New answers tagged

0

OK, I'm not super happy with what I have, but I'm posting in case it helps others find a better formula. $$\boxed{I(x)=\pi\log 2 -\frac 1 2 \int_x^1 \frac{K(\sqrt{1-t})-\frac \pi 2}{1-t}dt=\pi\log 2-\frac {\pi} 4 \sum_{n\geq 1}\left( \frac{(2n)!}{2^{2n}(n!)^2}\right)^2\frac{(1-x)^{n+1}}{n+1}}$$ where we identify the complete elliptic integral of the first ...


4

Before attacking the integral, I mention something about cubic theta function. The whole solution is quite self-contained if you accept the stated facts, the "footnote" contains more information. The three cubic theta functions are defined by $$\begin{aligned} a(q) &= \sum_{m,n} q^{m^2+mn+n^2}\\ b(q) &= \sum_{m,n} \zeta_3^{m-n} q^{m^2+mn+n^2}\\ c(q) ...


3

For c=1, If we consider THE following integral $I=\int_{0}^{\infty}x^{a}e^{x^{b}}e^{-\lambda(e^{x^{b}}-1)}dy$ then by letting $y=x^{b}$ we get I=$\frac{e^{\lambda}}{b^{a+1}}\int_{1}^{\infty} (log~y)^{\frac{a}{b}}e^{-\lambda y}dy$ if we consider $\frac{a}{b}$ is an integer then $I=\frac{e^{\lambda}}{b^{a+1}} (\frac{a}{b})! E_{0}^{\frac{a}{b}}(\lambda)$ ...


0

As pointed out, this is really straight forward (if say, $a\lt0$): $\int_0^1 ab^ac^{-a-1}\operatorname dc=-b^a[c^{-a}]_0^1=-b^a$. If $a\gt0$, it appears to not converge.


1

We're looking to compute the scalar line integral $$\int_\mathcal{C_1} (x+y) ds$$ where $\mathcal{C_1}$ is the straight line segment from $A=(1,0)$ to $B=(0,1)$. Optimal Solution In order to compute a line integral, we should parametrize our curve, say $$\vec{r}(t) = \langle 1-t, t \rangle \quad 0 \leq t \leq 1.$$ Then $ds = \Vert \vec{r}'(t) \Vert dt= \...


2

We can write $$ \ln(1-x)=-\int_0^x \frac{dt}{1-t},\hspace{7mm}\operatorname{Li}_3(x)=\int_{0\leq t_1\leq t_2\leq t_3\leq x}\frac{dt_1\,dt_2\,dt_3}{(1-t_1)t_2t_3}. $$ We can multiply out $\ln(1-x)^3\operatorname{Li}_3(x)$ and break the result into a sum over the different possible orderings of the variables of integration. This will allow us to write $I$ as ...


6

It's quite common for such integrals to use algebraic identities in order to solve them, see also here. We can use the first one in the link from above, namely: $$6ab^2=(a+b)^3+(a-b)^3-2a^3\Rightarrow I=\int_0^1 \frac{\ln(1-x) \ln^2(1+x)}{x} dx$$ $$=\frac16\int_0^1 \frac{\ln^3(1-x^2)}{x}dx+\frac16\int_0^1 \frac{\ln^3\left(\frac{1-x}{1+x}\right)}{x}dx-\...


0

that is a great question but there is a splendid Q , I will post for knowlodge $$I=\int_{0}^{\frac{\pi }{4}}x^2 cotxdx\ \ \ \ \ \ \ \ \ by\ parts\ we\ have\\ \\ I=x^2ln(sinx)\tfrac{\frac{\pi }{4}}{0}-2\int_{0}^{\frac{\pi }{4}}xln(sin(x))dx\\ \\ =-\frac{\pi ^{2}ln(2)}{32}+\int_{0}^{\frac{\pi }{4}}(2ln(2)+2\sum_{k=1}^{\infty }\frac{cos(2kx)}{k})xdx\\ \\ =\...


