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0

$=-\frac{d}{x^2} \cdot \Phi^{\prime}(\frac{d}{x^2}) \cdot \frac{-2d}{x^3}$ $=\frac{2d^2}{x^5} \cdot \Phi^{\prime}(\frac{d}{x^2})$


0

Upper and lower sums can only be computed explicitly for special examples, like the example in question, or for an exponential function. For the example at hand do the following: Partition the interval $[1,2]$ into $N$ equal parts $I_k:=[x_{k-1},x_k]$ with $$x_k:=1+{k\over N}\qquad(0\leq k\leq N)\ .$$ Looking at the graph of $f$ it is then easy to set up the ...


0

I suggest that you instead try to integrate using the differential equation $$ x^2J_\nu''(x)+xJ_\nu'(x)+(x^2-\nu^2)J_\nu(x)=0 $$ and the recurrence formulas $$ \frac{2\nu}{x}J_\nu(x)=J_{\nu-1}(x)+J_{\nu+1}(x) \quad \text{and} \quad 2J'_\nu(x)=J_{\nu-1}(x)-J_{\nu+1}(x). $$ I show you a first step, and you tell me if you need further steps. Using the ...


0

This is mechanized in Maple: restart; with(OrthogonalExpansions): BesselSeries(x^2, x = 0 .. 1, infinity, 0, 'Coefficients'); $$\sum _{i=1}^{\infty }2\,{\frac { \left( {{J}_{1}\left({\it BesselJZeros} \left( 0,i \right) \right)} \left( {\it BesselJZeros} \left( 0,i \right) \right) ^{2}-4\,{{ J}_{1}\left({\it BesselJZeros} \left( 0,i \right) \right)} ...


4

Let we set $A=x^2,B=y^2,C=z^2$ and: $$ S_n(A,B,C) = \frac{A^n}{(A-B)(A-C)}+\frac{B^n}{(B-A)(B-C)}+\frac{C^n}{(C-A)(C-B)} .\tag{1}$$ By partial fraction decomposition, it is straightforward to check that $S_0=S_1=0$ and $S_2=1$. For every $n>2$ induction gives: $$ S_n(A,B,C) = h_{n-2}(A,B,C)\tag{2} $$ where $h_k$ is a complete homogeneous symmetric ...


0

Using the integration by parts, one has \begin{eqnarray} \int_0^1\ln x\ln(1-x)dx&=&x\ln x\ln(1-x)|_0^1-\int_0^1x(\frac{\ln(1-x)}{x}-\frac{\ln x}{1-x})dx\\ &=&-\int_0^1(\ln x+\ln(1-x))dx+\int_0^1\frac{\ln x}{1-x}dx\\ &=&2-\frac{\pi^2}{6}. \end{eqnarray} Here $$ \int_0^1\frac{\ln x}{1-x}dx=-\frac{\pi^2}{6} $$ is well-known.


4

Before providing my solution, I'd must admit that Oliver Oloa provides the way to calculate this integral. I merely provide a different approach, using Fourier transforms. First a comment. I tried to use a symmetry argument saying that $$ \int_0^{+\infty}f(x+1/x)\arctan x\frac{dx}{x} =\frac{\pi}{4}\int_0^{+\infty} f(x+1/x)\frac{dx}{x}, $$ but I was not ...


1

The factors $n(n-1)$ and $2n(2n-1)$ indicate that we could try integration by parts twice. We obtain for $n\geq 2$ \begin{align*} V_n&=\int_0^1e^xU_n(x)dx\\ &=\int_0^1e^xx^n(1-x)^ndx\\ &=\left.e^xx^n(1-x)^n\right|_0^1-\int_0^1e^x\left[nx^{n-1}(1-x)^n-nx^n(1-x)^{n-1}\right]dx\tag{1}\\ ...


