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1

$$\frac{1}{1+x+x^2}=(1-x)\sum_{n\geq 0}x^{3n} = \sum_{n\geq 0}\left(x^{3n}-x^{3n+1}\right).$$


2

Hint: $$\frac 1{1+x+x^2}=\frac {1-x}{1-x^3}$$


0

Let $$\displaystyle I = \int\frac{5x^4+4x^5}{(x^5+x+1)^2}dx = \int\frac{\left[\left(5x^4+1\right)+(4x^5-1)\right]}{(x^5+x+1)^2}dx$$ So $$\displaystyle I = \underbrace{\int\frac{5x^4+1}{(x^5+x+1)^2}dx}_{J}+\underbrace{\int \frac{4x^5-1}{(x^5+x+1)^2}dx}_{K}$$ So for Calculation of $$\displaystyle J= \int \frac{5x^4+1}{(x^5+x+1)^2}dx$$ Let $(x^5+x+1) = ...


0

Hint: Let $~I(a)=\displaystyle\int_{-\infty}^\infty e^{-ax^2}~dx,~$ and then evaluate $I'\bigg(\dfrac12\bigg)$.


3

Let $$\displaystyle I = \int \frac{1}{(\alpha+\beta \cos x)^2}dx\;,$$ Now Let $$\displaystyle t = \frac{\beta+\alpha\cos x}{\alpha+\beta \cos x}$$ So $$\displaystyle \frac{dt}{dx} = \frac{\left(\alpha+\beta \cos x\right)\cdot -(\alpha \sin x)-(\beta+\alpha \cos x)\cdot (-\beta \sin x)}{(\alpha+\beta\cos x)^2} = \frac{(\beta^2 -\alpha^2)\sin ...


0

Let $$\displaystyle I = \int \sqrt{4a^2(y-b)^2+c^4}dy\;,$$ Let $(y-b) = t\;,$ Then $dy = dt$ So Integral $$\displaystyle I = \int \sqrt{4a^2t^2+c^4}dt = 2a\underbrace{\int \sqrt{t^2+k^2}dt}_{J}\;,$$ Where $\displaystyle k= \frac{c^2}{2a}$ For calculation of Integral $J\;,$ We Use Integration by parts. Now Let $$\displaystyle J = \int \sqrt{t^2+k^2}\cdot ...


1

$$\sqrt{4 a^2 (y-b)^2+c^4}=\dfrac1{2|a|}\sqrt{(y-b)^2+\left(\dfrac{c^2}{2a}\right)^2}$$ Using Trigonometric substitutions , set $y-b=\dfrac{c^2}{2a}\cdot\tan u$ and use How to integrate $\sec^3 x \, dx$? Or Indefinite integral of secant cubed


0

HINT let $c^2\sinh\theta=2a(y-b)$


2

$e^{-x^2/2}$ has no elementary antiderivative. To compute its definite integral over the line, we perform the trick: $$\left ( \int_{-\infty}^\infty e^{-x^2/2} dx \right )^2 = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-x^2/2-y^2/2} \, dx \, dy.$$ Then we convert to polar coordinates, obtaining $$\int_0^{2 \pi} \int_0^\infty e^{-r^2/2} r \, dr \, d ...


1

Notice, we have $$\int\frac{x}{1+\cos x}dx$$ $$=\int\frac{x}{1+\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}dx$$ $$=\int\frac{x\left(1+\tan^2\frac{x}{2}\right)}{1+\tan^2\frac{x}{2}+1-\tan^2\frac{x}{2}}dx$$ $$=\int\frac{x\sec^2\frac{x}{2}}{2}dx$$ $$=\frac{1}{2}\int x\sec^2\frac{x}{2}dx$$ $$=\frac{1}{2}x \int \sec^2\frac{x}{2}-\frac{1}{2}\int\left( \int ...


0

$\int \frac{x}{1+{cos}\, x} dx= \int \frac{x}{2{cos}^{2}\frac{x}{2}} dx = \frac{1}{2}\int x \, {sec} ^{2}\frac{x}{2}dx$ Then integrate by parts. You should get $x\, $tan$\frac{x}{2} + 2 \, $log$($cos$\frac{x}{2}) $ + constant


1

There are different ways. One way is to multiply top and bottom by $1-cosx$ This gives you a $x(1-cosx)$ in the numerator and a $sin^2x$ in the denominator. Now use partial fraction decomposition using the fact that both $sec^2x$ and $\frac{cosx}{sin^2x}$ can be anti derived twice in terms of elementary functions.