0

$$I=\int_{0}^{1}\int_{0}^{1}\frac{ln(x)-ln(y)}{x-y}dxdy\\ \\ \\$$ $$=\int_{0}^{1}\int_{0}^{1}\frac{ln(\frac{x}{y})}{x-y}dxdy\ \ \ , let\ \ \frac{x}{y}=t\\ \\$$ $$=\int_{0}^{1}\int_{x}^{\infty }\frac{ln(t)}{t(t-1)}dt=[\int_{x}^{\infty }\frac{xln(t)}{t(t-1)}dt]\tfrac{1}{0}+\int_{0}^{1}\frac{ln(x)}{x-1}dx\\ \\ =\int_{1}^{\infty }\frac{ln(t)}{t(t-1)}dt+\int_{0}^...


5

That substitution has no meaning since $x\mapsto x(x-a-b)$ is not injective. You can not express $x$ in uniqe way with $u$: $$x= {a+b\pm \sqrt{(a+b)^2+4u}\over 2}$$ So what is $$\int _{-ab}^{-ab} {f(x)\,du \over 2x-a-b}?$$ Is it choice for $x$ with + or -?


2

there is a nice question related with this problem , and we can evaluate it by using Real analysis $$I=\int_{-\infty }^{\infty }\frac{e^{ax}}{1-e^{x}}dx,\ \ \ \ \ \ \ \ (*)\ \ \ \ \ 0<a<1\\ \\ let \ x\rightarrow -x \ \ then\ \ I=\int_{-\infty }^{\infty }\frac{e^{-ax}}{1-e^{-x}}\ \ \ \ \ (**)\\ \\ adding\ (*)\ and\ (**)\ \ then\ 2I=\int_{-\infty }^...


2

I'll just show an idea that avoids those type of sum, but skip the calculations. You might also have better ideas to solve them. For start we will denote $a=\ln(1-x)$ and $b=\ln(1+x)$ and use the following identity: $$a^2=\frac12 (a+b)^2+\frac12(a-b)^2-b^2$$ $$\Rightarrow I=\frac12 \underbrace{\int_0^1 \frac{\ln x\ln^2(1-x^2)}{1+x}dx}_{I_1}+\frac12\...


5

Not an answer, but an extended comment for now. This hypergeometric function is a special case, and some complicated quadratic and cubic transformations apply to it. See this like for reference: https://dlmf.nist.gov/15.8. The formulas 15.8.25 and 15.8.26 both apply here. However, the most interesting one is so called Ramanujan’s Cubic Transformation (15....


1

Here is the best that we can hope for: Note that $(x)^+ = \max\{0, x\}$ is a convex function. Then by the Jensen's inequality, we have $$ \left( \frac{1}{b-a} \int_{a}^{b} f(x) \, \mathrm{d}x \right)^+ \leq \frac{1}{b-a} \int_{a}^{b} (f(x))^+ \, \mathrm{d}x. $$ Now using the fact that $x \mapsto (x)^+$ is positive homogeneous, we can cancel out the common ...


1

Note that $$\int_0^{2\pi}(\sin x)^+\,dx = \int_0^{\pi}\sin x\,dx =2,$$ while $$(\int_0^{2\pi}\sin x\,dx)^+ = 0^+ = 0.$$


0

Here is an independent solution: \begin{align} I&=\int_0^1\frac{\ln y\ln^2(1+y)}{y}\ dy\overset{y=\frac{1-x}{x}}{=}\int_{1/2}^1\frac{\ln(1-x)\ln^2x-\ln^3x}{x(1-x)}\ dx\\ &=\underbrace{\int_{1/2}^1\frac{\ln(1-x)\ln^2x}{x}\ dx}_{IBP}-\int_{1/2}^1\frac{\ln^3x}{1-x}\ dx-\underbrace{\int_{1/2}^1\frac{\ln^3x}{x}\ dx}_{-(\ln^42)/4}+\int_{1/2}^1\frac{\ln(1-x)...