3

This has to do with the fact that using integration by parts you get a dilogarithm, and $\text{Li}_2(1)=\zeta(2)$, since the two defining series happen to be the same at $x=1$. Why $\zeta(2)$ involves $\pi^2$ can be seen as a by-product of the series expansion of $\sin(x)$, as can be read about here, on the solution to the Basel problem. In short one shows ...


5

Since $x^2-x=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}$ we have, for $x\in[0,2]$, $$-\frac{1}{4}\le x^2-x \le 2$$ Then \begin{align*} \int_0^2e^{-\frac{1}{4}}\,dx&\le\int_0^2e^{x^2-x}\,dx\le\int_0^2e^2\,dx\\ 2e^{-\frac{1}{4}}&\le\int_0^2e^{x^2-x}\,dx\le 2e^2 \end{align*}


5

Differentiation under the integral sign gives: $$ \int_{0}^{1}\log(x)\log(1-x)\,dx = \left.\frac{\partial^2}{\partial\alpha\,\partial\beta}\int_{0}^{1}x^\alpha(1-x)^{\beta}\,dx\right|_{\alpha,\beta=0} $$ hence you just have to differentiate a beta function and $\zeta(2)$ arises as $\psi'(1)$. Differentiation is carried on through: $$ \frac{d}{dz}\,f(z) = ...


8

I just want to seek ways that have nothing to do with $\ln (\sin x)$. Hint. You may consider $$ I(a):=\int_0^\infty\frac{\arctan (ax)}{x(x^2+1)}\:\mathrm dx,\quad 0<a<1, \tag1 $$ and obtain $$ I'(a)=\int_0^\infty\frac1{(x^2+1)(a^2x^2+1)}\:\mathrm dx. $$ By using partial fraction decomposition, we have $$ ...


1

One little picture says more than a long speech ! And brute force calculus :


2

There is a way to deal with the integral using integration in the complex plane. For simplicity, let's consider the case $n_1=n_2=1$. In this case, we apply Cauchy's theorem along an appropriate contour to convert this troublesome Fourier integral into a finite definite integral. The finite definite integral may be evaluated in closed form. As a bonus, ...


0

It seems that we only need the first integral to dominate the negative term. We use the following general inequality (See for example exercise 7.17 in Apostol - Mathematical analysis): If $f$ and $g$ are both increasing or decreasing on $[a,b]$, then $$ (b-a)\int_a^b f(t)g(t)\,dt\geq \int_a^b f(t)\,dt\int_a^b g(t)\,dt. $$ Now, both $f(t)=t^{b-1}$ ...


1

You are indeed correct. It is a general fact that $$\frac{d}{dx} \int_c^{g(x)} f(t) dt=g'(x) f(g(x))$$ due to both the FTC and chain rule. Your book likely had a typo.


1

(A) is false: On $[0,1]$ take $f(x)= x^2-1/2$ and $P= \{0,1\}.$ Then $M(P,f) = 1/2, \int_0^1f = -1/6, \sup |f| = 1/2.$ So the left side equals $2/3,$ the right side equals $1/2.$ (C ) is true: We have $$M(P,f) - \int_a^b f = \sum_{k=1}^{n}\int_{x_{k-1}}^{x_k}(M_k - f) = \sum_{k=1}^{n}\int_{x_{k-1}}^{x_k}(f(c_k) - f(t))\ dt$$ $$ \implies |M(P,f) - ...


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 ...


5

This is not a novel result, but rather an alternate derivation of @Start wearing purple's result. I tried to keep everything simple and explained, which resulted in a slightly verbose solution. So if you are not interested in details, you may follow only the tagged equations until Step 3. Step 1 (Reduction of Integral). We begin with the following ...


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( ...


