New answers tagged

1

The change of variable $a\,t=s$ changes the integral into $$\int_{a\gamma_0}^\infty \frac{1}{s}e^{-s}\,ds = \Gamma(0,a\,\gamma_0),$$ where $\Gamma(z,s)$ is the incomplete $\gamma$ function. To solve the equation $\Gamma(0,z)=12.1$ you can use Wolfram's function FindRoot. $z$ is of the order of $10^{-6}$, which makes computations a little bit shaky. The ...


1

With your unusual notation (normally the Legendre polynomals are denoted with $P_n$ and $P_n^{(\alpha, \beta)}$ are the Jacobi polynomials), you have (see http://dlmf.nist.gov/18.7.E9 or Abramowitz/Stegun 22.5.35) $$L_n(x) = J_n^{(0,0)}(x)$$ Use the recurrence relations for Jacobi polynomials (http://dlmf.nist.gov/18.9.E5 or A/S 22.7.18 and 22.7.19) ...


1

In mathematics, the Gaussian or ordinary hypergeometric function 2F1(a,b;c;z) is a special function represented by the hypergeometric series, that includes many other special functions as specific or limiting cases. It is a solution of a second-order linear ordinary differential equation (ODE). Every second-order linear ODE with three regular ...


1

This is probably the hypergeometric function.


2

There is a closed form of this integral: $$\int_0^{\pi/2} (\log \sin x)^n\text{ d}x=\frac{1}{2^{n+1}}B^{(n)}\left(\frac{1}{2},\frac{1}{2}\right)$$


0

This is by no means an answer, but just too long to fit in a comment. Integration by parts yields a sort-of-interesting result. If I am doing this correctly, then you get that: $LG_3 = [\int^1_0(\ln(\Gamma(x)))dx]·[(\ln(\Gamma(x)))]^1_0-2·\int_0^1[(\ln(\Gamma(x)))·(\frac{d}{dx}(\ln(\Gamma(x))))·(∫(\ln(\Gamma(x))dx)]dx$ Inside the integral on the right is ...


1

Assuming you mean the version defined by Landau, namely $$\xi(s) = \frac{1}{2}s(s-1)\pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta\left(s\right)$$ then any computer algebra system will be able to do this for you. For example, in Maple, xi := s -> s/2*(s-1)*Pi^(-s/2)*GAMMA(s/2)*Zeta(s): nth := proc(n) limit(diff(xi(s),s$n),s=0) end will give you what ...


8

\begin{align}\int_0^{1} \frac{\arcsin^2 x^2}{\sqrt{1-x^2}}\,dx &= \frac{1}{2}\int_0^{1} \frac{\arcsin^2 x}{\sqrt{x}\sqrt{1-x}}\,dx \tag{1}\\&= \frac{1}{2}\int_0^{\pi/2} \frac{\theta^2\cos \theta}{\sqrt{\sin \theta - \sin^2 \theta}}\,d\theta \tag{2}\\&= \frac{1}{\sqrt{2}}\int_0^{\pi/2} \frac{\left(\frac{\pi}{2} - \theta\right)^2\cos ...


2

We have: $$ \arcsin^2(z^2)=\sum_{n\geq 0}\frac{2^{2n+1} n!^2}{(2n+2)!}z^{4n+4} \tag{1}$$ hence: $$ I = \frac{\pi}{4}\sum_{n\geq 0}\frac{n!^2 (4n+3)!}{2^{2n+1}(2n+2)!^2 (2n+1)!}=\frac{3\pi}{16}\cdot\phantom{}_5 F_4\left(1,1,1,\frac{5}{4},\frac{7}{4};\frac{3}{2},\frac{3}{2},2,2;1\right).\tag{2} $$


2

Hint The number of elements of length $k$ is $3^k$, so the number of elements of length $\leq k$ is $$3 + 3^2 + \cdots 3^k = \frac{3}{2} (3^k - 1) .$$ Thus, the $n$th element in the sequence $A, B, C, AA, AB, \ldots$ has length $k$ iff $$\tfrac{3}{2} (3^{k - 1} - 1) < n \leq \tfrac{3}{2}(3^k - 1) .$$


2

Let $\chi_{\Bbb Q}(x):=1 $ if $x \in \Bbb Q$ and $0$ otherwise.


