Hot answers tagged

22

We have the identity $$\frac{\sin\left(x\right)}{\cosh\left(ax\right)+\cos\left(x\right)}=2\sum_{n\geq1}\left(-1\right)^{n-1}\sin\left(nx\right)e^{-anx},\, a>0,\, x\geq0$$ so ...


9

By Frullani's theorem we have $$\int_{0}^{\infty}\frac{e^{-x/\sqrt{3}}-e^{-x/\sqrt{2}}}{x}dx=\frac{1}{2}\log\left(\frac{3}{2}\right).$$


8

We have $$\frac d{dx}\frac1{e^x+1}=\frac{-e^x}{(e^x+1)^2}$$ Also $$\frac{e^x}{(e^x+1)^2}=\frac{e^{-x}}{(1+e^{-x})^2}$$ So $$\begin{align}\int_{-\infty}^{\infty}x^2\frac{e^x}{(e^x+1)^2}dx&=2\int_0^{\infty}x^2\frac{e^x}{(e^x+1)^2}dx=-2\int_0^{\infty}x^2\frac d{dx}\frac1{e^x+1}dx\\ &=\left.-2x^2\frac1{e^x+1}\right|_0^{\infty}+4\int_0^{\infty}\frac ...


7

$$\int_{0}^{+\infty}\frac{dt}{\sqrt{(1+t^2)(1+x^2 t^2)}}=\frac{\pi}{2\,\text{agm}(1,x)}\tag{1}$$ hence $$ I(z)=\int_{0}^{1}\frac{x^z}{\text{agm}(1,x)}\,dx = \frac{2}{\pi}\int_{0}^{1}\int_{0}^{+\infty}\frac{x^z}{\sqrt{1+x^2\sinh^2(t)}}\,dt\,dx\tag{2}$$ that can be managed through integration by parts, getting the same recurrence relation of the Euler Beta ...


7

Here, we present an approach that uses "Feynmann's Trick" for differentiating under the integral along with Contour Integration. Let $I$ be the integral given by $$I=\int_{-\infty}^\infty \frac{x^2e^x}{(e^x+1)^2}$$ "FEYNMANN'S TRICK" Enforcing the substitution $x\to \log(x)$ reveals $$\begin{align} I&=\int_0^\infty ...


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: $$ ...


6

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


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} ...


5

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 ...


5

An alternative approach to Marco Cantarini's perfectly sound answer. If we set, for any $\alpha>1$, $$ I(\alpha) = \int_{0}^{+\infty}\frac{e^{-x}-e^{-\alpha x}}{x}\,dx $$ differentation under the integral sign/Feynman's trick gives $$ I'(\alpha) = \int_{0}^{+\infty} e^{-\alpha x}\,dx = \frac{1}{\alpha}, $$ and since $\lim_{\alpha\to 1^+}I(\alpha) = 0$, ...


5

First $$1+x+x^2=\frac{1-x^3}{1-x}$$ So rewrite the integrand as $$\int_{0}^{1} \frac{\ln(1-x^3)}{x}-\frac{\ln(1-x)}{x} dx.$$ But using u-subtitution $u=x^3,du=3x^2dx$ $$\int_{0}^{1} \frac{\ln(1-x^3)}{x}dx=\int_{0}^{1} \frac{\ln(1-u)}{3u}du$$, so this means $$\int_{0}^{1} \frac{\ln(1-u)}{3u}du-\int_{0}^{1}\frac{\ln(1-x)}{x} ...


4

Here is a hint: Suppose $A = \{x \in [a,b] : f(x) = 0\}$ is not dense. Then there is some pocket $(c,d)$ in the interval $[a,b]$ untouched by $A$, i.e., $f(x) \neq 0$ for all $x \in (c,d)$. (Why? What does density even mean?) Then since $f(x) \geq 0$ by assumption, and thus $f(x) > 0$ on $(c,d)$ (since it's not equal to $0$ at any point in this ...


4

\begin{align} \int_0^{\infty}e^{-yx}\ \mathrm dx &=\frac{1}{y}\\[9pt] \int_b^a\int_0^{\infty}e^{-yx}\ \mathrm dx\ \mathrm dy &=\int_b^a\frac{\mathrm dy}{y}\\[9pt] \int_0^{\infty}\int_b^ae^{-xy}\ \mathrm dy\ \mathrm dx&=\ln a-\ln b\\[9pt] \int_0^{\infty}\left[-\frac{e^{-xy}}{x}\right]_b^a\ \mathrm dx &=\ln\left(\frac{a}{b}\right)\\[9pt] ...