3

Hint: $$\begin{align} \int\frac{x}{1+\cos{\left(x\right)}}\,\mathrm{d}x &=\int\frac{x}{2\cos^{2}{\left(\frac{x}{2}\right)}}\,\mathrm{d}x\\ &=\frac12\int x\sec^{2}{\left(\frac{x}{2}\right)}\,\mathrm{d}x.\\ \end{align}$$ This form suggests that the probable next step is integration by parts.


1

Judging from the RHS side of your integral I see believe you are trying to notate a standard multiple integral. In general it is fine to write $f(x)dx$ when $x$ is an element of $\mathbb{R}^n$. However, then care must be taken with how you write the "limits" of integration. First, you don't want to specify a lower and upper limit of integration if you are ...


1

I would write something like $\int_{\prod_{i=1}^{n}[z_i+\Delta z_i,z_i]}f(x) dx $ or define $Q := \prod_{i=1}^{n}[z_i+\Delta z_i,z_i]$ and write $\int_{Q}f(x) dx $.


0

Experimenting with wolfram alpha lead me to: $$ \int {x^n \over 1+e^x} = (-1)^{n+1} x^{n+1} + (-1)^n * log(e^x+1)+ \sum_{k = 0}^{n-1} x^{n-i-1} (-1)^n * {n! \over (n-i)!} * Li_{i+2}(-e^x)$$ I got this by trying out small values for n so wolfram alpha manages to do the computations and then generalized it. Can someone verify/correct that?


1

Integrating by parts, it is not difficult to check that $\int_{0}^{1}\log(x)\sin(x)\,dx$ is given by the difference between the cosine integral of one and the Euler-Mascheroni constant; the other piece is trivial. Notice that: $$\begin{eqnarray*} \int_{0}^{1}\log(x)\sin(x)\,dx &=& \sum_{n\geq ...


0

(x2+1)/(x2-5x+6)= (x2+1)/{(x-2)(x-3)} By partial fraction (x2+1)/(x-3)(x-2)= (Ax+B)/(x-2) +C/(x-3) x2+1= Ax2+x(B-3A-C)-(B+2C) Solving above equation, We get A=1, B=-7, C=10 So (x2+1)/(x2-5x+6)=(x-7)/(x-2) +10/(x-3) => 1+10/(x-3)-5/(x-2) So integration will be = x+10ln(x-3)-5ln(x-2)


1

You probably won't find any meaningful necessary condition other than "all occuring terms need to be well-defined". The problem is that even if the function is awfully irregular, the identity might hold "by accident". The search for sufficient conditions on the other hand is a bit easier. Essentially you need to exchange the limit in the definition of the ...


0

Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $, $$\sin^2(n+1)x-\sin^2nx=\sin x\cdot\sin(2n+1)x$$ $$I_{n+1}-I_n=\int_0^\pi\dfrac{\sin(2n+1)}{\sin x}\ dx=J_n\text{(say)}$$ Now using Prosthaphaeresis Formulas, $$\sin(2m+1)x-\sin\{2(m-1)+1\}x=2\sin x\cos(2m)x$$ $$\implies J_m-J_{m-1}=\int_0^\pi2\cos2mx\ dx=\cdots=0$$ $$\implies J_{m+1}-J_m=0\iff ...


1

Let $$\displaystyle I = \int \frac{x^2}{(x\sin x+\cos x)^2}dx$$ Put $x=\tan \phi\;,$ Then $dx = \sec^2 \phi d\phi$ So Integral $$\displaystyle I = \int\frac{\tan^2 \phi \cdot \sec^2 \phi}{\left[\tan \phi\cdot \sin (\tan \phi)+\cos(\tan \phi)\vphantom{\frac11}\right]^2}d\phi $$ So $$\displaystyle I = \int\frac{\tan ^2\phi \cdot \sec^2 \phi\cdot \cos^2 ...