0

I've just found a physical problem (in classical mechanics) involving the trajectory of a particle in which I had to take the inverse of $E(\phi,k)$, the incomplete elliptic integral of the 2nd kind. For the elliptic integral of the 1st kind $F(x,k)$ this is an easy task beacuse the Jacobi elliptic function $sn(x,k)$ (or JacobiSN(x,k) in mathematical ...


0

The second half of Whittaker and Watson's A Course of Modern Analysis is devoted to special functions.


1

IMO this is not your problem, but the sign of the $l(l+1)$ term is positive in (a) but negative in (b). You can split and re-combine the sums with index shift because the $n(n-1)$ term is zero for $n=0,1.\;$ Here my manipulations $$\sum_0^\infty \Big( n(n-1)a_nx^{n-2} - n(n-1)a_n x^n -2na_nx^n + l(l+1)a_n x^n \Big)=$$ $$\sum_0^\infty n(n-1)a_nx^{n-2} + ...


1

The integral can be written as $$ \frac{1}{\sigma^2}\int_b^\infty dx\ x\ e^{-x^2/2}\int_0^\infty dy\ y e^{-y^2 \left(\frac{1}{2}+\frac{1}{2\sigma^2}\right)}I_0(xy)\ . $$ Now we can use the following formula 6.633.4 of Gradshteyn-Ryzhik $$ \int_0^\infty dy\ y\ e^{-\alpha y^2}I_\nu (\beta y)J_\nu(\gamma ...


1

Solution: \begin{equation} \frac{1}{a_{2}^{b}}\int_{0}^{\infty} x^{n}\, e^{-a_{1}x} \, \gamma(b, a_{2}(-p+x)) \ \mathrm{d}x \end{equation} Use this fact: $\frac{\partial}{\partial x}\, \gamma(a,f(x)) = e^{-f(x)}\, f(x)^{a-1} \, f'(x) \ \mathrm{d}x$ and do integration by parts. \begin{equation*} \begin{aligned} u &= \gamma(b, a_{2}(-p+x)) &&v = ...


1

From $$ \mathrm{B}(x,y)=\frac{\Gamma (x)\Gamma (y)}{\Gamma (x+y)} $$ you get $$ \mathrm{B}(\alpha,r\alpha +1)=\frac{\Gamma (\alpha)\Gamma (r\alpha+1)}{\Gamma ((r+1)\alpha+1)} $$ then use Stirling's formula, as $z \to \infty$, $$ \Gamma(z) = \sqrt{2\pi} z^{z - 1/2} e^{-z} (1 + O(1/z)) $$ to get, as $\alpha \to \infty$, $$ \mathrm{B}(\alpha,r\alpha ...


1

The integral definition of $\gamma(\alpha,X)$ leads to a double integral which can be expressed on closed form :


1

Wolfram alpha gives the result $$\frac{2 y^{a+m+\frac{3}{2}} \, _2F_2\left(a+m+\frac{3}{2},-k+m+\frac{1}{2};a+m+\frac{5}{2},2 m+1;y\right)}{2 a+2 m+3}.$$


1

The only way this can be true for bounded $f$ is when $f \equiv 0$. Letting $a = 0, h = 1$ in (3) gives that $\int_{-\infty}^\infty K(t)dt = \int_{-\infty}^\infty K(-t)dt = M$ for some $M < \infty$. So, making the substitution $u =\frac{a-t}h$, we get: $$\int_{-\infty}^\infty K\left(\frac{a-t}h\right) dt = h\int_{-\infty}^\infty K(u)du = Mh$$ If $f \le ...


