7 votes
Accepted

how to evaluate $ \int_0^1 \frac{\arctan \left(x \pm \sqrt{x^2+1}\right)}{x+1} d x $

Integrate by parts \begin{align} &\int_0^1 \frac{\tan^{-1} \left(x +\sqrt{x^2+1}\right)}{x+1} d x\\ =& \ \tan^{-1}\left(x +\sqrt{x^2+1}\right)\ln(1+x)\bigg|_0^1 -\frac12\int_0^1 \frac{\ln(1+x)}...
Quanto's user avatar
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6 votes

How to solve $\int_0 ^1 \left( \frac{x^2}{1+x^2} \right)\frac{1-x\tan(x)+ \tan(x)-x}{1 -x\tan(x)-\tan(x)-x}dx$?

The integral diverges at the singularity $s=0.40262817\dots$ which is the solution to $x+\arctan x=\pi/4$. The integral $$\begin{align}I=&\int \frac{x^2}{1+x^2}\left (\frac{1}{1-\tan(1+\arctan x)}-...
Eric Ley's user avatar
  • 539
4 votes

Evaluate $\int_{0}^{1}K(x)^2\text{d}x -\int_{0}^{1} \frac{x\sqrt{1-x^2} }{2-x^2}K(x)^2\text{d}x$

To calculate this integral we can utilize the following Ramanujan identity: $$\int_0^\sqrt{1-x^2} \frac{K(y)}{\sqrt{1-y^2}\sqrt{1-x^2-y^2}}dy=\frac{2}{1+x}K^2\left(\sqrt{\frac{1-x}{1+x}}\right)\tag{*}$...
Zacky's user avatar
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3 votes
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Solve trig integral $\int_{0}^{\pi/2} \left(\frac{\sin5x}{\sin x}\right)^2 \,dx $

Per symmetry $$I=\int_{0}^{\pi/2} \left(\frac{\sin5x}{\sin x}\right)^2 \,dx =\frac14 \int_{0}^{2\pi} \left(\frac{\sin5x}{\sin x}\right)^2 \,dx $$ With $\frac{\sin5x}{\sin x}=\frac{e^{i5x}- e^{-i5x}}{e^...
Quanto's user avatar
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3 votes

Remarkable logarithmic integral $\int_0^1 \frac{x \log ^2(x) \log (1-x)}{1+x^2} dx$

To continue the evaluation of the integral by using your work so far, you might first notice that $$\int_0^1 \frac{\log ^2(x) \log \left(1+x^2\right)}{1-x} \textrm{d}x=\int_0^1 \frac{\log ^2(x) \log((...
user97357329's user avatar
  • 5,324
3 votes

Integral $\int_0^{1/\sqrt{2}}\frac{K\left[\sqrt{1-k^2}\right]}{\sqrt{1-2k^2}\sqrt{1-k^2}}dk$

We may define $$ K_{4}(x) =\frac{K\left (\sqrt{\frac{2x}{1+x} } \right ) }{ \sqrt{1+x} } =\frac{\pi}{2}\,_2F_1\left ( \frac14,\frac34;1;x^2 \right ). $$ And consider the contour integral on the ...
Setness Ramesory's user avatar
3 votes

how to evaluate $ \int_0^1 \frac{\arctan \left(x \pm \sqrt{x^2+1}\right)}{x+1} d x $

Noting $$ (x+\sqrt{x^2+1})(\sqrt{x^2+1}-x)=1 $$ one has $$\begin{eqnarray} I+J &=& \int_0^1 \frac{\arctan \left(x+\sqrt{x^2+1}\right)+\arctan \left(x-\sqrt{x^2+1}\right)}{x+1} \,dx\\ &=&...
xpaul's user avatar
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3 votes

Evaluate $\int_{0}^{1}K(x)^2\text{d}x -\int_{0}^{1} \frac{x\sqrt{1-x^2} }{2-x^2}K(x)^2\text{d}x$