3

One method is to evaluate the integral $$ I(\alpha) = \int_{0}^{2 \pi} \ln(\alpha + \beta \, \cos x) \, dx.$$ Then take two derivatives with respect to $\alpha$. Alternatively take one derivative with respect to $\alpha$ of \begin{align} I_{1}(\alpha) = \int \frac{dx}{\alpha + \beta \, \cos x} = - \frac{2}{\sqrt{\beta^2 - \alpha^2}} \, \tanh^{-1}\left( ...


0

Here's an example that may convey some intuition. Suppose we consider how much it rains in a given year in Los Angeles. This random variable, which we shall call $R$, has a distribution, which is a PDF that we may denote $f_R(r)$. Suppose, for the sake of argument, that $f_R(15) = 0.1$. Now what does that mean? Obviously, it doesn't mean that there's a ...


2

The velocity of a moving particle is the rate of change of the displacement at that time. The probability density of a continuous real valued random variable is the rate of change of the cumulative probability at that point.


4

Strictly speaking, at any given point the pdf doesn't mean anything. You can change the value of the pdf at that particular point to be whatever you want, and the distribution of the random variable is the same. Still, "generically", $f(x)$ is essentially the probability to find the random variable in the interval $(x-\epsilon,x+\epsilon)$, divided by $2 ...


1

You could use that $(x\sin x+\cos x)^{\prime}=x\cos x$ to try integrating by parts with $\displaystyle u=x\sec x, \;\;dv=\frac{x\cos x}{(x\sin x+\cos x)^2}dx$ and $\displaystyle du=\sec x(x\tan x+1)dx, \;\;v=-\frac{1}{x\sin x+\cos x}$ to get $\displaystyle\int\frac{x^2}{(x\sin x+\cos x)^2}dx=-\frac{x\sec x}{x\sin x+\cos x}+\int\frac{\sec x(x\tan ...


3

Here's an alternative approach: If you see an integral which is made of a fraction with a squared term in the denominator and you have serious hopes that the integral is elementary then you might think of the quotient rule. Note that this tells you that $\int \frac{u'v-uv'}{v^2} dx=\frac{u}{v}+C$. So in this case you clearly have $v(x)=x\sin x+\cos x$ and ...


1

For fun I tried Yves' hyperbolic suggestion. $$ I = \int_{-\infty}^\infty (1+x^2)^{-3/2}dx $$ substitute $x=\sinh t$ so that $dx = \cosh t\,dt$ and $$ I = \int_{-\infty}^\infty (\cosh^2 t)^{-3/2} \cosh t\,dt= \int_{-\infty}^{\infty} (\cosh t)^{-2} dt = \tanh t \big|_{-\infty}^\infty = 2 $$


1

$$I_{n+1}-I_n = \int_0^{\pi}\frac{\sin^2(n+1)x}{\sin^2x}-\frac{\sin^2nx}{\sin^2x}dx$$ $$=\int_0^\pi \frac{\sin (2n+1)x}{\sin x}dx=\pi$$ This means that the values of these integrals form an AP, and we know that $I_1=\pi$. Thus we conclude that the value of $I_n=n\pi$.


2

Using Chebyshev polynomials of the second kind, $$ I_n = \int_{0}^{\pi} U_{n-1}(\cos x)^2\,dx = \int_{-1}^{1} \frac{U_{n-1}^2(z)}{\sqrt{1-z^2}}\,dz = \color{red}{\pi n}\tag{1} $$ since Chebyshev polynomials of the second kind are an orthogonal base of $L^2(-1,1)$ with the inner product $\langle f,g\rangle = \int_{-1}^{1}\frac{f(x)g(x)}{\sqrt{1-x^2}}\,dx $. ...


1

The error of the trapezoid method is given in terms of the second derivative of a function. Our function is smooth and fast-decaying outside any right neighbourhood of the origin, but in a right neighbourhood of the origin the second derivative of $f$ is unbounded and behaves like $\frac{1}{x^{1/2}}$. That is the reason for which $p\to 2+\frac{1}{2}$. You ...


