(65-T)=0$$you properly found$$T=65-ce^{-kt}$$So, now, applying the variation of the constant,$$T'=e^{-k t} c'-k c e^{-k t}$$and plugging in the original equation you find$$c'=65 e^{k t}-20 e^{k t} \sin \left(\frac{\pi t}{6}\right)$$which is not excatly what you wrote. So, integrating$$c=\frac {65}k e^{kt}-20\int ...

0

Thank you the replies. Basically I had already done what everyone here is saying but I was under the impression that the answer should be more elegant or insightful. However, it seems it is just an exercise in algebra and nothing more.

2

The facts below should allow you to compute the integral in question in terms of $l$ and $n$: The Legendre polynomials $P_0, \dots, P_n$ form a basis for the space of polynomials of degree at most $n$. The Legendre polynomials are orthogonal: $$\int_{-1}^{1} P_m(x) P_n(x)\,dx = {2 \over {2n + 1}} \delta_{mn}$$ $\dfrac{d^4P_l(x)}{dx^4}$ is a polynomial ...

0

The answer is yes. The method is detailed in the paper by S. K. Lucas. The approximate fractions are obtained from truncating the exact continued fraction of the respected numbers, so the signs are alternating. We first give a few examples. Results For $\pi$ The continued fractions are $3, 22/7, 333/106, 355/113, 103993/33102, \dots$. \begin{align} ...

1

Sanchayan, do you mean to be taking a definite integral over interval $[0,\pi]$? This integral $$I_n=\int_0^\pi \sec^n(x) \, dx$$ doesn't even always converge. For example, check out the case $n=2$. Standard calculus 2 techniques will prove my point.

0

The key to solving this is to notice that since $\sin x$ is symmetric about $x=\pi/2$, then so is $f(\sin x)$. This means that $$\int_0^\pi xf(\sin x) dx=\int_0^\pi (\pi-x)f(\sin (\pi-x)) dx=\int_0^\pi (\pi-x)f(\sin x) dx,$$ after which you can proceed as in the other answers.

2

Your substitution has to be a bijection, or else you lose bits of the integral, as you have noticed: draw a picture of the substitution, and notice that you've lost bits. Also, as @GudsonChou points out, $\sqrt{\sin^2{t}}$ is only equal to $\sin{t}$ half the time. If you draw the curve, you should notice that it's both symmetric about $t=\pi$ and covers the ...

3

I=$\int _{ 0 }^{ \pi }{ { e }^{ { \sin ^{ 2 } x } } } { \cos((2n+1)x) } dx$ Let $u=\pi -x$, $dx=-du$, $${ { e }^{ { \sin ^{ 2 } x } } }={ { e }^{ { \sin ^{ 2 }( {\pi -u} )} } }={ { e }^{ { \sin ^{ 2 } u } } }$$, ${ \cos((2n+1)x) }$= ${ \cos((2n+1)(\pi-u) }$= ${ \cos(2n\pi + (\pi - (2n+1)u)) }$= ${ \cos(\pi - (2n+1)u)) }$= $-{ \cos(2n+1)u)) }$ ...

2

Hint: You can make $x=t+\frac{\pi}{2}$, so \begin{align} \int_0^{\pi}{e^{\sin^2 x}\cos((2n+1)x) dx}&=\int_{-\pi/2}^{\pi/2}{e^{1-\cos^2 (t+\frac{\pi}{2})}\cos\left((2n+1)t+n\pi+\frac{\pi}{2}\right) dt}\tag{1} \end{align} Also, if $n\in\mathbb{Z}$, we have \begin{align} \cos\left((2n+1)t+n\pi+\frac{\pi}{2}\right)&=\cos [(2n+1)t]\cos\left(n\pi ...

0

If you accept the book's hint, you have \begin{align} \int_0^{\pi}xf(\sin(x))\,dx &=\int_{\pi}^0(\pi-u)f(\sin(\pi-u))\cdot(-1)\,du\\ &=\int_{0}^\pi(\pi-u)f(\sin(u))\,du\\ &=\int_{0}^\pi\pi f(\sin(u))\,du-\int_{0}^\pi uf(\sin(u))\,du\\ &=\pi\int_{0}^\pi f(\sin(u))\,du-\int_{0}^\pi uf(\sin(u))\,du\\ \end{align} Now set ... 0 If you make the substitution that the book suggests, you have x = \pi-u and dx = -du, so the integral becomes \begin{align} \int_{0}^{\pi} x\,f(\sin(x))\,dx &= -\int_{\pi}^0 (\pi-u)f(\sin(\pi-u))\,du \\ &= \int_0^{\pi} (\pi-u)f(u)\,du \\ &= \pi\int_0^{\pi} f(u)\,du - \int_0^{\pi} u\,f(u)\,du. \end{align} Collect terms and ... 0 We can define A = \inf_P U(P, f). From the assumption of your problem, for each \epsilon, you can find a P_N such thatU(P_N, f) - L(P_N,f) \leq \epsilon. \quad \quad (\star)$$Continue from your remark, we have$$ L(P_N,f) -A \leq S(P_N, f) - A \leq U(P_N, f) - A.$$From (\star),$$ U(P_N, f) - A \leq U(P_N, f) - L(P_N,f) \leq \epsilon$$and ... 0 Both f are Riemann integrable, with a proof below for the first one. For N \ge 1 integer, define$$s_N(x)=\begin{cases} 0 & \text{ for } x \in [0, \frac{1}{N+1})\\ f(x) & \text{ else} \end{cases} \ \ \ S_N(x)=\begin{cases} 1 & \text{ for } x \in [0, \frac{1}{N+1})\\ f(x) & \text{ else} \end{cases}$$s_N,S_N are step functions and for ... 1 Here is a picture of one of the washers obtained. This is what I meant in my comment. 0 Here is also another way which works with your formula$$\int_1^x {{1 \over {1 + {t^2}}}dt} = \int_1^x {{{{t^2}} \over {1 + {t^2}}}{1 \over {{t^2}}}dt} = \int_1^x {{1 \over {1 + {{\left( {{1 \over t}} \right)}^2}}}{1 \over {{t^2}}}dt} $$Now choose f(x) = {1 \over {1 + x}} and g(t) = {1 \over t}. Then it follows that$$\int_1^x {{1 \over {1 + ...