Not a complete answer. $$I=\int_0^1K^2dk-\int_0^1\frac{kk'}{2-k^2}K^2dk$$ Well as far as I have seen Closed Form of the First Integral is not known, so our first step should be to remove the same ...
Miracle Invoker's user avatar
3 votes

Calculating $\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^2 \, dx$

The following is another way to show that $$ \begin{align} \int_{0}^{\infty} \operatorname{Li}_{2}^{2} \left(-x^{-1/a} \right) \, \mathrm dx &= a\int_{0}^{\infty} \frac{\operatorname{Li}_{2}^{2}(-...
Random Variable's user avatar
2 votes

Evaluating the Definite Integral $\int_0^{\pi}\cos^{2n} \theta d\theta$

\begin{align} \int_{0}^{\pi}\cos^{2n}\theta\ dt =\sum_{k=0}^{2n} \frac{\binom {2n}k}{2^{2n}}\int_0^{\pi}\cos[2(n-k) \theta]d \theta =\frac{\pi \binom {2n}n}{2^{2n}} \end{align}
Quanto's user avatar
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2 votes

Show that $ \int_0^{\pi\over 2}\frac{\sin(2nx)}{\sin^{2n+2}(x)}\frac{1}{e^{2\pi \cot x}-1}dx =(-1)^{n-1}\frac{2n-1}{4(2n+1)} $

(Assuming $n$ is a positive integer.) As you have noted $$\frac{\sin 2nx}{\sin^{2n}x}=\Im(\cot x+i)^{2n}=\frac1{2i}\left((\cot x+i)^{2n}-(\cot x-i)^{2n}\right),$$ the given integral, after the ...
metamorphy's user avatar
  • 38.9k
2 votes
Accepted

Jeffrey's integrals

Denote \begin{align} I_n(t)&= \int_{-1}^1 \frac{P_n(x)}{\sqrt{1-2x t+t^2}}dx\\ J_n(t)& = \int_{-1}^1 \frac{P_n(x)}{(1-2x t+t^2)^{3/2}}dx \end{align} Then, per the generating function $\frac1{\...
Quanto's user avatar
  • 95.7k
2 votes

Fourier transform of $\exp(-\cosh(x))$

The modified Bessel function of the second kind, $K_\alpha$ has an integral representation given by $$K_\alpha(x)=\int_0^\infty e^{-x\cosh(t)}\cosh(\alpha t)\,dt$$ for $\text{Re}(x)>0$. Now, let $\...
Mark Viola's user avatar
  • 179k
2 votes

Improper Integral $\int_{0}^{\infty} \log(t) t^{-\frac{1}{2}} \exp\left\{-t\right\} dt$

Too long for a comment Using the Frullani' integral $\,\,\displaystyle \ln (t)=-\int_0^\infty\frac{e^{-xt}-e^{-x}}xdx\,\,$ and changing the order of integration $$\int_0^\infty \ln (t) \,t^{-\frac12} ...
Svyatoslav's user avatar
  • 15.1k
2 votes
Accepted

How do I combine a PDF with another continuous function and then do a sum product?

Let random variable $X$ represent a student's test score, with pdf $f(x)$ for $a \le x \le b$. Let $g(x) = \mathbb P[C \mid X = x]$ be the conditional probability of making it to college ($C$) given ...
Robert Israel's user avatar
1 vote

How to solve $\int_0 ^1 \left( \frac{x^2}{1+x^2} \right)\frac{1-x\tan(x)+ \tan(x)-x}{1 -x\tan(x)-\tan(x)-x}dx$?

The integrand has the form $$\frac{x^2}{1+x^2}\ \frac {(1-x)(1-\tan x)}{(1-x)-(1+x)\tan x}$$ It has a simple pole at $$x=\frac{1-\tan x}{1+\tan x} = \frac{\cos x-\sin x}{\cos x+\sin x}$$ with $x=0.4\...
Roland F's user avatar
  • 1,627
1 vote

Jeffrey's integrals

$$\sum_{n=0}^{\infty}t^nI_n(3/2)=\int_{-1}^1\frac{dx}{\sqrt{1-2x+t^2}(2\cosh u -2x)^{3/2}}=\frac{\sqrt{2}e^{-|u|/2}}{\sinh |u|}\times \frac{1}{1-te^{-|u|/2}}$$ is computable by the change of variable $...
Letac Gérard's user avatar
1 vote

