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3

$$\begin{align} \int_0^1\frac{x^3-x^2}{\ln x }\mathrm{d}x &=\int_0^1\frac{x^3-1-x^2+1}{\ln x }\mathrm{d}x\tag{1}\\ &=\int_0^1\frac{x^\color{blue}{3}-1}{\ln x }\mathrm{d}x-\int_0^1\frac{x^\color{red}{2}-1}{\ln x}\mathrm{d}x\tag{2}\\ &=\ln(\color{blue}{3}+1)-\ln(\color{red}{2}+1)\tag{3}\\ &=\ln\frac43\tag{4}\\ \end{align}$$ ...


1

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3

Hint: These are two curves of the form $f(x)$ and $g(x)$ and we know that the area between $f(x)$ and $g(x)$ is $$ \displaystyle\int_{a}^{b} \left| f(x) - g(x) \right| \, \mathrm{d}x, $$ where $x=a$ and $x=b$ represent the $x$-coordinates of the points of intersection of $f(x)$ and $g(x)$. Can you finish from here?


3

By series expansion $$\displaystyle \int^1_0 \frac{\text{Li}_2(x)^2}{x}\,dx=\sum_{k,n\geq 1}\frac{1}{(nk)^2}\int^1_0x^{n+k-1}\,dx =\sum_{k,n\geq 1}\frac{1}{(nk)^2(n+k)}$$ By some manipulations $$\sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{k}{n^2(n+k)}= \sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{1}{n^2}-\sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq ...


1

Writing $f_n\colon x\in(0,1]\mapsto \frac{n x^p+x^q}{n x^q+x^p}$ pointwise convergence to $f\colon x\in(0,1]\mapsto x^{p-q}$ (with $p-q > -1$), which is integrable on $(0,1]$. for all $n\geq 1$ and $x\in(0,1]$, $$ 0 \leq f_n(x) \leq \frac{n x^p+x^q}{n x^q} = \frac{1}{x^{q-p}}+\frac{1}{n} \leq \frac{1}{x^{q-p}}+1= g(x) $$ where $g$ is integrable on ...


1

Hint: $$\frac{nx^p+x^q}{nx^q+x^p}-x^{p-q}=\frac{x^q-x^{2p-q}}{nx^q+x^p}.$$ If you prove that the last term is small, then: $$\int_{0}^{1}\frac{nx^p+x^q}{nx^q+x^p}\,dx \sim \int_{0}^{1}x^{p-q}\,dx = \frac{1}{p-q+1}.$$


3

$$\int_{-1}^1f(x)dx=\int_{-1}^0f(x)dx+\int_0^1f(x)dx$$ By the ranges of definition, this $$=\int_{-1}^0x\ dx+\int_0^1 x^2\ dx$$


0

Here's an idea for even n; I haven't looked at any of the links above, so hope this is not the same as that in Strehlke. [Ok, after finishing this, took a look at Strehlke. Approach below is not the same as his.] The sinc function is the Fourier transform of a boxcar. Here I'll work with a boxcar that is 1 for x in -1/2,1/2 and zero otherwise. The ...


2

The $\sqrt{2}$ is a complete red herring. In fact consider $$I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\alpha}(x)}\,dx$$ where $\alpha$ is any nonnegative real number. Then we have \begin{align} I&=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\alpha}(x)}\,dx \\&=\int_0^{\frac{\pi}{2}}\frac{1}{1+\frac{\sin^{\alpha}(x)}{\cos^\alpha(x)}}\,dx ...


5

I finally figured it out due to a hint by Sameer Kailasa. $$\int_{0}^{\frac{\pi}{2}}{\frac{1}{1+\tan^\sqrt{2}(x)}} \ dx$$ $$u=\frac{\pi}{2}-x \implies du=-dx$$ $$= -\int_{\frac{\pi}{2}}^{0}{\frac{1}{1+\cot^\sqrt{2}(x)}} \ dx = \int_{0}^{\frac{\pi}{2}}{\frac{\tan^\sqrt{2}(x)}{\tan^\sqrt{2}(x)+1}} \ dx $$ ...