0

Do polynomial long division (I will not do that here) to get $$ \int \left(\frac{2x-1}{x^2+4}+\frac{4}{x}+1\right)dx \\ 2\int\frac{x}{x^2+4}dx-\int\frac{1}{x^2+4}dx+4\int\frac{1}{x}dx+\int dx. $$ The last two integrals are easy to integrate. The first two integrals need substitutions. For the first integral, use $s=x^2+4$ and $t=\frac{x}{2}$ for the second ...


0

Some random babblings: Let $f(x) =\frac{x^{3/2}}{\cosh(x)} $. For large $x$, $f(x)$ is essentially zero, so you might be losing many significant digits in subtraction, especially if $I_{2n}$ is quite close to $I_n$. Also, at the origin, $f(x) \approx x^{3/2} $, so there is no problem there. $f'(x) =\frac{(3/2)x^{1/2}\cosh(x)-x^{3/2}\sinh(h)}{\cosh^2(x)} ...


5

Just let $$ u=\cot x. $$ You will get the integral $$ \int_1^{+\infty}\frac{u}{u^2+\sqrt{u}}\frac{1}{1+u^2}\,du, $$ which I'm sure you can handle with your skills (just let $s=\sqrt{u}$ and you have a rational function).


1

$$f(z)=z^3 e^{-z^4}$$ is an entire function having $F(z)=-\frac{1}{4}e^{-z^4}$ has a primitive. That implies: $$ \int_{\mathcal{C}} f(z)\,dz = F(b)-F(a),$$ where $a=0$ and $b=1-i$ are the endpoints of $\mathcal{C}$.


1

We have: $$\begin{eqnarray*} \sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{(k+1)^{s-1}} &=& \int_{0}^{+\infty}\frac{x^{s-2}}{(s-2)!}\sum_{k=0}^{n}\binom{n}{k}(-1)^k e^{-(k+1)x}\,dx\\&=&\int_{0}^{+\infty}\frac{x^{s-2}}{(s-2)!}e^{-x}(1-e^{-x})^n\,dx\end{eqnarray*}$$ hence: $$\begin{eqnarray*} \sum_{n\geq ...


2

Let $x=\dfrac{c\theta}t$. Then $$\int_{\theta}^{1} x \exp \left(- \frac{c\theta}{x}\right) dx =-(c\theta)^2\int_{c}^{c\theta} t^{-3} \exp \left(-t\right) dt=(c\theta)^2(\Gamma(-2, c\theta)-\Gamma(-2,c)).$$ Depending on the values of $c$ and $\theta$, you can use some asymptotic approximation of $\Gamma$.


1

Theorem: Define $S_r(n)=1^r+2^2+...+n^r$, $$\frac{n^{r+1}}{r+1}\leq S_r(n)\leq\frac{(n+1)^{r+1}}{r+1}$$ for any positive $r\geq0$ Then $$\frac{12n^{k+1}}{(k+1)(n+1)^3(n+1)^4}\leq{\frac{1^k + 2^k +...+n^k}{(1^2 + 2^2 +...+n^2)\cdot(1^3 + 2^3 +...+n^3)}}\leq \frac{12(n+1)^{k+1}}{(k+1)n^3n^4}$$ So, $$F(6)=\frac{12}{7}$$ and $F(k)=0$, if $0\leq k<6$ and ...


1

You only need the largest terms in the fraction, the rest converge to 0. What are the leading constants in $\sum_{j=1}^{n} k^5, \sum_{j=1}^{n} k^2, \sum_{j=1}^{n} k^3$?


0

You can simplify the sums: $$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}\\ \sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}{4}\\ \sum_{i=1}^n i^5= \frac{n^2(n+1)^2(2n^2 + 2n - 1)}{12}\\$$


4

Let $f(z)=e^{iz^2}$. Then, $f$ is entire and by the residue theorem we have $$\oint_C f(z)\,dz=0 \tag 1$$ for any sufficiently smooth closed contour $C$ in the complex plane. Now, suppose $C$ is comprised of the three segments; $C_1$, $C_2$, and $C_3$, where $(i)$ $C_1$ is the line segment on the real axis from $(0,0)$ to $(R,0)$. $(ii)$ $C_2$ is the ...