3

What one could do fairly easy (pure heuristics) is to obtain an expansion for large $x$. The Riemann Zeta function can be approximated in this limit by $\zeta(z)\sim 1+\frac{1}{2^z}$ Therefore our sum reads $$ S(x)=\sum_{k=1}^{\infty}\frac{\zeta(kx)}{k!}\sim_{x\rightarrow\infty}\sum_{k=1}^{\infty}\frac{1}{k!}+\frac{1}{k! ...


3

Hint The asymptotics of Airy function (see here and here ) is given by $$\text{Ai}(x)=e^{-\frac{2 x^{3/2}}{3}} \left(\frac{\sqrt[4]{\frac{1}{x}}}{2 \sqrt{\pi }}-\frac{5 \left(\frac{1}{x}\right)^{7/4}}{96 \sqrt{\pi }}+O\left(\left(\frac{1}{x}\right)^{13/4}\right)\right)$$ So, $$\exp\left ( \frac{2x^{3/2}}{3} \right )\sqrt[4]{x}\mathrm{Ai}\left ( x ...


6

The integral can be rephrased, using the transformation $x=e^x$ in the same spirit as the general analogy of Taylor series to Laplace transforms, as $$ \int_0^\infty \frac{x^t \mathrm dt}{\Gamma(t+1)} =\int_0^\infty \frac{e^{st}\mathrm dt}{\Gamma(t+1)}. $$ This integral is known as the nu function, denoted $\nu(x)=\nu(e^s)$. Wikipedia and MathWorld have the ...


1

The problem you pose stems from an implicit false dichotomy introduced from the various meanings of "solve". Any equation can be solved, whether polynomial, linear or transcendental. The algebraic process of solving usually involves algebraic transformations which preserve or do not preserve the equivalence of the solutions to the original one. When it does ...


2

Looonnng-winded hint: The work below is admittedly a lot of algebra away from actually reaching an explicit final value for the desired integral, but I do think it covers the trickiest part of the derivation with the remainder being straightforward, though tedious. Given $z>y>p>a>b>c>d$, define the elliptic integral ...


0

$K_\alpha(x)=\int_0^\infty e^{-x\cosh t}\cosh\alpha t~dt$ $\alpha K_\alpha(x)=\alpha\int_0^\infty e^{-x\cosh t}\cosh\alpha t~dt$ $=\int_0^\infty e^{-x\cosh t}~d(\sinh\alpha t)$ $=[e^{-x\cosh t}\sinh\alpha t]_0^\infty-\int_0^\infty\sinh\alpha t~d(e^{-x\cosh t})$ $=\int_0^\infty e^{-x\cosh t}x\sinh t\sinh\alpha t~dt$ ...


1

Why special functions ? $$\int_0^\infty\frac{dt}{1-t^4}=\int_0^\infty\left(\frac1{4(1-t)}+\frac1{4(1+t)}+\frac1{2(1+t^2)}\right)dt =\left.\left(\frac14\ln\left|\frac{1+t}{1-t}\right|+\frac12\arctan t\right)\right|_0^\infty=\frac\pi4.$$


1

Hint: Evaluate $I(a)=\displaystyle\int_{-\infty}^\infty e^{-ax^2}~dx,$ and then differentiate both sides with regard to a.


1

Integration by parts brings that integral to the usual gaussian integral: $$\int_{\mathbb{R}}t^2 e^{-t^2}\,dt = \int_{0}^{+\infty} t\cdot\left(2t e^{-t^2}\right)\,dt = \int_{0}^{+\infty}e^{-t^2}\,dt = \frac{\sqrt{\pi}}{2}.$$


2

Hint: Let $t^2 = u$ then $du = 2tdt$ thus we get $$2\int_{0}^{\infty}t e^{-t^2} t\,dt = \int_{0}^{\infty} u^{1/2}e^{-u}du = \color{red}{\Gamma\left(\frac{3}{2}\right)}$$


4

(Assuming principal value) Let $t=u^4$ as you suggested, then we get $$ PV\int_{0}^{+\infty}\frac{du}{4u^{3/4}(1-u)}= \lim_{\varepsilon\to0}\left[\int_{0}^{1-\varepsilon}\frac{du}{4u^{3/4}(1-u)}+\int_{1+\varepsilon}^{+\infty}\frac{du}{4u^{3/4}(1-u)}\right] $$ and now an idea could be letting $u=1/(1-y)$ in the second integral to get $$ ...