4

$\sqrt[3]{2x^2-x^3} = x\sqrt[3]{\dfrac{2}{x}-1} \to u = \sqrt[3]{\dfrac{2}{x} - 1} \to u^3 = \dfrac{2}{x} - 1 \to x = \dfrac{2}{u^3+1}$. Can you take it from here?


9

Here is an elementary way to evaluate the integral without involving any special functions or advance formulas. Notice that $$\int_0^\infty e^{-y}\sin(xy)\;\mathrm dy=\frac{x}{1+x^2}$$ Hence we have \begin{align} \int_0^\infty \frac{x}{\left(1+x^2\right)\left(e^{2\pi x}+1\right)}\,\mathrm dx&=\int_0^\infty\int_0^\infty \frac{e^{-y}\sin(xy)}{e^{2\pi ...


9

A Generalisation: \begin{align} \int^\infty_0\frac{x}{(x^2+w^2)(1+e^{2\pi x})}{\rm d}x\tag1 =&\int^\infty_0\frac{xe^{-x}}{(x^2+4\pi^2w^2)(1+e^{-x})}{\rm d}x\\ \tag2 =&\int^\infty_0\frac{x}{x^2+4\pi^2w^2}\left(\sum^\infty_{n=1}(-1)^{n-1}e^{-nx}\right){\rm d}x\\ \tag3 ...


1

This is a partial answer and I need more work. Denote the original integral by $I$. From M.N.C.E's idea, define $$ I_1=\int_0^1\frac{\arctan x\ln^4x}{x}dx, I_2(a)=\int_0^1\frac{x\arctan(a x)\ln^4x}{1+x^2}dx $$ and then $I_2(0)=0$ and $I=2I_1-4I_2(1)$. Now \begin{eqnarray} I_1&=&-\frac15\int_0^1\frac{1}{1+x^2}\ln^5xdx\\ ...


6

Hint: make use of the Binet's second formula http://mathworld.wolfram.com/BinetsLogGammaFormulas.html. $$\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}+1)} dx$$ $$=\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}-1)} dx-2\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{4\pi x}-1)} dx$$ $$=\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}-1)} ...


1

You are calculating the area in the following graph and your calculation is correct. But the question asks for the area "inside" both graphs which can be seen below: This area is then $$A=\int_0^{\pi/3}(1-\cos x)^2dx+\int_{\pi/3}^{\pi/2}\cos^2xdx$$ which is what the back of your book says.


0

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3

We want to figure out when $e^{-x} = x$. Multiplying by $e^x$ gives $$ xe^x = 1 $$ The solution to this equation is defined as the $\Omega$-constant, and shares many interesting properties. So we have $\Omega e^\Omega=1$ or $e^{\Omega} = 1/\Omega$ or $\Omega=e^{-\Omega}$. Several fast approximations can be found at the link above. The integral is then ...


0

A typical way to prove that a function $f$ is constant is to show that $$\int_0^\pi f(x)g(x)\,dx=0\tag{1}$$ for every function $g\in C[0,\pi]$ with zero mean (i.e., $\int_0^\pi g(x)\,dx=0$). Indeed, if $f(x_1)\ne f(x_2)$, then let $g$ be the function that is zero on most of the interval and has two triangular peaks at $x_1,x_2$: one positive, one negative. ...


0

$$\lfloor 1 - \log_2 x \rfloor = \lfloor \log_2 2 - \log_2 x \rfloor = \left\lfloor \log_2 \frac{2}{x} \right\rfloor.$$ Now note $$n \le \log_2 \frac{2}{x} < n+1$$ whenever $2^{n-1} \le \frac{1}{x} < 2^n$, or equivalently, $$2^{-n} < x \le 2^{1-n}.$$ So we immediately have $$\int_{x=0}^1 \frac{dx}{\lfloor 1 - \log_2 x \rfloor} = \sum_{n=1}^\infty ...