Evaluate $\lim_{x\to0^+}\int_{ax}^{bx}\frac{\tanh y\cdot\tan^2y}{y^4}\,\mathrm{d}y$

Since $x\to 0^+$, you can consider the expansion of the integrand function, keeping in mind that $\tanh y \sim y$ and $\tan y \sim y$: $$\frac{\tanh y\tan^2y}{y^4}\sim \frac1y $$ Hence you have: $$\...
Sine of the Time's user avatar
1 vote

Let f be an infinitely differentiable function such that f(1) = 0, f(5) = ln 4, f'(1) = 2 and f'(5) = −2. Using the given equation(below),Find f(x).

Since we have an $\ln$ in the initial conditions, assume that $f(x) = \ln g(x)$. Then $f'(x) = \frac{g'(x)}{g(x)}$ and $f''(x) = \frac{g''(x)g(x) - g'(x)^2}{g(x)^2}$. Plugging this into the given ...
Dan's user avatar
  • 14.7k
1 vote

How to evaluate $\int_0^{\frac{\pi}{2}} \frac{\ln(\cos(x))}{1+\sin^2(x)} \, dx$

Let $t=\tan x$, then $$ \begin{aligned} \int_0^{\frac{\pi}{2}} \frac{\ln (\cos x)}{1+\sin ^2 x} d x = & \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x \ln (\cos x)}{\sec ^2 x+\tan ^2 x} d x \\ = & -\...
Lai's user avatar
  • 19.9k
1 vote

Log-tangent integral using exponential generating function $\int_0^{\pi/2}x\log^n(\tan x)\,dx$

For the even case \begin{align} I_{2m}=& \int_0^{\pi/2}x\ln^{2m}(\tan x)\ \overset{t=\tan x}{dx}\\ =& \int_0^{\infty}\frac{\tan^{-1}t\ln^{2m}t}{1+t^2}\ \overset{t\to 1/t}{dt} =\int_0^{\infty}\...
Quanto's user avatar
  • 95.7k
1 vote

The indefinite integral $\int\frac{dx}{x\ln(x+1)}$

If you are interested in the asymptotics of the function $$ \int_1^x \frac{1}{t\log(1+t)} dt $$ for large $x$, one good way to approximate it is to observe that $\frac1t \approx \frac1{1+t}$. In fact $...
Dark Malthorp's user avatar
1 vote

How to evaluate $\int_0^\pi \cos(x) \cos(2x) \cos(3x) \cos(4x)\, dx$

Let $$I=\int_0^\pi \cos(x) \cos(2x) \cos(3x) \cos(4x)\, dx,$$ since $\cos(x+\pi) \cos(2x+2\pi) \cos(3x+3\pi) \cos(4x+4\pi)=\cos(x) \cos(2x) \cos(3x) \cos(4x)$, $$I=\int_{-\frac\pi2}^\frac\pi2 \cos(x) \...
Riemann's user avatar
  • 6,685
1 vote

Evaluation of $\int \limits _0 ^{2 \pi} \frac {(r \cos \phi +x) \cos(n\phi)} {r^2+2xr \cos \phi +x^2} d\phi$

Note that, for $|a|>1$ \begin{align} I(a) =& \int_0^{2\pi} \cos(n\phi)\ln(1+2a\cos\phi+a^2)\ d\phi\\ = & \int_0^{2\pi} \cos(n\phi)\bigg(2\sum_{k=1}^\infty \frac{(-1)^{k+1}}{ka^k}\ \cos(k \...
Quanto's user avatar
  • 95.7k
1 vote

Calculating $\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^2 \, dx$

A potentially baffling solution is to rewrite in terms of Meijer's $G$ function, use the formula for integrating a pair of $G$s, then evaluate. Convert the dilogarithm to a generalized hypergeometric ...
Eric Towers's user avatar
  • 66.6k

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