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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


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

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