3

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2

Let's see, if you do $u=1-\log_2 x$ then $$-2^{1-u}(\log 2)\, du=dx$$ Then you get $$(\log 2)\cdot \int_{1}^\infty{2^{1-u}\over \lfloor u\rfloor}\,du$$ This is clearly $$2(\log 2)\sum_{n=1}^\infty \int_n^{n+1}{2^{-u}\over \lfloor u\rfloor}\,du$$ But on the interval $[n,n+1)$ by definition the floor function is equal to $n$, so the integrand is just ...


4

Here is one line proof $$\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\cos^4x}dx=\int_0^{\pi/2}\frac{\tan(x) (\tan(x))'}{\tan^4(x)+1}\ dx=\int_0^{\infty} \frac{x}{x^4+1}\ dx=\left[\frac{\arctan(x^2)}{2}\right]_0^{\infty}=\frac{\pi}{4}$$ Q.E.D.


4

\begin{align} \int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\cos^4x}\mathrm dx&=\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\left(1-\sin^2x\right)^2}\mathrm dx\\[7pt] &=\int_0^{\pi/2}\frac{\sin x\cos x}{2\sin^4x-2\sin^2x+1}\mathrm dx\\[7pt] &=\frac14\int_0^1\frac{\mathrm dt}{t^2-t+\frac12}\qquad\color{blue}{\implies}\qquad t=\sin^2x\\[7pt] ...


0

You can use $\displaystyle V=\int_0^{2\pi}\int_{\frac{\pi}{2}}^{\frac{3\pi}{4}}\int_0^{\sqrt{6}}\rho^2\sin\phi\; d\rho\; d\phi\; d\theta$


4

First, let's split up the integral: $$ \begin{align} \int_0^\infty\frac{x^n}{1+e^{x-t}}\mathrm{d}x &=\int_{-t}^\infty\frac{(x+t)^n}{1+e^x}\mathrm{d}x\\ &=\color{#C00000}{\int_{-t}^0\frac{(x+t)^n}{1+e^x}\mathrm{d}x}+\color{#00A000}{\int_0^\infty\frac{(x+t)^n}{1+e^x}\mathrm{d}x}\tag{1} \end{align} $$ Note that the first integral on the right side of ...


8

$1 - \dfrac{\sin^2(2x)}{2} = \dfrac{1+\cos^2(2x)}{2}$, and $\sin x\cos x = \dfrac{\sin (2x)}{2} \Rightarrow \displaystyle \int \dfrac{\sin x\cos x}{\cos^4x+\sin^4x}dx = \displaystyle \int -\dfrac{1}{2}\dfrac{d(\cos(2x))}{1+\cos^2(2x)}dx = -\dfrac{1}{2}\arctan(\cos (2x)) + C$


3

We have: $$ F_0(t) = \log(e^t+1) = t+\log(1+e^{-t}) = t+o(1) $$ and since your identity gives: $$ F_n(t) = n\int_{0}^{t} F_{n-1}(u)\,du $$ we have: $$ F_1(t) = \frac{t^2}{2}+o(t), $$ $$ F_2(t) = \frac{t^3}{3}+o(t^2), $$ and so on, so the claim holds by induction for any $n\in\mathbb{N}_{>0}$. Moreover, since: $$ F_n(t) = ...


3

This is not an answer, but I would like to share what I did on this. Let $s=1-w^2$ and $y=z^2$ to have $$I=4\int_0^1\int_0^1\frac{wz\arcsin wz}{\sqrt{1-z^2}(1-w^2z^2)}dwdz$$ Now using $wz=u$ and $z=v$ we obtain $$I=4\int_0^1\int_0^v\frac{u\arcsin u}{v\sqrt{1-v^2}(1-u^2)}dudv$$ Approach $1$: write both $\arcsin u$ and $(1-u^2)^{-1}$ in terms of Taylor's ...


