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18

Let $u=x^2, du=2xdx$ Then the Integral becomes $$\int_{0}^{\infty}\frac{1}{u^4+2u^2+1}du$$ Now notice $u^4+2u^2+1=(u^2+1)^2$, so the integral becomes: $$\int_{0}^{\infty}\frac{1}{(u^2+1)^2}du$$ Let $u=\tan(z),du=\sec^2(z)dz$ So the integral becomes : ...


11

Your conjecture is a trivial consequence of Frullani's theorem after the substitution $x=e^{-z}$.


9

Assume $x>1$ and write $${1\over\sqrt{x^2-1}}={1\over\sqrt{x^2-1}}\>{\sqrt{x^2-1}+x\over \sqrt{x^2-1}+x}={1+{x\over\sqrt{x^2-1}}\over x+\sqrt{x^2-1}}={u'(x)\over u(x)}$$ with $u(x):=x+\sqrt{x^2-1}>0$. It follows that $$\int{dx\over\sqrt{x^2-1}}=\log\bigl(u(x)\bigr)+C=\log\bigl(x+\sqrt{x^2-1}\bigr)+C\ .$$


7

Try integration by parts, differentiating $z$ and integrating $ze^{-\frac{z^2}{2}}$. You'll need the fact that $$ \int_{-\infty}^{\infty}e^{-\frac{z^2}{2}}\;dz=\sqrt{2\pi}$$


6

Let the function $I(a)$ be the integral given by $$\begin{align} I(a)&=\int_{-\infty}^\infty e^{-az^2}\,dz\\\\ &=\sqrt{\frac{\pi}{a}} \end{align}$$ Then, note that $I'(1/2)$ is $$\begin{align} I'(1/2)&=-\int_{-\infty}^\infty z^2e^{-\frac12 z^2}\,dz\\\\ &=-\sqrt{2\pi} \end{align}$$ Therefore, we find $$\int_{-\infty}^\infty ...


6

By subbing $x=\sqrt{z}$, then $\frac{1}{z^2+1}=u$ $$\int_{0}^{+\infty}\frac{2x}{(x^4+1)^2}\,dx = \int_{0}^{+\infty}\frac{dz}{(z^2+1)^2}=\frac{1}{2}\int_{0}^{1}u^{1/2}(1-u)^{-1/2}\,du$$ that by Euler's beta function and $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ equals: $$ \frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)}{2\,\Gamma(2)} = ...


6

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} ...


5

An easy way to prove this (while also generalising your statement to any real number) is by using the technique of differentiating under the integral sign. Let $$f(\alpha)=\int_0^1 \frac{x^\alpha -1}{\log x} dx.$$ By differentiation under the integral we see that $$f'(\alpha)=\int_0^1 \frac{\partial}{\partial \alpha}\left(\frac{x^\alpha -1}{\log ...


4

Admitting that $s$ is a positive integer, $$\frac{x^s-1}{x-1}=\sum_{k=0}^{s-1}x^k$$ So, $$\int_{0}^{\infty}{x^s-1 \over x-1}\cdot{1-e^{-ax} \over 1-e^{ax}}dx=\sum_{k=0}^{s-1}\int_{0}^{\infty}x^k{1-e^{-ax} \over 1-e^{ax}}dx$$ Now, assuming $a>0$, $$\int x^k{1-e^{-ax} \over 1-e^{ax}}dx=-\int x^k e^{-ax}dx=\frac{1}{a^{k+1}}\Gamma (k+1,a x)$$ where appears ...


4

Hint. One may write $$ \begin{align} \int_{0}^{\infty}\frac{4x^2}{(x^4+2x^2+2)^2}dx&=4\int_0^{\infty}\frac1{\left(\left(x-\frac{\sqrt{2}}x\right)^2+2+2\sqrt{2}\right)^2}\:\frac{dx}{x^2} \\\\&=2\sqrt{2}\int_0^{\infty}\frac1{\left(\left(x-\frac{\sqrt{2}}x\right)^2+2+2\sqrt{2}\right)^2}\:dx \quad (x \to \sqrt{2}/x) ...