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 $$ I = \int_{-\infty}^\infty (\cosh^2 t)^{-3/2} \cosh t\,dt= \int_{-\infty}^{\infty} (\cosh t)^{-2} dt = \tanh t \big|_{-\infty}^\infty = 2 $$


0

Another strategy: Consider a variable slope straight line passing through the point $(2,-2)$: $\;y+2=t(x-2)$ and form the equation for the abscissae of the intersection points of the straight line with the parabola: $$y=t(x-2)-2=x^2\iff x^2-tx+2(1+t)=0$$ The line is tangent to the parabola if and only if this equation has a double root, i.e. ...


0

Suppose you want to find the tangent line from $(\alpha,\beta)$ to a function $f(x)$. You should be able to prove that the point $(u,f(u))$ where the line meets the function $f(x)$ satisfy $$f'(u)=\frac{f(u)-\beta}{u-\alpha}.$$


0

Here is a hint: you need to find the points $(a,a^2)$ on the parabola such that the tangent line through that point also passes through $(2,-2)$.


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

You can integrate $U_n \cdot e^x$, $U_{n-1} \cdot e^x$ and $xU_{n-1} \cdot e^x$ by parts then look for a suitable linear combination. Alternatively, you can find directly that $$ (U_n +2n(2n-1)U_{n-1}-n(n-1)U_{n-2})\cdot e^x = \Big((U_n-n(1-2x)U_{n-1})\cdot e^x\Big)'. $$


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 $$ ...


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} ...


6

Here is how one can compute $\mathcal S_m$ for arbitrary $m$. Your formula (3) can be rewritten as $$\mathcal{S}_m=-m\int_0^1\frac{\ln\frac{1+z}{2}}{z^m-1}z^{m-1}dz=-m\sum_{k=0}^{m-1}\alpha_{km}\int_0^1\frac{\ln\frac{1+z}{2}}{z-e^{2\pi i k/m}}dz,$$ where $$\alpha_{km}=\lim_{\quad z\to\; \exp{\frac{2\pi i k}m}}\frac{z^{m-1}\left(z-e^{2\pi i ...


2

Hint: Let $I(a)=\displaystyle\int_{-\infty}^\infty e^{-ax^2}~dx$, and then evaluate $I'(1)$.


3

Ok i want to tackle the problem from a different starting point: Recall that the $\text{Li}_2(x)$ has the following integral representation which is closely related to the Debye functions, well known in condensed matter physics: $$ \text{Li}_2(z)=\frac{z}{\Gamma(2)}\int_{0}^{\infty}\frac{t}{e^t-z}dt \quad \quad(1) $$ This leads us to the following ...


2

This is not the full solution, it's too long for a comment, and gives some intuition of @user153012's observation about the appearance of $\operatorname{Li_2(\frac12)}$ and $\ln^2 2$. We can see that $$S_m=\sum_{n,k=1}^{\infty}\frac{(-1)^n}{k(mk+n)}=\sum_{n,k=1}^{\infty}\frac{(-1)^n}{k}\int_0^1 x^{mk+n-1}dx=\int_0^1 \frac{\ln(1-x^m)}{1+x}$$ Now we can divide ...


4

Another way to evaluate the integral $$I_{2} = \int_{0}^{\infty} \frac{\arctan (\frac{1}{x}) \log(1+x^{2})}{x} \, dx$$ in Jack D'Aurizio's answer (which according to Wolfram Alpha evaluates to $\frac{\pi^{3}}{12}$) is to consider the complex function $$f(z) = \frac{\arctan \left(\frac{1}{z} \right) \log(1-iz)}{z} = \frac{\text{arccot}(z) \log(1-iz)}{z}.$$ ...


3

And what about this one? $$\mathcal{S}_6 \stackrel{?}{=} \ln^2(2) + \ln(2)\ln(3) -\frac{5\pi^2}{36} .$$ Furthermore, you probably know that $$ \operatorname{Li}_2\left(\frac{\sqrt5 - 1}{2}\right) = \frac{\pi^2}{10} - \ln^2\left(\varphi\right) = \frac{\pi^2}{10} - \ln^2(2) -\ln^2\left(1+\sqrt{5}\right) + 2\ln(2)\ln\left(1+\sqrt{5}\right), $$ ...


