New answers tagged

0

First we can write down $$\int_0^3\int_{\sqrt[3]x}^1e^{y^3}dydx=\int_0^1\int_{\sqrt[3]x}^1e^{y^3}dydx-\int_1^3\int_1^{\sqrt[3]x}e^{y^3}dydx$$ and now reversing the integration order: $$\int_0^1\int_0^{y^3}e^{y^3}dxdy-\int_1^{\sqrt[3]3}\int_{y^3}^3e^{y^3}dxdy$$


8

We have the identity $$\frac{\sin\left(x\right)}{\cosh\left(ax\right)+\cos\left(x\right)}=2\sum_{n\geq1}\left(-1\right)^{n-1}\sin\left(nx\right)e^{-anx},\, a>0,\, x\in\mathbb{R}$$ so ...


0

In THIS ANSWER, I showed that the function $I(x)$ as given by $I(x)=\int_x^\infty \frac{e^{-t}}{t}\,dt$ satisfies the inequalities $$\frac{1}{2}e^{-x}\ln\left(1+\frac{2}{x}\right)<\int_{x}^{\infty}\frac{e^{-t}}{t}dx<e^{-x}\ln\left(1+\frac{1}{x}\right) \tag 1$$ for $x>0$. Note from the left-hand side inequality in $(1)$ that ...


2

As David Mitra commented, the integral diverges at zero. More precisely, for any $c > 0$, $\begin{array}\\ \int_{c}^{1}\frac{dx}{xe^x} &\gt \int_{c}^{1}\frac{dx}{ex} \qquad\text{since }e^x \le e \text{ for }0 \le x \le 1\\ &= \frac1{e}\int_{c}^{1}\frac{dx}{x}\\ &= -\frac1{e}\ln(c)\\ &= \frac1{e}\ln(\frac1{c})\\ &\to \infty \text{ as ...


1

\begin{align} \int_{-\infty}^{\infty} x^2\exp\Big(-\frac{(x-x_0)^2}{2a^2}\Big)dx&=\sqrt{2\pi}a\int_{-\infty}^{\infty} x^2 \times \frac{1}{\sqrt{2\pi}a}\exp\Big(-\frac{(x-x_0)^2}{2a^2}\Big)dx\\ &=\sqrt{2\pi}aEX^2 \end{align} where $X$ is normally distributed with mean $x_0$ and variance $a^2$. Since $Var(X)=EX^2-E^2X$ we have that $EX^2=a^2+x_0^2$ ...


1

$$\Gamma(s)=\int_{0}^{\infty}x^{s-1}e^{-x}dx $$ Your integral results for $s\to 0$, which is not convergent.


2

We define $$c:=\int_{0}^{1}g\left(x\right)dx$$ and consider $$\phi\left(x\right)=\int_{0}^{x}\left(g\left(x\right)-c\right)dx.$$ We can see that $\phi\left(x\right)$ has the the required assumption. Now we can observe that $$\int_{0}^{1}g\left(x\right)\phi'\left(x\right)dx=0=c\int_{0}^{1}\phi'\left(x\right)dx $$ so we have ...


1

Suppose $f(x)$ is not constant on $[0, 1]$. Let $m$ and $M$ be the minimum and maximum value attained by $f$ on $[0, 1]$. These values are attained on $[0, 1]$ because $f$ is continuous and $[0, 1]$ is compact. By Darboux property there exists $\gamma \in (0, 1)$ such that $f(\gamma)=\overline{f}=\frac{m+M}{2}$ and a sufficiently small $\epsilon > 0$ such ...


3

Hints: The conditions $\phi(0) = \phi(1) = 1$ imply $$ \int_{0}^{1} \phi'(x)\, dx = 0; \tag{1} $$ that is, "the continuous function $\phi'$ is orthogonal to the constants with respect to the standard inner product in $C([0, 1])$". Consequently, the condition $$ \int_{0}^{1} g(x) \phi'(x)\, dx = 0\quad\text{for all $C^{1}$ functions $\phi$ satisfying (1)} ...