3

Consider $$I=\int_{-\infty}^{\infty}x^2e^{\frac{-x^2}{2}}dx$$ $$I^2=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x^2y^2e^{\frac{-x^2-y^2}{2}}dydx$$ Convert to polar $x=r\cos(\theta),y=r\sin(\theta)$ so that $\frac{\partial(x,y)}{\partial(r,\theta)}=r$ The double integral becomes: ...


3

A real function $f(x)$ may be analytic at $x=0$ with a finite radius of convergence, due to the existence of singularities in the complex plane: take, for instance, $f(x)=\frac{1}{x^2+1}$ or its primitive $\arctan x$. In such a case the Taylor series actually converges to $f(x)$ only if $x$ is sufficiently close to the origin, so, even if $$\forall ...


2

Here is an approach using contour integration in case anyone is interested. Suppose we seek to verify that $$\int_0^\infty \frac{2x}{x^8+2x^4+1} dx = \frac{\pi}{4}$$ or alternatively $$\int_0^\infty \frac{x}{x^8+2x^4+1} dx = \frac{\pi}{8}.$$ We use a quarter pizza slice contour with the straight components $\Gamma_0$ and $\Gamma_1$ on the ...


2

Since $f\in\mathcal{S}(\mathbb{R})$ we have that both $x\cdot f(x)$ and $|x|^{3/2}\cdot f(x)$ belong to $L^2(\mathbb{R})$, so: $$ \left|\int_{\mathbb{R}} f(x)\left(1-e^{-\pi\delta x^2}\right)\,dx\right| \leq \|x\cdot f(x)\|_2 \cdot \sqrt{2\pi(2-\sqrt{2})\sqrt{\delta}}=\color{red}{C\cdot \delta^{1/4}} \tag{1} $$ or: $$ \left|\int_{\mathbb{R}} ...


2

You proof looks basically correct to me. $\lim_{n\to\infty}\frac{e^x-1}{x}=1$ should be $\lim_{x\to 0+}\frac{e^x-1}{x}=1$. I would write (1) as $$(1)\int_0^1\frac{e^x-1}{x}dx=\int_0^1g(x)dx.$$


2

Substitute $$\frac1u=t\implies dt=-\frac{du}{u^2}\implies$$ $$\int_0^1 t^{x-1}(1-t)^{y-1}dt=\int_\infty^1u^{1-x}\left(1-\frac1u\right)^{y-1}\left(-\frac{du}{u^2}\right)=\int_1^\infty u^{1-x}u^{1-y}(u-1)^{y-1}u^{-2}du=$$ $$=\int_1^\infty u^{-(x+y)}(u-1)^{y-1}du=:I$$ and now substitute $$u=z+1\implies du=dz\implies ...


2

Take little circles around $\;z=0,\,1\;$ , say $\;C_0,\;C_1\;$ of radius $\;0.1\;$, so the integral equals the sum of the integrals around these two circles . Now Laurent series around these points, using the well known ones for the exponential function and the geometric series: ...


2

Getting rid of the syntactic sugar, $$ I(s,a)=\int_{0}^{+\infty}\frac{x^s-1}{x-1}\cdot\frac{1-e^{-ax}}{1-e^{ax}}\,dx = -\int_{0}^{+\infty}\frac{x^s-1}{x-1}e^{-ax}\,dx$$ hence the problem is equivalent to finding a Laplace transform. If $s\in\mathbb{N}^+$, since $$ \frac{x^s-1}{x-1} = \sum_{k=0}^{s-1} x^k $$ and $\int_{0}^{+\infty}x^k e^{-ax}\,dx = ...


2

I will use $t$ in this answer as opposed to $\epsilon$ because of limits. We have that $\lim_{t\to 0} \frac{g(x,t)}{y(x,t)}=1$, that is: $$\forall \epsilon >0, \, \exists \delta>0,\, \forall t \text{ with } 0 < t < \delta,\\\left | \frac{g(x,t)}{y(x,t)}-1\right |\leq \epsilon \Longleftrightarrow (1-\epsilon)\cdot y(x,t)\leq g(x,t)\leq ...


