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6

No, for $$ f(t)=t, $$ we have $f(0)=0$ and $$ \int_0^t s^{-1}f(s)\,ds=t=f(t),\quad\forall t>0, $$ so (your constant is $1$, and you actually have equality) $$ f(t)\leq 1\int_0^t s^{-1}f(s)\,ds,\quad\forall t>0. $$


6

HINT: rewrite \begin{align*} x^5\sqrt{x}+x\sqrt[4]{x} &=x^5\cdot x^{1/2} + x\cdot x^{1/4}\\ &= x^{11/2}+x^{5/4} \end{align*} Now use the power rule.


5

It should be fairly easy to follow the path outlined by John Hughes' comment, but even quicker to see that the integral must be $0$ due to symmetry. Along the diagonal $x+y=\pi$, we have $\sin(x+y)=0$. A point above the diagonal has a negative value that corresponds exactly to the positive value at the point you get by reflecting across the diagonal. (And ...


5

There are several problems with integration by $d(rx)$. The first is the meaning, as usually taught in first-year calculus. The $d(xr)$ in an integral means that the variable of integration is $xr$. However, you have no $xr$ in your integrand, even though your variable $r$ does change when $xr$ changes. In an integral, you need your $d$ term to point out the ...


5

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 $$ ...


4

If $n < t < n+1 \to \dfrac{1}{n+1} < \dfrac{1}{t} < \dfrac{1}{n} \to \dfrac{1}{n+1} = \displaystyle \int_{n}^{n+1}\dfrac{1}{n+1} dt < \displaystyle \int_{n}^{n+1} \dfrac{1}{t}dt<\displaystyle \int_{n}^{n+1}\dfrac{1}{n}dt=\dfrac{1}{n}$


4

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) = ...


4

Possible approach: Let $0<a$. $$\begin{align} \mathcal{I}{\left(a\right)} &=\int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{x}+\sqrt{x+1}+\sqrt{x+2}}\\ &=\int_{1}^{a+1}\frac{\mathrm{d}y}{\sqrt{y-1}+\sqrt{y}+\sqrt{y+1}};~~~\small{\left[x+1=y\right]}\\ ...


4

Swap the order of integration: $$\int_{-2}^2\left(\int_{0}^{\sqrt{4-x^2}}y\,dy\right)\,dx$$ But $\int_{0}^{z}y\,dy = \frac{z^2}{2}$, so the above is: $$\int_{-2}^2 \frac{4-x^2}{2}\,dx$$ Which is easy to do. Without swapping: $$\int_{0}^2\left(\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}}y\,dx\right)\,dy$$ With the inner integral equal to $2y\sqrt{4-y^2}$. We ...


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 ...


3

For $t \in (n, n+1)$, $$\frac{1}{n +1} < \frac{1}{t} < \frac{1}{n}.$$ Integrate from $n$ to $n + 1$ to get the desired result. (well done improving your question)


2

To show that $F$ is not conservative on $\mathbb{R}^2 \setminus \{ 0 \}$, compute the curve integral of $F$ along the unit circle. It will turn out to be non-zero, which shows $F$ can't be conservative. For b), the domain $\Omega$ is simply connected, so the fact that $\operatorname{rot}(F) = 0$ on $\Omega$ is enough to guarantee that $F$ is convervative ...


2

In polar coordinates, $r$ is going from $0$ to $2$ and $\theta$ is going from $0$ to $\pi$.


2

If we set $g(t)=\frac{f(t)}{t}$ and $G(t)=\int_{0}^{t}g(u)\,du$, we have the inequality: $$ t\cdot g(t)\leq C\cdot G(t)\tag{1} $$ or: $$ \frac{g(t)}{G(t)}\leq \frac{C}{t}\tag{2} $$ hence integrating both sides over $[\varepsilon,x]$ we have: $$ \log(G(t))-\log(G(\varepsilon))\leq C\left(\log(t)-\log(\varepsilon)\right)\tag{3}$$ or: $$ G(t)\leq ...


2

Make the change of variables $$x=\frac12\left(t^2+t^{-2}-2\right),$$ under which $$\sqrt{x}=\frac{1}{\sqrt2}\left(\frac1t-t\right),\qquad \sqrt{x+2}=\frac{1}{\sqrt2}\left(\frac1t+t\right),\qquad \sqrt{x+1}=\frac{1}{\sqrt2} \frac{\sqrt{t^4+1}}{t}.$$ The integral thus becomes $$\sqrt2\int\frac{t^2-t^{-2}}{2+\sqrt{t^4+1}}dt.$$ The integrand being a rational ...


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, ...


1

APPROACH 1: There were two errors. First, $dx=a\cos z$ and not $1/(a\cos z)$. Second, the upper limit transforms from $x=a$ to $z=\pi/2$. Then, the integral of interest becomes $$\frac83 abc\int_0^{\pi/2}\cos^4 z\,dz=\frac83 abc\times \frac{3\pi}{16}=\frac{\pi}{2}abc$$ APPROACH 2: Here, the substitution $x=az\implies dx=adz$ and when the upper limit ...


1

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


1

You can also do this graphically by drawing $y=\frac1x$ , and the appropriate rectangles as shown below. ie. Area of AFDC < area under the curve < Area of EBDC


1

Hint: $$\int_0^1 (-1)^{[1/x]} \, dx = \lim_{n \to \infty}\sum_{k=1}^n\int_{1/(k+1)}^{1/k} (-1)^{[1/x]} dx = \lim_{n \to \infty}\sum_{k=1}^n\int_{1/(k+1)}^{1/k} (-1)^k dx \\ = \sum_{k=1}^{\infty} (-1)^k \left[\frac1{k}-\frac1{k+1} \right]=\sum_{k=1}^{\infty} \frac{(-1)^k}{k(k+1)}$$ Alternatively consider a partition $P = (0, 1/n, \ldots, 1/2, 1)$. The ...



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