2

This integral sounds a little cumbersome, but the following hint may help. Partial Hint $$(1-e^{-t \cdot \lambda_P})^{n-x}=\sum_{l=0}^{n-x}\binom{n-x}{l}(-1)^{l}e^{-l\lambda_P t}$$therefore$$\int_0^{\infty} e^{-t \cdot \lambda_S} \cdot e^{-t \cdot \lambda_P \cdot x} \cdot (1-e^{-t \cdot \lambda_P})^{n-x} dt=\sum_{l=0}^{n-x}\binom{n-x}{l}(-1)^{l}\int_0^\...


0

Wolfram Alpha gives: $$\int_{0}^{\infty} xe^{ax-b/x}\,dx =\frac{2b}{a}K_2(2\sqrt{ab})$$ where $K_2$ is a "modified Bessel function." But $K_2$'s definition is fairly messy. For example, Wikipedia first defines $$I_\alpha(x)=\sum_{m=0}^{\infty} \frac{1}{m!\Gamma(m+\alpha+1)}\left(\frac{x}{2}\right)^{2m+\alpha}$$ Then $$K_2(x)=\frac{\pi}2\lim_{\alpha\to 2} \...


0

Maple does this in terms of modified Bessel functions of the second kind. $$ \int_{0}^{\infty }\!x{{\rm e}^{-{a}^{2}x-{\frac {{b}^{2}}{x}}}} \,{\rm d}x = \frac{2b^2}{a^2} {{{\rm K}_{0}\left(2\,ab\right)} } +\frac{2b}{a^3}\,{{{\rm K}_{1}\left(2\,ab\right)}} $$


1

$$I=\int_{0}^{\pi} \frac {\sin x \sin n x}{1-2a \cos x+a^2} dx ~~~~(1)$$ Integrate it by parts taking $\sin x$ as the first function, we get $$I=-\frac{n}{2a}\int_{0}^{\pi} \cos n x \ln[1-2a \cos x +a^2] dx ~~~~(2).$$ Let $y=\cos x+i \sin x$, then $$f(x)=\ln(1-2a\cos x+a^2)=ln(1-ay)+\ln(1-a/y).$$ If $a^2<1,$ then $$f(x)=-2[a \cos x + \frac{a^2\cos 2 x}{...


2

Partial Answer This function is even so we can write$$\int_0^\pi {\sin x\sin nx\over 1-2a\cos x+a^2} dx={1\over 2}\int_{-\pi}^\pi {\sin x\sin nx\over 1-2a\cos x+a^2} dx$$by defining $z=e^{ix}$ we have$$\int_{-\pi}^\pi {\sin x\sin nx\over 1-2a\cos x+a^2} dx{=\oint_{|z|=1} {{1\over 2i}(z-z^{-1}){1\over 2i}(z^{n}-z^{-n})\over 1-a\left(z+{1\over z}\right)+a^2} {...


4

We can do something more general using a result proven here, namely: $$\int_0^\pi\frac{\cos(mx)}{a^2-2ab\cos x+b^2}dx=\frac{\pi}{a^2-b^2}\left(\frac{b}{a}\right)^m$$ $$\int_0^\pi \frac{\sin(kx)\sin(n x)}{a^2-2ab\cos x+b^2}dx=\frac12\int_0^\pi \frac{\cos((k-n)x)-\cos((k+n)x)}{a^2-2ab\cos x+b^2}dx$$ $$=\frac{\pi}{2(a^2-b^2)}\left(\left(\frac{b}{a}\right)^{n-...


0

\begin{align} I&=\int_0^1\frac{\operatorname{Li}_2(1-x)\ln^2(1-x)}{x}\ dx=\int_0^1\frac{\operatorname{Li}_2(x)\ln^2x}{1-x}\ dx\\ &=\sum_{n=1}^\infty H_n^{(2)}\int_0^1x^n\ln^2x\ dx=2\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)^3}\\ &=2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}-2\zeta(5)\\ &=2\left(3\zeta(2)\zeta(5)-\frac92\zeta(5)\right)-2\zeta(5)\\ &...