1

$$\frac{d}{dx}\frac{e^x}{(x-1)} = \frac{(x-2)}{(x-1)^2}e^{x} $$ hence if $$ \frac{x^2+ax+b}{(x-1)^2} = 1+k\frac{x-2}{(x-1)^2} $$ i.e. $\color{red}{2a+b = -3}$, we have an elementary integral, $\color{red}{\left(1+\frac{a+2}{x-1}\right)e^x}$. Some differential Galois theory is required to prove that the previous if is in fact an iff.


1

$\sqrt{1-x^2-y^2}=x^2+y^2-1$ ${-(-1+x^2+y^2)}=(x^2+y^2-1)^2$ $-1=x^2+y^2-1$ Your last line is wrong. Let $A=x^2+y^2-1$. Your second line gives $$ -A=A^2 $$ which implies $A(A+1)=0$ and thus $A=-1$ or $A=0$. You just picked the wrong one since $A\geq 0$ by your first line.


1

You're looking for the intersection of: $$z = \rho^2-1$$ $$z^2 = 1 - \rho^2$$ where $\rho^2 = x^2 + y^2$. Add the two equations to get $z^2 + z = 0$, so $z = -1,0$. For the $z=0$ solution, $\rho^2 = 1$ which is what you suspected.


1

You can write $$ \dfrac{e^z}{z(1-z)^2} = \dfrac{1}{z} + \dfrac{e^z-1}{z} + \dfrac{d}{dz} \left(\dfrac{e^z}{1-z}\right) $$ The last term is explicitly the derivative of a meromorphic function, so its integral around any contour that avoids $z=1$ is $0$. The singularity of the second term is removable, so its integral around any contour is $0$. That leaves ...


1

A floor function is rather annoying to integrate (in my opinion), but you have a very easy area over which you integrate: the triangle with vertices $(-2,0), (0,0)$ and $(0,2)$. You can work out what the value of the floor function $\lfloor x+y\rfloor$ will be on this triangle: $\lfloor x+y\rfloor = 0 \iff 0 \leq x+y < 1 \iff -x \leq y < -x+1$, in ...


1

Draw the Region of the plane $$\{(x,y)\mid y\in [0,2],x\in[0,2-y]\}$$ Then note that \begin{align*} & [\dots]\\\text{if }\ & 0\ge x+y<1\quad \lfloor x+y\rfloor=0\\ \text{if }\ & 1\ge x+y<2\quad \lfloor x+y\rfloor=1\\ & [\dots] \end{align*} and so on for the various integer values. Can you identify these regions in the picture? What is ...


1

There is a beautiful (in my opinion) formula, that can be derived from the coarea formula (see here): $$\int_{\mathbb{R}^n}fdx=\int_{0}^{\infty}\left ( \int_{\partial B(0,r))}fdS \right )dr$$ Applying it to the case of n-dimensional ball, we get ($r=|x|$): $$\int_{B(0,R)}\frac{1}{|x|^{n-1}}dx=\int_{\mathbb{R}^n}fdx=\int_{0}^{R}\left ( \int_{\partial ...


1

$f(w)e^{-π \delta w^2}\xrightarrow[\delta\to0]{}f(w) $ almost everywhere in $w$ and we have the dominating function $f$. By Lebesgue dominated convergence theorem, the result follows.


1

It is because $0\leq y\leq x$ is the support if $Y$. When you factor in the support of $X$, the support of the joint distribution is , $0\leq y, \max(1,y)\leq x$ $$\begin{align}f_{Y}(y) =&~ \int_{y}^\infty f_{X,Y}(x,y)\operatorname d x~\big[0\leq y\big]\\[2ex] =&~ \int_{\max(1,y)}^\infty \tfrac 1{x^3}\operatorname d x~\big[0\leq ...


1

In the general case, neither of these is right. You have two inequalities for $x$ in $f(x,y)$, namely $x\ge1$ and $x\ge y$, so the lower limit of the integral is $\max(1,y)$. You appear to be implying that the lower limit $y$ was specified in some book or lecture or the like. If so, I suspect that this was done because you need the value for $Y=\frac32$, and ...


1

$\int\left(\int\phi(a-z)~dz\right)dz$ $=z\int\phi(a-z)~dz-\int z~d\left(\int\phi(a-z)~dz\right)$ $=z\int\phi(a-z)~dz+\int z~\phi(a-z)~dz$



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