11

$$ I=\int_{0}^{\infty} \frac{\sin x-x \cos x}{x^2+\sin^2 x} dx= - \int_{0}^{\infty}\frac{\frac {x\cos x -\sin x}{x^2}}{1+(\frac{\sin x}{x})^2}dx= -\int_{1}^{0} \frac{dt}{1+t^2}=\frac{\pi}{4}.$$


9

$$I =\int_{-2}^{0}\frac{xdx}{\sqrt{e^x+(x+2)^2}}=\int_{-2}^{0} \frac {xe^{-x/2} dx}{\sqrt{1+(x+2)^2e^{-x}}}=\int_{0}^{2} \frac{-2 dt}{\sqrt{1+t^2}}.$$ $$\Rightarrow I =-2 ~\mbox{arcsinh} t|_{0}^{2} =-2~\mbox{arcsinh} 2= -2 \ln(2+\sqrt{5})$$. Here we have used $t=(x+2)e^{-x/2}$.


8

I prefer to discuss Riemann sums in pictures. Here's the region whose area we are supposed to approximate: You need to use $5$ rectangles. This means that you should subdivide the domain into $5$ equal pieces: Next, we draw perpendicular lines up to the graph, ready to become the sides of rectangles: Now, we we need to choose the height of the rectangles. ...


6

A little late to the party here, but I finally managed to get a solution. From the answer here, and using a trivial rescaling of the integration variable, we can see that: $$f(a,t)=\int_{0}^{\infty}\frac{dx}{x^2+a}e^{-\frac{x^2}{2t}}=\frac{\pi}{2\sqrt{a}}\text{erfc}\Big(\sqrt{\frac{a}{2t}}\Big)\exp\Big(\frac{a}{2t}\Big)$$ As a function of $a$, the above ...


6

Both the function that you are integrating as the region over which you are integrating it get unchanged if you exchange $x$ with $y$. Therefore, your integral is equal to$$2\int_0^a\int_0^x\frac1{(a^2+x^2+y^2)^{3/2}}\,\mathrm dy\,\mathrm dx.$$You can compute this integral using polar coordinates: $\theta$ can take values in $\left[0,\frac\pi4\right]$ and, ...


5

Let $f(s)$ be defined by $$ f(s) = \int_{0}^{\infty} \frac{1-\cos x}{x^s} \, \mathrm{d}x $$ for $1 < s < 3$. By writing $\frac{1}{x^s} = \frac{1}{\Gamma(s)} \int_{0}^{\infty} t^{s-1}e^{-xt} \, \mathrm{d}t$, we get \begin{align*} f(s) &= \int_{0}^{\infty} \frac{t^{s-1}}{\Gamma(s)} \int_{0}^{\infty} (1 - \cos x)e^{-xt} \, \mathrm{d}x \mathrm{d}t \\...


5

Not a closed form, but quite a few curious series. $$I=\int_{0}^{1}\frac{(1-x)\log(1-x)}{\log x}dx=-\int_0^1 \int_0^1\frac{x(1-x)dx dt}{\log x (1- x t)}$$ Now make a substitution $x=e^{-u}$: $$I=\int_0^1 \int_0^\infty \frac{e^{-2u}(1-e^{-u})du dt}{u (1- t e^{-u} )}$$ Now let's expand the bracket in the numerator as a series: $$I=\sum_{k=1}^\infty \frac{...


5

We may apply a discrete Fourier transform to the following generating function $$\sum_{n=1}^\infty \frac{x^n}{n^2}H_n=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\operatorname{Li}_2(1-x)\ln(1-x)+\frac{1}{2}\ln x \ln^2(1-x)+\zeta(3)$$ since $$ I(p) = \sum_{n\geq 1}\frac{H_{p n}}{pn^2}. $$ The only term leading to a non-elementary contribution is the sum ...


4

Just use linearity: $$\int\limits_0^3 (x+f(x))\, dx=\int\limits_0^3 x\, dx+\int\limits_0^3f(x)\, dx,$$ and compute them separately. It looks to me like you integrated $f$ correctly and $x$ incorrectly. For the $x$ integral, you should instead have that $$\int\limits_0^3 x\, dx=\frac{x^2}{2}\Big|_{x=0}^{x=3}=\frac{9}{2},$$ like your answer key says.