0

HINT: Let $$\sqrt x-1=y\implies\dfrac{dx}{2\sqrt x}=du$$


3

Set $x=z^2$, so that $dx=2z\,dz$ and $$ \int\frac{dx}{\sqrt{x}(\sqrt{x}-1)} = \int\frac{2\,dz}{z-1} = C+2\log(z-1) = C+2\log(\sqrt{x}-1).$$


4

We can write the integrand as $$\begin{align} \frac{x^2}{x^4+2a^2x^2+b^4}&=\frac{x^2}{(x^2+a^2+\sqrt{a^4-b^4})(x^2+a^2-\sqrt{a^4-b^4})}\\\\ &=\frac{A}{x^2+a^2-\sqrt{a^4-b^4}}+\frac{B}{x^2+a^2+\sqrt{a^4-b^4}} \end{align}$$ where $A$ and $B$ are given respectively by $$A=\frac{-a^2+\sqrt{a^4-b^4}}{2\sqrt{a^4-b^4}}$$ and ...


6

Let we set $\lambda=\frac{b}{a}$ . We want: $$ I(a,b)=\int_{-\infty}^{+\infty}\frac{x^2\,dx}{(x^2+a^2)^2+(b^4-a^4)} = \frac{1}{a}\int_{-\infty}^{+\infty}\frac{x^2\,dx}{(x^2+1)^2+(\lambda^4-1)}$$ and assuming that $\zeta_1(\lambda),\zeta_2(\lambda)$ are the roots of $(x^2+1)^2=1-\lambda^4$ in the upper half-plane, the residue theorem gives: $$ ...


1

Integrating by parts we obtain $$ \int_{\mathbb{R}}\cos(x\,\xi)\cos((1-x)^{\alpha})\chi_{[0,1]}(x)\,dx=\frac{\sin\xi}{\xi}-\frac{\alpha}{\xi}\int_0^1\sin(x\,\xi)\sin((1-x)^{\alpha})(1-x)^{\alpha-1}\,dx. $$ Since $\alpha-1>0$, the last integral converges to $0$ as $\xi\to\infty$. Then $$ ...


2

To change the order, you should first sketch the region of integration. Now you can see that if we turn the order of integration around, the smallest $y$ in the region is $y_{\min}=0$ and the biggest $y$ is $y_{\max}=\frac{\pi}2$. For a given $y$, draw a horizontal line through the figure (a curve of constant $y$) and we see that it enters the region at ...


1

Intuitively, no matter what happens with $f(x)$, only the region near $x=1$ will contribute to the integral. This isn't very rigorous, but this gives you the idea: Assume $f(x)$ is integrable. Divide the interval $[0,1]$ into $M$ intervals. On each interval $[a,b]$ take the average value of $f(x)$, so that: $$G_n = \sum_{m=1}^M ...


1

If $g$ is non negative, and if $\int_a^b g(x)dx>0$, then the result is true. To prove it, note that for all $x\in(a,b)$, $xg(x)<bg(x)$ whenever $g(x)>0$. Thus $$ \int_a^b xg(x)dx <b\int_a^bg(x)dx=b\int_b^cg(x)dx<\int_b^cxg(x)dx. $$ If there is no assumption on the sign of $g$, then it is not necessarily true. For instance, take $a=0,b=1,c=2$ ...


2

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6

By setting $x=au$ and $y=bv$ the problem boils down to computing $$ I(a,b) = ab\iint_{\mathbb{R}^2}\sqrt{u^2+v^2} e^{-(u^2+v^2)}\,du\,dv = 2\pi ab \int_{0}^{+\infty} \rho^2 e^{-\rho^2}\,d\rho = \pi a b\cdot\Gamma\left(\frac{1}{2}\right).$$


3

Lemma. Let $q(x,y)=Ax^2+2Bxy+Cy^2$ a positive definite quadratic form, associated with the symmetric matrix $M=\begin{pmatrix}A & B \\ B & C \end{pmatrix}$. We have: $$ \iint_{\mathbb{R}^2}\exp\left(-q(x,y)\right)\,dx dy = \frac{\pi}{\sqrt{\det M}}.$$ In our case, the quadratic form has coefficients $A=C=\frac{1}{2}$ and $B=-\frac{t}{2}$, ...