4

I'd rather claim that $$ \int_{-\infty}^{+\infty}e^{it^2}\,dt=(1+i)\sqrt{\frac{\pi}{2}}. $$ Note that $$ e^{it^2}=\cos(t^2)+i\sin(t^2), $$ so this really follows from the Fresnel integrals $$ \int_{-\infty}^{+\infty}\cos(t^2)\,dt = \sqrt{\frac{\pi}{2}} \quad\text{and}\quad \int_{-\infty}^{+\infty}\sin(t^2)\,dt = \sqrt{\frac{\pi}{2}}. $$ Edit Since $$ ...


5

I had read the similar problem, but it doesn't work. Of course it doesn't ! The integral you posted is nothing else than the formula for the arc length of the $($co$)$sine function, which is rather famous for giving rise historically to the study of elliptic integrals ! In particular, ...


1

Suppose $f$ is a $C^1$ function on some open interval $I$. Then applying the quotient rule we find that $$ \frac{d}{dx}\left(\displaystyle\frac{\int_0^v f(x)\,dx}{f(v)}\right) = \frac{[f(v)]^2 - f'(v)\int_0^v f(x)\,dx}{[f(v)]^2} $$ Therefore, this function is increasing on $I$ if $$ [f(v)]^2 - f'(v)\int_0^v f(x)\,dx \geq 0 ~~\forall v\in I $$ This ...


3

Here is a solution only using dilogarithm identities: $$ \operatorname{Li}_2(z) + \operatorname{Li}_2(1-z) = \zeta(2) - \log z \log(1-z). \tag{1} $$ $$ \operatorname{Li}_2(1-z) + \operatorname{Li}_2(1-\tfrac{1}{z}) = -\tfrac{1}{2}\log^2 z. \tag{2} $$ $$ \operatorname{Li}_2(z) + \operatorname{Li}_2(-z) = \tfrac{1}{2}\operatorname{Li}_2(z^2). \tag{3} ...


1

Rephrasing the conjecture: Let $\alpha,\beta>0$ with $\alpha+\beta=1$. Then there exists $C>0$ such that $$ s_{\alpha,\beta}(k) = \sum_{n=1}^k (k+1-n)^{-\alpha} n^{-\beta} < C $$ uniformly in $k$. Split the sum around $n=\frac k2$. For the smaller values of $n$, $$ \sum_{n=1}^{k/2} (k+1-n)^{-\alpha} n^{-\beta} \le (k/2)^{-\alpha} \sum_{n=1}^{k/2} ...


0

(Edited with a solution for $\alpha <\frac 12$) We have equation of the form $\int_0^{\alpha b}f(x)\ dx = \int_{\alpha b}^b f(x)\ dx$ and that is equivalent to $$2F(\alpha b) = F(b) + F(0)$$ where $F'=f$. Since we know that $(a^x)'=a^x\ln a$, it easily follows that $$\int (a^x-1)\ dx = \frac{a^x}{\ln a}-x + C$$ so we need to solve equation ...


1

You can write $a^x$ as $e^{x\ln a}$. Using this trick you can integrate $a^x$ as follows $$ \int a^x\,dx = \int e^{x\ln a}\,dx = \frac{e^{x\ln a}}{\ln a} + C = \frac{a^x}{\ln a} + C $$ Now apply the fundamental theorem of calculus to evaluate both integrals. You will end up with an equation involving $a$, however, I don't believe it can be solved ...


0

An antiderivative of $$ \sec\theta\tan\theta=\frac{\sin\theta}{\cos^2\theta}= -\frac{-\sin\theta}{\cos^2\theta} $$ is $1/\cos\theta$. Therefore, integrating by parts, $$ \int\theta\frac{\sin\theta}{\cos^2\theta}\,d\theta= \frac{\theta}{\cos\theta}-\int\frac{1}{\cos\theta}\,d\theta $$ The remaining integral can be computed with the substitution $$ ...


4

Your answer is the same. Note that $$\log|\cos(\theta/2)-\sin(\theta/2)|-\log|\cos(\theta/2)+\sin(\theta/2)|=\log |\frac{\cos(\theta/2)-\sin(\theta/2)}{\cos(\theta/2)+\sin(\theta/2)}|\\ =\log |\frac{\cos^2(\theta/2)-2\sin(\theta/2)\cos(\theta/2)+\sin^2(\theta/2)}{\cos^2(\theta/2)-\sin^2(\theta/2)}|\\=\log ...



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