0

Contrary to what you might find in many contemporary textbooks (especially if they're elementary), you can well think of an infinitesimal as an entity and manipulate it according to its own rules (this has been rigorously set down in the last century in the so-called nonstandard calculus, which in fact I think is the more intuitive way to do calculus as ...


2

Let $F(y,t)$ be given by the integral $$F(y,t)=\int_0^\infty \frac{e^{-y^2/4\omega-t^2\omega}}{\sqrt \omega}\,d\omega\tag1$$ First, enforcing the substitution $\omega\mapsto\omega^2$ reveals $$\begin{align} F(y,t)&=2\int_0^\infty e^{-y^2/4\omega^2-t^2\omega^2}\,d\omega\tag2 \end{align}$$ Second, making the substitution $x=\sqrt{\frac{2t}{y}}\,\omega$...


1

Using King's rule, we have, $$I= \int_0^1\sqrt{\frac{x}{1-x}}dx =\int_0^1\sqrt{\frac{1-x}{x}}dx$$ $$2I = \int_0^1\left(\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}\right)dx = \int_0^1\frac{1}{\sqrt{x-x^2}}dx=\int_0^1\frac{1}{\sqrt{\frac{1}{4}-(x-\frac{1}{2})^2}}dx$$ $$2I = \big[\arcsin(2x-1)\big]^1_0 = \frac\pi2+\frac\pi2 $$ $$\implies I = \frac\pi2$$


1

Let $u = \sqrt{1-x}$ Then the integral transforms into: $$2 \int_0^1 \sqrt{1-u^2} \, \mathrm{d} u = \int_{-1}^1 \sqrt{1-u^2} \, \mathrm{d} u$$ Which has a nice geometric interpretation.


1

$$I=\int_0^\infty\operatorname{Li}_2^2\left(-\frac1{x^2}\right)\ dx\overset{IBP}{=}-4\int_0^\infty\operatorname{Li}_2\left(-\frac1{x^2}\right)\ln\left(1+\frac1{x^2}\right)\ dx$$ By using $$\int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$$ setting $n=1$ and replacing $x$ with $-\frac1{x^2}$ we can write $$\operatorname{Li}_2\left(-\...


7

In order to give some insight I will show a derivation for $a=1$, where a closed form in terms of the Inverse Tangent Integral is: $$\bbox[10pt, border:2px, lightblue]{\int_1^s \frac{\operatorname{arccosh}x}{\sqrt{(x-1)(s-x)}}dx=4\operatorname{Ti}_2\left(\sqrt{\frac{s-1}{2}}\right),\ s>1}$$ We'll start off with the substitution $\frac{s-x}{s-1}=t$ to get: ...


4

Finally I got the answer: Using the generalized integral expression of the polylogrithmic function which can be found in the book (Almost) Impossible Integrals, Sums and series page 4. $$\int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$$ and by setting $n=2$ we get $$\operatorname{Li}_{3}(x)=\frac12\int_0^1\frac{x\ln^2 u}{1-xu}\ du$$ ...


2

Reverse the $x$ and the $y$ coordinate axis (which amounts to a symmetry, an operation that doesn't change the absolute value of an area ; see figure below), giving equations $$y=x^2 \ \text{and} \ y+2x=8 \ \ \ \iff \ \ \ y=x^2 \ \text{and} \ y=-2x+8 \tag{1}$$ In this way, taking into account the fact that the intersection points of the new curves are $(...


3

Your first integral is slightly incorrect. The correct setup and solution would be: $$A=\int_{-4}^{2} \left(8-2y-y^2\right) \,dy=36\ \text{sq. units}.$$ If you now want to do integration with respect to $x$, begin by expressing your two functions as functions of $x$: $$ y=\pm\sqrt{x},\\ y=-\frac{x}{2}+4. $$ And then find where the curves intersect each ...