3

Hint. Assume $-1<t<1$. One may just integrate with respect to $u$, using the classic gaussian result, $$ \int_{-\infty}^\infty e^{tuv} e^{-u^2/2} \ du=\sqrt{2\pi} \:e^{t^2v^2/2} $$ then with respect to $v$, $$ \int_{-\infty}^\infty e^{t^2v^2/2} e^{-v^2/2} \ dv=\int_{-\infty}^\infty e^{-(1-t^2)v^2/2} \ dv=\frac{\sqrt{2\pi}}{\sqrt{1-t^2}} $$ obtaining ...


2

Why (3) is so different from (1) ? Because if we rewrite the double integrals as iterated ones, for (1) we get $$\int_0^1 \left(1-x_2\right)^{\alpha_3-1}\left({\color{blue}{\int_0^{x_2}x_1^{\alpha_1-1}\left(x_2-x_1\right)^{\alpha_2-1}dx_1}}\right)dx_2,$$ and for (3) we find $$\int_0^1 x_1^{\alpha_1-1}\left({\color{red}{\int_0^{x_1} ...


0

Hint: We will assume that $x$ and $x^{\prime}$ are real (as opposed to complex) variables. OP's distribution $$ \sum_{n\in \mathbb{N}_0} \left\{ \cos[n \pi(x-x')] - \cos[n \pi(x+x')] \right\} ~=~ \frac{1}{2}\sum_{n\in \mathbb{Z}} \left\{ \exp[in \pi(x-x')] - \exp[in \pi(x+x')] \right\} ~=~III_2(x-x')-III_2(x+x')$$ is a linear combination of Dirac ...


2

You just need to integrate $$ \int_{0}^{\infty}e^{-xy}\sin[nx]dx=\frac{1}{2i}\int^{\infty}_{0}\left(e^{(ni-y)x}-e^{-(ni+y)x}\right)dx $$ And you use the fact $$ \int^{\infty}_{0}e^{cx}dx=\frac{1}{c}\Big|^{\infty}_{0}e^{cx}=\frac{-1}{c} $$ Thus you have $$ \frac{1}{2i}\left(\frac{1}{y-ni}-\frac{1}{y+ni}\right)=\frac{n}{n^2+y^2} $$ I am sure there are other ...


5

You should have obtained $$\int_{x=0}^\infty e^{-yx} \sin nx \, dx = \frac{n}{n^2 + y^2}.$$ There are a number of ways to show this, such as integration by parts. If you would like a full computation, it can be provided upon request. Let $$I = \int e^{-xy} \sin nx \, dx.$$ Then with the choice $$u = \sin nx, \quad du = n \cos nx \, dx, \\ dv = e^{-xy} ...


1

Consider first completig the square $$a x^2+A x=\left(\sqrt{a} x+\frac{A}{2 \sqrt{a}}\right)^2-\frac{A^2}{4 a}$$ Now, change variable $$\sqrt{a} x+\frac{A}{2 \sqrt{a}}=y\implies x=\frac{2 \sqrt{a} y-A}{2 a}\implies dx=\frac{dy}{\sqrt{a}}$$ This makes $$I=\int e^{-a x^2-Ax}dx=\frac{e^{\frac{A^2}{4 a}}}{\sqrt{a}} \int e^{-y^2} dy=\frac{\sqrt{\pi ...


2

Let me consider the two integrals $$I = \int \mathrm{e}^{-v}\int_0^{a - bv} \mathrm{e}^{-u^2} du\,dv$$ $$J= \int \int_0^{a + bv} \mathrm{e}^{-u^2} du\,dv$$ First $$\int_0^{a - bv} \mathrm{e}^{-u^2} du =\frac{\sqrt{\pi } }{2} \text{erf}(a-b v)$$ which makes $$I=\frac{ \sqrt{\pi }}{2} \int e^{-v} \text{erf}(a-b v)\,dv$$ This one can be integrated by parts ...


0

Too long for a comment: The easiest way to prove this would be by using the properties of Beta and Gamma functions: $$\int_0^{\infty} \frac{x^{p-1}}{(1+x)^{p+q}}dx=B(p,q),~~~~p,q>0$$ Thus, the integral is equal to: $$\int_0^{\infty} \frac{x^{n-1}}{1+x}dx=B(n,1-n),~~~~n>1$$ By the properties of the Beta function: $$B(p,q)=\frac{\Gamma(p) ...


