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Through the substitution $x=e^{-t}$ the original integral equals: $$I=\int_{0}^{+\infty}\left(2\frac{e^{-t}-1}{t^2}+\frac{e^{-t}+1}{t}\right)e^{-t}\,dt \\=2\color{purple}{\int_{0}^{+\infty}\frac{e^{-t}-1+t}{t^2}\,e^{-t}\,dt}+\color{blue}{\int_{0}^{+\infty}\frac{e^{-t}-1}{t}\,e^{-t}\,dt}$$ where the blue integral is yet manageable through Frullani's theorem (...

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Hint How do you define the set of points in the blue region in terms of inequalities? $$\text{BlueRegion}=\{(x,y)|0 \le x \le 6 , 0 \le y \le \color{red}{\text{?}}\}$$ There is also another way to do it $$\text{BlueRegion}=\{(x,y)|0 \le y \le \frac 23 , 0 \le x \le \text{?}\}$$

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Your intuition is very true. You are adding apples up. For a non-wild function $\int_a^b f(x)dx$ means adding up (infinitely many) values $f(x)$. But why does it become finite? Because each value $f(x)$ only occupies the single point $x$ on the interval $[a,b]$. So, it is adding up very many values but each value being multiplied by an almost zero segment. ...

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I think that your confusion stems from a misunderstanding of how summations relate to integrals. Think of an integral as the limit of a Riemann Sum, or $Lim_{\Delta x\rightarrow 0} \Sigma f(x)\Delta x$. Let's think for a second about how we derive this. We know an integral is the area under a curve. But how can we calculate this without calculus? Well, we ...

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Assuming it is $\;y=\frac19x\;,\;\;y=0\;,\;\;x=6\;$ , the integral can be put in the form $$\int_0^6\int_0^{\frac19x}(x+y)dydx$$

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You start with the inner integral: $\int_0^{1/9x}(x+y)dy=[xy+1/2y^2]_0^{1/9x}=10/9x+1/162x^2$ and this function you integrate with dx on the limits of $0,6$.

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Assuming that $d$ and $n$ are positive integers, you can just use the binomial theorem to expand the integrand, and integrate term by term, since it's a polynomial in $y$. $$(1 - u)^n = \sum_{i = 0}^n (1)^{n-i}{n \choose i} (-u)^i = \sum_{i = 0}^n (-1)^i{n \choose i} u^i$$ In your case, $u = y^d$, so this becomes $$(1 - y^d)^n = \sum_{i = 0}^n (-1)^i{n \... 8 Here is yet another approach. We first note that we can write \frac{x-1}{\log(x)} as$$\frac{x-1}{\log(x)}=\int_0^1 x^t\,dt$$Therefore, we can write$$\begin{align} \int_0^\infty \left(\frac{x-1}{\log^2(x)}-\frac{1}{\log(x)}\right)\frac{1}{1+x^2}\,dx&=\int_0^\infty \int_0^1 \frac{x^t-1}{\log(x)}\,\frac{1}{1+x^2}\,dt\,dx\\\\ &=\int_0^1 \int_0^\...

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$e^{ix}$ is a vector of length 1 and argument $x$. $e^{ix}dx$ is a vector of length $dx$ and argument $x$. The sum of all these vectors is, intuitivelly speaking (or even rigorously speaking, if one is allowed to use non standard analisys), a closed polygon with infinitely many edges of lenght $dx$. This is an example figure with $n=10$ edges: \$e^{ikx}...

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