2

The first should be $$\int\limits_{-4}^2(8-2y-y^2)dy.$$ The second should be $$\int\limits_0^4\left(\sqrt{4-\frac{x}{2}}-\left(-\sqrt{4-\frac{x}{2}}\right)\right)dx+\int\limits_4^{16}\left(4-\frac{x}{2}-\left(\sqrt{4-\frac{x}{2}}\right)\right)dx$$


3

Let $$ y=sin^{-1}x\ \ \Rightarrow x= \sin y,\ \ dx= \cos y dy,$$ then $$ I=\int_{0}^{1}\frac{sin^{-1}x}{x}dx=\int_{0}^{\frac{\pi }{2}}\frac{y}{siny}.cosy=\int_{0}^{\frac{\pi }{2}}\frac{y}{tany}dy.$$ Let $$I(a)=\int_{0}^{\frac{\pi }{2}}\frac{\arctan(a \tan y)}{\tan y}dy \Rightarrow I'(a)=\int_{0}^{\frac{\pi }{2}}\frac{dy}{1+(a \tan(y))^2}.$$ One can easily ...


0

Let $f(x)=x$ for $0 \leq x \leq 1$. Let $g(x)=x$ for $0 \leq x \leq 1, x \neq \frac 1 2$ and $g(\frac 1 2 )=1$. Then $f$ and $g$ are both Riemann integrable and $\int_0^{x}f(t)dt=\int_0^{x}g(t)dt$ for all $t$. [Changing the value of the function at one point does not change the value of the integral]. Thus $\frac d {dt} \int_0^{x} g(t)dt=\frac d {dt} \int_0^{...


4

Note that the domain of second integral is only a quarter circle, so both of the answers are correct.


1

The integration step was correct now let $J$ the remaining, we have: $J=[\frac{1}{4w}\sin(2wt) + \frac{t}{2}]_{0}^{\pi/2} =\frac{1}{4w}\sin(w\pi)+\pi/4 -0=\frac{\sin(w\pi)}{4w}+\frac{\pi}{4}$ If $w\in\mathbb{Z} \Rightarrow \sin(w\pi)=0$ Therefore $J=0+\pi/4=\pi/4$


1

Hint: $$\int\cos(2wt)dt=\frac{\sin(2wt)}{2w}+C$$


1

Desmos is your friend. $$\int_{0}^{2}\frac{x^{2}}{2}dx+\int_{2}^{4}16x^{-3}dx=\frac{4}{3}+\frac{3}{2}=\frac{17}{6}$$


3

Hint: We get from $$\frac{x^2}{2}=\frac{16}{x^3}$$ the equation $$x^5=32$$ And we have to integrate $$\int_{0}^{2}\frac{x^2}{2}dx+\int_{2}^{4}\frac{16}{x^3}dx$$


2

In the paper Zujin Zhang Mean-value property of the heat equation the related calculations are carried out in detail.


2

The integrand in $y(t)$ is always defined ($2\cos u+5\ge3$), but the expression under the square root in $x(t)$ becomes zero at $\cos u=\frac12$ or $u=\pi/3$, going upwards from $0$. Thus $t_0=\pi/3$ and the length of the curve is $$\int_0^{\pi/3}\sqrt{x'(t)^2+y'(t)^2}\,dt=\int_0^{\pi/3}\sqrt{(\sqrt{2\cos t-1})^2+(\sqrt{2\cos t+5})^2}\,dt$$ $$=\int_0^{\pi/3}\...


1

Well, when you get the antiderivative, you can put in the limits and get $ \begin{array}{l} \lim\limits _{x\rightarrow \infty }[ x-ln( cosh( x))] =\lim\limits _{x\rightarrow \infty }\left[\frac{( x+ln( cosh( x)))( x-ln( cosh( x)))}{( x+ln( cosh( x)))}\right] =\lim\limits _{x\rightarrow \infty }\left[\frac{x^{2} -ln^{2}( cosh( x))}{x+ln( cosh( x))}\right]\\ =...