10

With a substitution and a step of integration by parts, the problems boils down to computing: $$ I = -4\int_{0}^{+\infty}\frac{\log(1+x^2)\,\text{Li}_2(-x^2)}{x^2}\,dx. \tag{1}$$ Integrating by parts again, the problem boils down to computing: $$ I_1 = \int_{0}^{+\infty}\frac{\log(1+x^2)^2}{x^2}\,dx,\qquad ...


1

A variant: $$ \int_1^{\infty}\frac{\sin(x-1)}{x}dx=\Im\int_1^{\infty}\frac{e^{i(x-1)}}{x}dx= \quad (1)\\ \Im\int_1^{\infty}\int_0^{\infty}e^{i(x-1)-xt}dtdx=\Im\int_0^{\infty}\int_1^{\infty}e^{i(x-1)-xt}dxdt=\quad (2) \\ \Im\int_0^{\infty}\frac{e^{-t}}{t-i}dt=-\int_0^{\infty}\frac{e^{-t}}{1+t^2}dt=\quad (3) \\ \int_0^1\frac{1}{1+\log^2(x)}dx\quad (4) $$ ...


12

Hint. You may observe that $$ \frac1{1+\ln^2 x}=-\Im \frac1{i-\ln x}=-\Im \int_0^{\infty}e^{-(i-\ln x)t}dt,\quad x \in (0,1), $$ giving $$ \begin{align} \int_0^1 \frac{1}{1+\ln^2 x}\,dx&=-\Im \int_0^1\!\!\int_0^{\infty}e^{-(i-\ln x)t}dt\:dx\\\\ &=-\Im \int_0^{\infty}\!\!\left(\int_0^1x^t dx\right)e^{-it}dt\\\\ &=-\Im ...


13

Define $I(a)=\int_0^{\infty} e^{-(x+1)a}\frac{\sin(x)}{x+1}dx$. Then $\displaystyle I'(a)=-\int_0^{\infty} e^{-(x+1)a}\sin(x)dx=-\frac{e^{-a}}{a^2+1}$, and since $\lim_{a\to \infty} I(a)=0$, we have $\ \displaystyle I(0)=\int_0^{\infty} \frac{\sin(x)}{x+1}=\int_0^{\infty}\frac{e^{-a}}{a^2+1}da=\int_0^1\frac{dx}{1+\ln^2 x}.$


2

If$$F(x)=\int_{g(x)}^{h(x)}f(t)\,dt$$then $$F'(x)=f(h(x)).h'(x)-f(g(x)).g'(x)$$ I think you are wrong to calculate $g'(x)$. It will be identically zero. In your problem , $g'(x)=2x\sin\{(1+x^2)^3\}-2x\sin\{(1+x^2)^3\}=0$. So $g''(x)=0.$


1

You have nice answers already. I add this, since (to me) it is more elementary. Integrating by parts, $$ \int \frac{1}{2}x\times 2x e^{-x^2}\,dx = -\frac{1}{2}xe^{-x^2}+\int \frac{1}{2}e^{-x^2}\,dx. $$ With the bounds the "out-integrated" part vanishes, and we find that $$ ...


2

Hint. A potential problem for convergence is near $x_1$. Since $x \mapsto V(x)$ is smooth, then as $x \to x_1^-$, by the Taylor expansion we have $$ V(x)=V(x_1)+(x-x_1)V'(x_1)+\mathcal{O}\left( x-x_1\right)^2 $$ or $$ V(x)=E_0-(x_1-x)V'(x_1)+\mathcal{O}\left( x_1-x\right)^2. \tag1 $$ Case 1. $V'(x_1)\neq0.$ Clearly, since $V(x)<E_0$ for any ...