1

I'm going to assume you meant $\int_0^1 (1-x)^{1/n}dx=\frac{n}{n+1}$ because the formula in your question is incorrect, but this formula is correct. Now, first, we want to use $u$-substitution in order to get the integral in terms of $u=1-x$. Thus, we have $dx=-du$ and the bounds change from $0 \to 1$ to $1 \to 0$, so we get: $$\int_1^0 -u^{1/n}du=\int_0^1 ...


2

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1

$$ \int_{\tau}^{\infty}\tfrac{\exp(u)}{[1+\exp(u)]^2}du=\left.-\tfrac{1}{1+\exp(u)}\right|_\tau^\infty=\frac{1}{1+\exp(\tau)}=\tfrac{\exp(-\tau)}{\exp(-\tau)}\cdot\frac{1}{1+\exp(\tau)}=\frac{\exp(-\tau)}{1+\exp(-\tau)} $$


1

It is not helpful and is source of confusion to write $\frac d{dx} \int_0^1 f(x)dx$. You should write this as $\frac d{dx} \int_{x'=0}^{x'=1} f(x')dx'$. Then you are taking $\frac{d}{d x}$ of a constant which is zero. However, you do have, $\frac d{dx} \int^{x'=x} f(x')dx'= f(x)$.


6

You are missing the fact that there are two variables in the expression $\frac{x^a-1}{\log x}$. There's $x$, and then there's $a$. The expression $$\int_0^1 \frac{x^a-1}{\log x}\;dx$$ is independent of $x$, but it is not independent of $a$, and the derivative with respect to $a$ could be non zero. Basically, you don't have $\int_0^1 f(x)dx$, you have ...


1

See Gradsteyn&Ryzhik, 4.634. With the notation $D=\left\{0\le x_i\le 1,0\le i\le n, x_1+x_2+\ldots+x_n\le 1\right\}$ one has \begin{align} &\int_0^1dx_1\int_0^{1-x_1}dx_2\int_0^{1-(x_1 + x_2)}dx_3 \cdots \int_0^{1 - \sum_{i = 1}^{n -1}x_i} \prod_{i = 1}^n x_i^{k_i} dx_n=\\ &=\int\limits_D \prod_{i = 1}^n ...


3

Let $x=e^t$. Then $$ I = \frac{2}{\sqrt{\pi}}\int_{-\infty}^{+\infty}\int_{0}^{+\infty}\exp\left(2t-(u+a+bt)^2\right)\,du\,dt $$ is just the integral of $\exp(-q(u,t))$ with $q$ being a quadratic form in $u,t$, hence it can be computed by completing the square or just switching the order of integration: $$ I = ...


1

Consider as a counterexample, the function $$f(x)=e^{-x^2}$$ defined in $[0,+\infty)$ which is a constraint of a Gaussian Function, and is integrable but its antiderivative cannot be written in terms of elementary functions. For another example consider $$\sqrt{1-x^4}$$ defined in $[0,1)$ for which the integral can be approached this way.


1

I don't think there is a general formula using only elementary functions. Just consider $n=2$. Partial integration leads to $$\int f(y)^2\mathrm{d}y=f(x)g(x)-\int f'(y)g(y)\mathrm{d}y,$$ and only if $f$ is differentiable and $f'$ integrable. But from there, you won't get to far. On the other hand, let's consider $f$ to be infintely differentiable and let ...


1

Transform the equation into polar coordinates: $$ D = \left\{ (r,\theta) \middle| r^2 \ge 1, r^2 - r(\cos\theta+\sin\theta) \le 0 \right\} $$ The two inequalities can be combined as $$1\le r\le\cos\theta+\sin\theta=\sqrt2\sin(\theta+45^\circ),$$ which is solvable only when $\sin(\theta+45^\circ)\ge\frac1{\sqrt2}$, which means $45^\circ\le ...


5

Here is another approach for evaluating the integral (3). Using @achille hui's transformation, we can write $$ I=\int_0^1\frac{dx}{\sqrt[3]{x}\sqrt{1-x}\sqrt[6]{9-x}}= \frac{4\pi^2 2^{\frac{1}{3}}}{\Gamma\left(\frac{1}{3}\right)^3}\int_0^\frac{1}{\sqrt{2}}\frac{1}{\sqrt{1+x^6}}dx $$ Using the substitution $x=\sqrt{t}$, we get $$ I=\frac{\pi^2 ...