1

Substitute $\alpha = \frac{1}{a}$ ($\alpha\rightarrow \infty$ as $a\rightarrow 0$)and $t=\frac{1}{x}$ so that the integral transforms as $$I(\alpha)=\frac{1}{\alpha ^n}\int_{\alpha}^{\infty}\big(t-\alpha\big)^nt^{-n}e^{-t}dt.$$ Expanding the first term in the integrand $$I(\alpha)=\frac{1}{\alpha ^n}\Big[\sum_{r=0}^{n}{n\choose ...


0

HINT: $$\int_0^a \frac{(a-x)^n e^{-1/x}}{x^2}\ dx$$ Assume $u=-\frac{1}{x}$ And $du=\frac{1}{x^2}dx$ Hence we have that as $x=0 , u \to -\infty$ and $x=a , u=-\frac{1}{a}$ Thus the integrand becomes $$\int^{-\frac{1}{a}}_{-\infty} (a+\frac{1}{u})^n e^u\ du$$ Can you look for some recursion?


0

$\int_0^1\dfrac{t^{\mu-1}(1-t)^{\nu-\mu-1}}{(1-wt)^\rho}{_2F_1}(\alpha,\beta;\gamma;zt)~dt$ $=\int_0^1\sum\limits_{n=0}^\infty\dfrac{(\alpha)_n(\beta)_nz^nt^{n+\mu-1}(1-t)^{\nu-\mu-1}}{(\gamma)_nn!(1-wt)^\rho}dt$ $=\sum\limits_{n=0}^\infty\dfrac{(\alpha)_n(\beta)_nz^nB(n+\mu,\nu-\mu){_2F_1}(\rho,n+\mu;n+\nu;w)}{(\gamma)_nn!}$ ...


4

By definition? $P_\ell^m(x) = (-1)^m (1-x^2)^{m/2} P_\ell^{(m)}(x)$ where $f^{(m)}$ stands for the $m^{th}$ derivative of $f$. We have $$P_\ell^2(x) = (1-x^2)\frac{d^2}{dx^2}P_\ell(x) = \frac{d}{dx}\left[(1-x^2)\frac{d}{dx}P_\ell(x)\right] + 2x P'_\ell(x) $$ Since $P_\ell(x)$ satisfy the Legendre DE: $$\frac{d}{dx}\left[(1-x^2)\frac{d}{dx}P_\ell(x)\right] ...


2

As noted in the comments by tired the problem is easy once you establish $$\frac{1}{\sqrt{1-x}} = \sqrt{2}\sum_{n=0}^\infty P_n(x)\tag{1}$$ With this in hand the ortogonality relation $\int_{-1}^1 P_m(x)P_n(x){\rm dx} = \frac{2\delta_{nm}}{2m+1}$ does the rest of the job $$\color{red}{\int_{-1}^1\frac{P_m(x)}{\sqrt{1-x}}{\rm d}x = \sqrt{2}\sum_{n=0}^\infty ...


2

This is not an answer but it is too long for a comment. Considering $$I_{n,l}=\int_{-1}^{+1} P_l^2(x) P_n(x)\,dx$$ this integral seems to show interesting patterns I give you below (this is just based on numerical evaluation and observation). For positive values of $n,l$ for $l<n \implies I_{n,l}=0$ for $l=n+(2k-1)\implies I_{n,l}=0$ for ...


1

I believe the value of the constant (finite limit) should be $$ \int_1^\infty dt \frac{e^{-a t}\sqrt{t^2-1}}{t}\ . $$ [checked with Mathematica for a few values] You can just use the integral representation $$ K_1(r)=r \int_1^\infty e^{-rt}\sqrt{t^2-1}dt $$ and swap the integrals. The integral in $r$ is easy $$ \int_a^R dr e^{-r t}=\frac{e^{-a t}-e^{-R ...



Top 50 recent answers are included