1

Note that $$\left( \int_{-\infty}^{\infty} x^2e^{-x^2}\, dx \right)^2 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} x^2y^2e^{-(x^2+y^2)}\, dx\, dy$$ In polar coordinates, that is $x=r\cos \theta$ and $y=r\sin \theta$, this integral becomes $$\int_0^{2\pi} \int_0^{\infty} r^4 \sin^2\theta \cos^2\theta e^{-r^2}\, \cdot r\, dr\, d\theta$$ which in turn is ...


1

$$ \int_{a} ^p f(\theta) d \theta = \int_{a/p} ^{p/p} f \left( p \theta \right) p d \theta = p \int_{a/p} ^{1} f \left( p \theta \right) d \theta $$ is the correct equality.


3

Another chance is given by the substitution $x=\sqrt{u}$ and the definition of the $\Gamma$ function: $$ \int_{\mathbb{R}}x^2 e^{-x^2}\,dx = 2\int_{0}^{+\infty}x^2 e^{-x^2}\,dx = \int_{0}^{+\infty}u^{1/2}e^{-u}\,du = \Gamma\left(\frac{3}{2}\right)=\frac{1}{2}\Gamma\left(\frac{1}{2}\right)=\frac{\sqrt{\pi}}{2}.$$


2

Since the complete answer has been given, I will complete my hint. $$ \begin{align} \int_{-\infty}^\infty e^{-x^2}x^2\,\mathrm{d}x &=\frac12\int_{-\infty}^\infty e^{-x^2}x\,\mathrm{d}x^2\tag{1}\\ &=-\frac12\int_{-\infty}^\infty x\,\mathrm{d}e^{-x^2}\tag{2}\\ &=\frac12\int_{-\infty}^\infty e^{-x^2}\,\mathrm{d}x\tag{3}\\ ...


5

Maybe it's interesting to note that we can also solve the integrals using complex analysis, noting that $$\int_{0}^{\infty}\frac{\sin\left(u\right)}{u\left(1+u^{2}\right)}du=\frac{1}{2}\textrm{Im}\left(\int_{-\infty}^{\infty}\frac{e^{iu}-1}{u\left(1+u^{2}\right)}du\right) $$ and ...


1

To continue the answer posted by @JackD'Aurizio, recall that $\sin(nx)\sin(mx)=\frac12(\cos((n-m)x)-\cos((n+m)x))$. Then, we have $$\int_{-\pi}^{\pi}\sin(nx)\sin(mx)\,dx=\frac12 \int_{-\pi}^{\pi}\cos((n-m)x)\,dx-\frac12 \int_{-\pi}^{\pi}\cos((n+m)x)\,dx$$ For $n=m$ $$\int_{-\pi}^{\pi}\cos((n-m)x)\,dx=2\pi$$ while for $n\ne m$ we have $$\begin{align} ...


2

You just have to use the main property (orthogonality) of the Fourier base. Provided that $a,b\in\mathbb{N}$, $$ \int_{-\pi}^{\pi}\sin(ax)\cos(bx)\,dx=0\tag{1}$$ as well as: $$\int_{-\pi}^{\pi}\cos(ax)\cos(bx)\,dx=\int_{-\pi}^{\pi}\sin(ax)\sin(bx)\,dx= \pi\cdot\delta(a,b)\tag{2} $$ hence your integrals equal $\frac{\pi}{10}$ and $0$ by termwise integration. ...


6

$$I = \int_0^{\frac{\pi }{2}} {\sin (\tan x)\cot xdx} - \int_0^{\frac{\pi }{2}} {{{\cot }^2}x[1 - \cos (\tan x)]dx} $$ For the first integral, making $\tan x = u$ gives $$\int_0^{\frac{\pi }{2}} {\sin (\tan x)\cot xdx} = \int_0^{ + \infty } {\frac{{\sin x}}{{x(1 + {x^2})}}dx} = \int_0^{ + \infty } {\frac{{\sin x}}{x}dx} - \int_0^{ + \infty } ...



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