1

$$ \begin{align} \int_0^{2\pi}\frac{\cos(\theta)\sin(\theta-\phi)}{1-\cos(\theta-\phi)}\,\mathrm{d}\phi &=\cos(\theta)\int_0^{2\pi}\frac{\sin(\theta-\phi)}{1-\cos(\theta-\phi)}\,\mathrm{d}\phi\tag{1}\\ &=\cos(\theta)\int_{\theta-2\pi}^\theta\frac{\sin(\phi)}{1-\cos(\phi)}\,\mathrm{d}\phi\tag{2}\\ ...


-1

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1

The graph below shows the integrand $f(\phi)$ for case $\theta=0$. We have $f(-\phi)=-f(\phi)$, so we can claim it is reasonable to take the integral over the range as 0. The same is true for any value of $\theta$ because $f(\phi)$ has period $2\pi$ and we are integrating over a complete period. The snag is that near 0 the function is approx ...


1

Note that for some $n$, $2n\pi\le \theta<2(n+1)\pi$. Then, interpreting the integral in the sense of a Cauchy Principal Value, we can write $$\begin{align} \int_0^{2\pi}\frac{\cos(\phi)\sin(\theta-\phi)}{1-\cos(\theta-\phi)}\,d\phi&=\lim_{\epsilon \to ...


0

$$\int_{0}^{2\pi }\frac{\cos\theta \cdot \sin(\theta -\phi )}{1-\cos(\theta -\phi )}\text d\phi =\cos\theta \int_{0}^{2\pi }\frac{\sin(\theta -\phi )}{1-\cos(\theta -\phi )}\text d\phi $$ and by periodicity of the integrand one can evaluate as if $\theta=0$ and get $$-\cos\theta\int_{0}^{2\pi }\frac{\sin(\phi )}{1-\cos(\phi )}\text ...


0

Ok let's star $$I(y)=\int_{0}^{1}\frac{\arctan(yx)}{x\sqrt{1-x^2}}\,dx$$ for use the formula above then use differentiation under integral sign and integral methods.


6

this is $\int_0^\infty e^{-bt}t^{(1-\alpha)-1}dt$ where $b=-\log (1-a)>0$. Substitute $bt=u$ you will get Gamma function.


2

Here's an easy way to solve this, pretty algorithmic - not the fastest by far, but easy to follow and carry out in general $$\pi \int _0^{\pi }\cos\left(\frac{x}{2}\right)\sqrt{4+\sin^2\left(\frac{x}{2}\right)}\,dx$$ Let $\frac{x}{2} = u \implies dx = 2du$ $$2\pi \int _0^{\frac{\pi}{2} }\cos\left(u\right)\sqrt{4+\sin^2\left(u\right)}\,du$$ Let $\sin u = v ...


1

$$\int_0^{\pi} \cos \left(\frac x2 \right) \sqrt{4+\sin^2 \left(\frac x2 \right)} dx$$ Let $\sin \left( \frac x2 \right) = 2\tan (\theta)$ $\frac 12 \cos \left( \frac x2 \right) dx = 2\sec^2 (\theta) d\theta$ Then the integral becomes $$\int_0^{\arctan \left(\frac 12 \right)} 8\sec^3 \theta d\theta$$


4

At least your integral turns out to have a closed form: $$ \int_{0}^{\pi/2} x \, \frac{\sqrt{\sin x} - \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, \mathrm{d}x = G + \pi \left( \frac{1+2\sqrt{2}}{4} \log 2 - \log (1+\sqrt{2}) \right), $$ where $G$ is the Catalan's constant. So it seems to me that the 'almost rationality' of this integral is just a ...


0

Consider any function $f$ such that $\int_{a}^{b} f(x)dx$ does not have a closed form, now choose $b-a$ small enough that the integral is very close to zero. Eg: $\int_{0}^{1} e^{-x^{100}} dx$ perhaps does not have closed form and is surely very close to zero. Any such function can be modified to make the integral very close to any given rational integer.



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