4

substitute $x$ as $3\sin \theta$, so you get $x=3\sin\theta\implies dx=-3\cos\theta \,d\theta$ and plug this in and you will get$$\int^{\pi/2}_03\cos^2\theta \,d\theta$$ and then you can use $\cos^2\theta=\dfrac{\cos2\theta+1}{2}$ and then proceed


4

Here's a quick method: If $y = \sqrt{9-x^2}$ then $y^2 = 9-x^2$ so $x^2+y^2=9$. If you know that $x^2 + y^2 = 9$ is the equation of a circle of radius $3$ centered at $(0,0)$, that means $y^2=9-x^2$ is also an equation of that circle, as is $y = \pm\sqrt{9-x^2}$. So $y = \sqrt{9-x^2}$ (without $\text{“}\pm\text{''}$) is the top half of the circle. So the ...


4

First, the obvious substitution is: $$t=x^p$$ Not thinking about the conditions for convergence for now, we have the integral: $$I=\int_0^\infty \frac{\sin(x^p)}{x^p}\mathrm{d}x=\frac{1}{p}\int_0^\infty \frac{\sin(t)}{t^{2-1/p} }\mathrm{d}t$$ Now let's put $2-1/p=q$. The trick is to turn this into a double integral. Notice that: $$\int_0^{\infty} ...


3

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


3

$\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}} ...


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)} ...


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 ...


3

Because the integral $$ \int_{5}^{1} \frac {dt}{1-t^2} $$ is not convergent. Doing partial fractions $$ \frac {1}{1-t^2}=\frac12\Bigl(\frac {1}{1-t}+\frac {1}{1+t}\Bigr), $$ and $$ \lim_{a\to1^+}\int_5^a\frac {dt}{1-t}=\lim_{a\to1^+}\log(t-1)\,\Bigr|_5^a=\lim_{a\to1^+}(\log(a-1)-\log4)=-\infty. $$


3

Let $u = \sqrt{x+1+x^{-1}}$, we have $2 u du = (1 - x^{-2})dx$ and $$ \int_0^1 \frac{1-x}{1+x}\frac{dx}{\sqrt{x+x^2+x^3}} = \int_\infty^\sqrt{3} \frac{1-x}{1+x}\frac{2u du}{(x-x^{-1})u} = 2 \int_\sqrt{3}^\infty \frac{du}{x+2+x^{-1}}\\ = 2 \int_\sqrt{3}^\infty \frac{du}{u^2+1} = 2 \left[\tan^{-1}u\right]_{\sqrt{3}}^\infty = 2 \left(\frac{\pi}{2} - ...


3

I would start by trying to find the indefinite integral, as follows: $$ \frac{d}{dx} \frac{x2e^x}{(1+e^x)^2} = \frac{x^2e^x}{(1+e^x)^2}+2\frac{x e^x}{1+e^x} $$ Now let's try to rid ourselves of the $2\frac{x e^x}{1+e^x}$: $$ \frac{d}{dx} \left( 2x \log(1+e^x) \right) = 2\frac{x e^x}{1+e^x} + 2 \log(1+e^x) $$ Well, we now have to add back $2\int \log(1+e^x) ...


3

$$\int_{0}^{3}\sqrt{9-x^2}\space\text{d}x=$$ Substitute $x=3\sin(u)$ and $\text{d}x=3\cos(u)\space\text{d}u$. Then $\sqrt{9-x^2}=\sqrt{9-9\sin^2(u)}=3\cos(u)$ and $u=\arcsin\left(\frac{x}{3}\right)$. This gives a new lower bound $u=\arcsin\left(\frac{0}{3}\right)=0$ and upper bound $u=\arcsin\left(\frac{3}{3}\right)=\frac{\pi}{2}$: ...


3

Let's do the job as suggested by @Dr. MV. Integrate by parts with $u=x^{m-1}$ and $dv=x\left(x^2-1\right)^{5}dx$. This means $du=(m-1)x^{m-2}$ and $v={\left(x^2-1\right)^6\over 12}$. And so $$\begin{align}I_m=&=\left[uv\right]_0^1-\int_0^1vdu\\ &=-{m-1\over 12}\int_0^1\left(x^2-1\right)^6x^{m-2}dx\\ ...


3

Note that $\ln(1+x+x^2)=\ln(1-x^3)-\ln(1-x)=-\sum_{k=1}^{\infty}(x^{3n}/n)+\sum_{k=1}^{\infty}(x^{n}/n)$. Change the order of integration and summation (it is valid, why?), and we have $$\begin{align}\int_0^1 \frac{\ln(1+x+x^2)}{x} \mathrm{d}x&=\int_0^1\left(-\sum_{k=1}^\infty\frac{x^{3n-1}}{n}+\sum_{k=1}^\infty\frac{x^{n-1}}{n}\right)\mathrm{d}x\\ ...


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 ...


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 ...


2

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 $a=0,b=1,c=2$ ...



Only top voted, non community-wiki answers of a minimum length are eligible