0

The change of variables should be in the form not $x=g(θ)$, but $\theta=g(x)$, so you should fix a branch $g$ of the multivalued function $\arccos$.


1

The conversion between cylindrical and Cartesian coordinates is $x = r\cos\theta$, $y = r\sin\theta$, $z = z$. Thus, the density $\rho(x,y,z) = 3-z$ (Cartesian) becomes $\rho(r,\theta,z) = 3-z$ (cylindrical).


0

Normally, just like $$\iiint_\Omega dV = \text{Volume}(\Omega),$$ so too $$\iiint_\Omega \rho(x,y,z) dxdydz = \text{Mass}(Omega),$$ where $\rho$ denotes density and $\Omega$ denotes your region.


3

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2

$$ \int \arctan(\frac{y}{x})\;dx, \;\text{ Let } u=\arctan\frac{y}{x}\; \Rightarrow \; du = \frac{1}{(\frac{y}{x})^2+1}\cdot \frac{-y}{x^2}dx \; \Rightarrow \; du = \frac{-y}{x^2+y^2}dx \\ \text{and let } dv=dx \; \Rightarrow \; v=x \\ \text{Therefore} \\ \int u \;dv = uv \; - \; \int v\;du \;\; \Rightarrow \; \; \int \arctan\frac{y}{x}dx = \; x\cdot ...


4

Main Idea In short, reverse the order of integration. Then it works very simply.


0

You are missing the $\text{d}x$ on your integrals. Aside from that, the area is \begin{align} \int_1^a \pi (1 - x^{-5/2})^2 \ \text{d} x &= \pi \left[ \frac{4}{3x^{3/2}} - \frac{1}{4x^4} + x \right]_1^a \\ &= \pi \left[\frac{4}{3a^{3/2}} - \frac{1}{4a^4} + a - \frac{25}{12} \right] \\ \end{align} Hence, $$ \lim_{a \to \infty} \int_1^a \pi (1 - ...


4

All you need is that $f$ is continuous with $f(1) = 1$. Note that for any $\delta \in (0,1)$, $$\int_0^{1-\delta} y x^y\; dx = \dfrac{y}{y+1} (1-\delta)^{y+1} \to 0$$ while $$ \int_0^1 y x^y\; dx = \dfrac{y}{y+1}\to 1$$ Take $\delta$ so that $|f(x) - f(1)| < \epsilon$ for $1-\delta < x \le 1$, and use the fact that $f$ is bounded...


0

Let $$I=\int_{-1}^1 \frac{\cos (x)}{a^x+1}\mathrm dx\tag1$$ Using identity $$\int_a^bf(x)\;\mathrm dx=\int_a^bf(a+b-x)\;\mathrm dx$$ we get $$I=\int_{-1}^1 \frac{\cos (-x)}{a^{-x}+1}\mathrm dx=\int_{-1}^1 \frac{a^{x}\cos (x)}{1+a^{x}}\mathrm dx\tag2$$ Adding $(1)$ and $(2)$, we get $$2I=\int_{-1}^1\cos x\;\mathrm dx=\sin x\,\Bigg|_{-1}^1=2\sin ...


3

I assume $m$ is supposed to be an integer. Let $$I_m = \int_\alpha^\infty \dfrac{e^{-At}}{(1+Bt)t^m}\; dt$$ Note that $$B I_m + I_{m+1} = \int_\alpha^\infty \dfrac{e^{-At}}{t^{m+1}}\; dt = J_{m+1}$$ where $$ J_m = \int_{\alpha}^\infty \dfrac{e^{-At}}{t^m}\; dt = \alpha^{1-m} \int_{1}^\infty \dfrac{e^{-A\alpha s}}{s^m}\; ds = \alpha^{1-m} E_m(A\alpha)$$ ...