4

You need to evaluate the limit $$\lim_{x\rightarrow \infty} (x - \ln(\cosh(x)))$$ Note that $$e^{\ln(\cosh(x)) -x} = e^{-x} \cosh(x) = \frac{1}{2}(1 + e^{-2x})\rightarrow \frac{1}{2}$$ Therefore $$\lim_{x\rightarrow \infty} (x - \ln(\cosh(x))) = -\ln(1/2) = \ln(2)$$ Hence $$\int_0^\infty (1-\tanh(x))dx = \lim_{y\rightarrow \infty} [x - \ln(\cosh(x))]_{x=0}...


0

Let me first make a little change to the parameters and put them according to the sketch below. Then let me introduce the Iverson bracket $$ \left[ P \right] = \left\{ {\begin{array}{*{20}c} 1 & {P = TRUE} \\ 0 & {P = FALSE} \\ \end{array} } \right. $$ which in this type of problems is of profitable use, and which is easy to program. So, ...


5

The Answer I have used the substitution $(x+1)(y+1)=2$ before to good effect because $$ \int_0^1f(x)\,\frac{\mathrm{d}x}{x+1}=\int_0^1f\!\left(\tfrac{1-y}{1+y}\right)\frac{\mathrm{d}y}{y+1}\tag1 $$ If $f(x)=\log\left(\frac{x+3}{(x+2)(x+1)}\right)$, then $f\!\left(\frac{1-y}{1+y}\right)=\log\left(\frac{(y+2)(y+1)}{y+3}\right)$. Therefore $$ \int_0^1\log\left(...


1

$$\int_0^\infty e^{cx^b}e^{-ax}\ \mathrm{dx}$$ $$=\int_0^\infty \sum_{k=0}^\infty \frac{c^kx^{bk}}{k!}e^{-ax}\ \mathrm{dx}$$ $$=\sum_{k=0}^\infty \frac{c^k}{k!}\int_0^\infty x^{bk}e^{-ax}\ \mathrm{dx}$$ $$=\sum_{k=0}^\infty \frac{c^k\Gamma(bk+1)}{k!a^{bk+1}}$$ $$=\frac1a\sum_{k=0}^\infty \frac{\Gamma(bk+1)}{k!}\left(\frac{c}{a^b}\right)^k$$ $$=\frac1a\ _1\...


7

That's quite an impressive method to show that the integral vanishes. For the first part I'll show using a different approach that your integral vanishes. $$\mathcal J=\int_0^1 \ln\left(\frac{x+3}{(x+2)(x+1)}\right)\frac{\mathrm dx}{x+1}\overset{x+1=t}= \color{blue}{\int_1^2\ln\left(\frac{t+2}{t+1}\right)\frac{\mathrm dt}{t}}-\color{red}{\int_1^2 \frac{\ln ...


3

$$ \int_0^1 \sqrt{\frac{x}{1-x}} \mathrm{d}x $$ Let $x = 1 - u^2$ and thus $\mathrm{d}x = -2u\mathrm{d}u$ . Note that the bounds are exchanged by this substitution, i.e. $u(0) = 1$ and $u(1) = 0$ $$ -2\int_{1}^0 \sqrt{\frac{1 - u^2}{u^2}} \cdot u\mathrm{d}u $$ $$ -2\int_{1}^0 \frac{\sqrt{1 - u^2}}{u} \cdot u\mathrm{d}u $$ $$ -2\int_{1}^0 \sqrt{1 - u^2}\;\...


3

\begin{align} I&=\int_0^1\sqrt{\frac{x}{1-x}}\ dx=\int_0^1x^{1/2}(1-x)^{-1/2}\ dx=\text{B}\left(\frac32,\frac12\right)\\ &=\frac{\Gamma(\frac32)\Gamma(\frac12)}{\Gamma(2)}=\frac12\Gamma^{\ 2}\left(\frac12\right)=\frac{\pi}{2} \end{align}


Top 50 recent answers are included