# Tag Info

## New answers tagged definite-integrals

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


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\,dvJ= \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


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


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.

Top 50 recent answers are included