-1

It's easy. Differentiate $$\frac{1}{4 (-1 + 2 x^2)} - \frac{\sqrt{2 + x^2 - x^4}}{6 (-1 + 2 x^2)} - {1\over 4} \log[3 + 2 \sqrt{2 + x^2 - x^4}]$$ To get your integral.


2

I want to point that the integrals $$I = \int e^{ax}\cos{bx} \qquad J = \int e^{ax}\sin{bx}$$ can be solved by using Euler's formula, as follows Consider the integral $$A = \int e^{ax}\cos{bx}dx -i\int e^{ax}\sin{bx}dx = \int e^{(a+ib)x} dx$$ Therefore $I = Re\{A\}$ and $J = Im\{A\}$ This approach is suitable for this problem


6

Just for references, I remark that the following proposition was proved in my answer: Proposition. If $0 < r < 1$ and $r < s$, then $$ I(r, s) := \int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}} \log \left( \frac{1 + 2rsx + (r^{2} + s^{2} - 1)x^{2}}{1 - 2rsx + (r^{2} + s^{2} - 1)x^{2}} \right) \, dx = 4\pi \arcsin r. \tag{*} $$ Recently, I ...


2

Both integrals are the same except the minus sign and the end points of the intervals. Let's use integration by part to find the anti-derivative for $e^{-x}\sin x$. $\displaystyle \int e^{-x}\sin xdx = -\displaystyle \int \sin xd(e^{-x}) = -\left(\sin xe^{-x} - \displaystyle \int e^{-x}\cos xdx\right) = -\sin xe^{-x} + \displaystyle \int e^{-x}\cos xdx = ...


1

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6

By clever substitutions your professor probably meant Euler Substitutions. $$\begin{align} I &=\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx\tag{1}\\ &=\frac12\int_0^1 \frac{t}{2(2-t)(1+t) + 3\sqrt{(2-t)(1+t)}}\,\mathrm dt\tag{2}\\ &=\frac13\int_{\sqrt2}^{1/\sqrt2} \frac{u^2-2}{(1+u)^2(u^2+1)}\,\mathrm du\tag{3}\\ ...


3

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1

Separating your variables appropriately, we have $$\int_0^T \delta(t-t_j)e^{-(s+\mu\lambda^2)t}\,dt\int_0^l \delta(x-R)\varphi(x)\,dx.$$ Since $0 < R < l$, the sifting property of the Dirac delta gives us $\varphi(R)$ for the second integral. The first is very similar since $0 < t_j < T$.


6

Here is a complex-analytic method: Notice that $$ \int_{0}^{\infty} \log\left( \frac{x^{2} + 2x\cos b + 1}{x^{2} + 2x\cos a + 1} \right) \frac{dx}{x} = 2 \int_{0}^{\infty} \Re \frac{\log(1 + e^{ib}x) - \log(1 + e^{ia}x)}{x} \, dx. \tag{1} $$ Let $R$ be a positive large number. Then the function $z \mapsto \log(1+z)/z$ is analytic on $\Bbb{C} \setminus ...


5

Another approach is to use the Fourier series $$\sum_{k=1}^{\infty}\frac{x^{k} \cos(ka)}{k} = - \frac{1}{2} \log \left(x^{2} - 2 x \cos(a) +1 \right) \ , \ |x| <1 $$ which can be derived from the Maclaurin series of $\log(1-z)$ by replacing $z$ with $xe^{ia}$ and equating the real parts on both sides. $$ \begin{align} ...


5

Let us first substitute $y = x^{2} + 1$. Then \begin{align*} I &:= \int_{0}^{1} \frac{x^{3}}{2(2-x^{2})(1+x^{2}) + 3\sqrt{(2-x^{2})(1+x^{2})}} \, dx \\ &= \frac{1}{2} \int_{1}^{2} \frac{y - 1}{2y(3-y) + 3\sqrt{y(3-y)}} \, dy \tag{1} \end{align*} Using the substitution $y \mapsto 3-y$, it follows that $$ I = \frac{1}{2} \int_{1}^{2} ...



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