4

Surprise ! Considering $$I=\int_{0}^{\infty} \frac{dx}{(x-\log x)^2}$$ and exploring simple linear combinations of a few basic constants, I found (be sure it took time !) $$\color{blue}{I\sim\frac{189}{4}(C+2\pi)+61 \pi \log (3)-\frac{1}{4} \left(57+101 \pi ^2+523 \pi \log (2)\right)}$$ which differs in absolute value by $10^{-18}$.


4

Hint #1: That function is periodic with period $\pi$. So, your integral is$$140\times\int_0^\pi\bigl\lvert\sin^2(x)\cos(x)\bigr\rvert\,\mathrm dx.$$Hint #2: $\displaystyle\int_0^\pi\bigl\lvert\sin^2(x)\cos(x)\bigr\rvert\,\mathrm dx=\int_0^{\frac\pi2}\sin^2(x)\cos(x)\,\mathrm dx-\int_{\frac\pi2}^\pi\sin^2(x)\cos(x)\,\mathrm dx$.


4

Write $x=\tan t$ so your integral is $$\int_0^{\pi/2}\tan^{1/2}tdt=\frac12\operatorname{B}\left(\frac14,\,\frac34\right)=\frac{\pi}{2}\csc\frac{\pi}{4}=\frac{\pi}{\sqrt{2}}.$$See here and here if any of the theory I used is unfamiliar.


3

In $\int_3^8 |g(x)| {\rm d}x$, the value which you are integrating is $|g(x)|$. As Hendrix says in a comment, this is always non-negative. As such, based on what integration means, you need to always take the absolute values of anything you're using first and then add those over the region of integration, i.e., $3$ to $8$. Doing it the other way around ...


3

Using a Riemann Sum (With Right Endpoints): $$\lim_{n \to \infty}(\frac{b-a}{n}\sum_{i=1}^nf(a +i(\frac{b-a}{n})))$$ $$\lim_{n \to \infty}(\frac{2}{n}\sum_{i=1}^n(i(\frac{2}{n}))^2)$$ $$\lim_{n \to \infty}(\frac{2}{n}\sum_{i=1}^n(\frac{4i^2}{n^2}))$$ Factor out the constants $$\lim_{n \to \infty}(\frac{8}{n^3}\sum_{i=1}^n(i^2))$$ Using the summation ...


3

Without complex analysis? Okay... Let $$I=\int_0^\infty\frac{\sqrt{x}}{x^2+1}dx$$ Enforce $x:=y^2\implies dx=2y \space dy$. So $$I=\int_0^\infty\frac{\sqrt{x}}{x^2+1}dx=2\int_0^\infty \frac{y^2}{y^4+1}dy=2\int_0^\infty\frac{y^2}{y^2\left(y^2+\frac{1}{y^2}\right)}dy$$ $$=\int_0^\infty \frac{1-\frac{1}{y^2}+1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy=\int_0^\...


3

I am not sure that I could find a closed formula for the result. Beside numerical integration, I should use the classical series expansion of $\log(1-x)$ and use the long division to get $$\frac 1 {\log(1-x)}=-\frac{1}{x}+\frac{1}{2}+\frac{x}{12}+\frac{x^2}{24}+\frac{19 x^3}{720}+\frac{3 x^4}{160}+\frac{863 x^5}{60480}+\frac{275 x^6}{24192}+O\left(x^7\...


3

Partial solution \begin{align} I&=\int_0^1\frac{1-x}{\ln x}\ln(1-x)\ dx=\int_0^1\left(-\int_0^1x^y\ dy\right)\ln(1-x)\ dx\\ &=\int_0^1\left(-\int_0^1x^y\ln(1-x)\ dx\right)\ dy=\int_0^1\left(\sum_{n=1}^\infty\frac1n\int_0^1x^{n+y}\ dx\right)\ dy\\ &=\int_0^1\left(\sum_{n=1}^\infty\frac{1}{n(n+y+1)}\right)\ dy=\sum_{n=1}^\infty\frac1n\int_0^1\frac{...


3

The lower Riemann sum would be $$\sum_{i=1}^5 \dfrac 1{ x} \Delta x$$ with $\Delta x=\dfrac15$ and $x=1+i\Delta x$. In other words, $\dfrac15\left(\dfrac1{1.2}+\dfrac1{1.4}+\dfrac1{1.6}+\dfrac1{1.8}+\dfrac12\right)$.