0

Try the substitution $$t = {1 \over u}$$ then it follows that \eqalign{ & t = x\,\,\,\,\,\, \to \,\,\,\,\,\,\,u = {1 \over x} \cr & t = 1\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,u = 1 \cr} and consider that \eqalign{ & dt = - {1 \over {{u^2}}}du \cr & {1 \over {1 + {t^2}}} = {1 \over {1 + {1 \over {{u^2}}}}} = {{{u^2}} ... 0 Hint: Start with I = \int_x^1 \frac{dt}{1 + t^2}. Substitute u = 1/t. Then du = -1/t^2 \, dt or dt = (-1/u^2)\, du andI = \int_{1/u}^1\frac{1}{1 + 1/u^2} \cdot\frac{-1}{u^2} du$$Can you finish from here? 0 You're right, X is a random element of (S,\mathcal{S}) means that X is a random variables with values in S, i.e. a measurable map from (\Omega,\mathcal{F}) to (S,\mathcal{S}). \int_Sf(y)\mu(dy) is the same thing as \int_Sfd\mu.It's just another notation that makes clear the variable of integration. 2 Here's another way to think about where the minus sign "went". Recall that n\ln a = \ln(a^n). Therefore, -\ln(x) = \ln(x^{-1}) = \ln(1/x) . 4 There are two things to this. The first thing is that you should know that$$\ln(1/x) = -\ln(x).$$Why does this happen? It follows from the rule \ln(a/b) = \ln(a) - \ln(b), and if we let a=1 and b=x, this rule gives us \ln(1/x) = \ln(1) -\ln(x) = 0-\ln(x) = -\ln(x), where I use that \ln(1) = 0. The second thing is that your integral has a minus ... 0 If f is integrable in [\frac{1}{n},1] for every n, so the set of descontinuities C_n, on [\frac{1}{n},1] has zero measure. Therefore, the set of descontinuities on [0,1] has zero measure and since f is bounded on [0,1] you have that f is integrable. 0 Notice that your initial region \Omega is the box (x,y) \in [0,1]^2. Your event Y/X \leq w introduces the constraint Y = wX (for each fixed w), which is a line passing through the origin. Define a region A which is the part of \Omega below the line Y = wX, and find the area of the region, e.g. by double integral over it (it should depend on ... 0 We simply use the "sifting" property of the Dirac Delta to obtain$$\begin{align} \int_{-\infty}^{\infty}\frac{\cos(\pi(x+1))}{x}\sum_{n=1}^{\infty}\delta(x-n)\,dx&=\sum_{n=1}^{\infty}\frac{\cos(\pi(n+1))}{n}\\\\ &=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\\\\ &=\log 2 \end{align}$$2 The answer is positive: if f is integrable over [\delta,1] for all \delta \in (0,1] then f is integrable over [0,1]. Following is a proof using Riemann integration. Take \epsilon > 0. By hypothesis f is bounded on [0,1], so you can find M > 0 with \vert f(x) \vert \le M for x \in [0,1]. Take \delta= \frac{\epsilon}{4M}. As ... 0 We have$$F(\omega)=\int_{-\infty}^{\infty}f(t)e^{i2\pi\omega t}\,dt \tag 1$$and$$f(t)=\int_{-\infty}^{\infty}F(\omega)e^{-i2\pi\omega t}\,d\omega \tag 2$$From (1) we find$$F(0)=\int_{-\infty}^{\infty}f(t)\,dt $$If we integrate (2), we find that$$\begin{align} ...

4

Let $M$ such that $|f(x)|\leq M$ for all $x\in [0,1]$. Let $F(x)=\int_x^1 |f(x)|dx$ with $0<x<1$. The function $F$ is nonincreasing and bounded since $$0\leq F(x)\leq M\int_0^1dx=M,$$ therefore it converges.

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