3

Hint: Note that the integrand is periodic. You can reduce this problem to finding $$ \lim_{t\to\infty} \frac{1}{t}\int_0^t\cos^3(x)dx.$$


3

First we prove $$\displaystyle\int_0^{\pi/2} \log \left(\sin x\right)\,dx =-\frac12\pi\log2$$ Proof $\ $ From the substitution $x \to \frac{\pi}{2}-x$ , we get $$ \displaystyle\int_0^{\pi/2} \log( \sin x )\,dx = \displaystyle\int_0^{\pi/2} \log ( \cos x )\,dx $$ Thus \begin{align} 2\displaystyle\int_0^{\pi/2} \log( \sin x )\,dx &= \...


3

$$\small \boxed{\int_0^1 \frac{\ln(1-x) \ln(1-x^5)}{x} dx=4\zeta(3)-\frac{\pi}{5}\operatorname{Cl}_2\left(\frac{4\pi}{5}\right)-\frac{3\pi}{5}\operatorname{Cl}_2\left(\frac{2\pi}{5}\right)+\frac32\operatorname{Cl}_3\left(\frac{4\pi}{5}\right)+\frac32\operatorname{Cl}_3\left(\frac{2\pi}{5}\right)}$$ $$\operatorname{Cl}_2\left(x\right)=\sum_{n=1}^\infty \frac{...


3

Let $a>-1$ be a real number. Then $$\int_0^1 \frac{\log(1+a^2x^2)}{1+x^2}\textrm{d}x=\frac{\pi}{2}\log(1+a)-G+\text{Ti}_2\left(\frac{1-a}{1+a}\right),$$ where $G$ is the Catalan's constant and $\displaystyle \text{Ti}_2(x)=\int_0^x \frac{\arctan(t)}{t}\textrm{d}t$ is the inverse tangent integral. Thanks to Cornel for this way of writing the closed-form ...


2

Hint: $$\sum_{i=0}^{n-1}(i\Delta x)^2\Delta x= (\Delta x)^3\sum_{i=0}^{n-1}i^2=\left(\dfrac2n\right)^3\dfrac{(n-1)n(2n-1)}{6}$$


2

Not a full answer, but I hope to add more once I get to Mathematica in a few hours. Using an obvious substitution $x=e^y$, we can transform the integral to: $$I=\int_{-\infty}^\infty \frac{e^y dy}{(e^y-y)^2}=\int_0^\infty \frac{e^y dy}{(e^y-y)^2}+\int_0^\infty \frac{e^{-y} dy}{(e^{-y}+y)^2}$$ Some simple algebra gives us: $$I=2 \int_0^\infty \frac{(1+y^2)...


2

As @saulspatz already pointed out in the comments: There is a difference between multiple integrals and iterated integrals. The term multiple integral refers to integration over multidimensional shapes. Multiple integrals are defined by splitting the shape you want to integrate over into pieces and by building a Riemann sum out of these pieces. If the ...


2

Let $(\Omega, \mathcal{M}, P)$ be a probability space and $X: \Omega\to\mathbb{R}$ (or perhaps to $\mathbb{C}$ or another suitable space) be measurable. The integral of concern is $$ \int_{0}^{\infty}P(\{|X|^{2m} > t \}) \,dt$$ As also suggested by @Vishnuram, we use substitution $t = u^{2m} $, so $dt = 2mu^{2m-1}du$. This doesn't change the bounds of ...


2

Hint: Solve the equation $$\cos(x)=\frac{1}{2}$$ for $$0\le x\le 2\pi$$


2

Long comment: Following the original comment by @automaticallyGenerated it is astounding that $$I=\int_0^{\infty } \frac{1}{(x-\log (x))^p} \, dx= \frac{ 1}{\Gamma(p)} {\int_0^{\infty } \frac{\Gamma(x+1)}{x^{x-p+2}} \, dx}$$ seems to hold true for all real $p>1$. Since two entirely different functions are being integrated, it seems likely that the ...


2

Note that $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\log(x-i\sin(x)) &=\frac{1-i\cos(x)}{x-i\sin(x)}\\ &=\frac{x+\sin(x)\cos(x)}{x^2+\sin^2(x)}+i\frac{\sin(x)-x\cos(x)}{x^2+\sin^2(x)} \end{align} $$ Taking the imaginary part of both sides $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\frac1{2i}\log\left(\frac{x-i\sin(x)}{x+i\sin(x)}\right) &...


2

When you look at the picture of the trapezoid in question, to find its area you are actually integrating a line segment from $(0,3)$ to $(3,6)$ in $dx$. The slope of this is line is $(6-3)/(3-0) = 1$ and plugging in the first point yields the equation $y=x+3$. Thus we will integrate $(x+3)$. As for the limits, you need to go $(0,3) \to (3,6)$ in $x$, so $x=...


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