Linked Questions

136
votes
37answers
12k views

Is there any integral for the Golden Ratio?

This is a curiosity. I was wondering about math important/famous constants, like $e$, $\pi$, $\gamma$ and obviously $\phi$. The first three ones are really well known, and there are lots of integrals ...
151
votes
4answers
14k views

Some users are mind bogglingly skilled at integration. How did they get there?

Looking through old problems, it is not difficult to see that some users are beyond incredible at computing integrals. It only took a couple seconds to dig up an example like this. Especially in a ...
62
votes
4answers
7k views

Integral $\int_0^1\frac{\ln\left(x+\sqrt2\right)}{\sqrt{2-x}\,\sqrt{1-x}\,\sqrt{\vphantom{1}x}}\mathrm dx$

Is there a closed form for the integral $$\int_0^1\frac{\ln\left(x+\sqrt2\right)}{\sqrt{2-x}\,\sqrt{1-x}\,\sqrt{\vphantom{1}x}}\mathrm dx.$$ I do not have a strong reason to be sure it exists, but I ...
21
votes
3answers
7k views

Evaluate $\int_0^{\pi/2}\frac{x^2\log^2{(\sin{x})}}{\sin^2x}dx$

How evaluate this integral? $$I=\int_0^{\pi/2}\frac{x^2\log^2{(\sin{x})}}{\sin^2x}\,dx$$ Note: $$\int_0^{\pi/2}\frac{x^2\log{(\sin x)}}{\sin^2x}dx=\pi\ln{2}-\frac{\pi}{2}\ln^22-\frac{\pi^3}{12}.$$
15
votes
9answers
1k views

Definite integrals with interesting results [closed]

I just stumbled across the fact that $\int_{-\infty}^{+\infty}{e^{-x^2}dx}=\sqrt{\pi}$. This intrigued my already-existing interest in integrals. It made me wonder, are there other integrals with ...
50
votes
2answers
3k views

Integral $\int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{(r-1)x^{2} + sx + 1}{(r-1)x^{2} - sx + 1} \right) \, \mathrm dx$

Regarding this problem, I conjectured that $$ I(r, s) = \int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{(r-1)x^{2} + sx + 1}{(r-1)x^{2} - sx + 1} \right) \, \mathrm dx = 4 \pi \...
25
votes
4answers
2k views

Yet another log-sin integral $\int\limits_0^{\pi/3}\log(1+\sin x)\log(1-\sin x)\,dx$

There has been much interest to various log-trig integrals on this site (e.g. see [1][2][3][4][5][6][7][8][9]). Here is another one I'm trying to solve: $$\int\limits_0^{\pi/3}\log(1+\sin x)\log(1-\...
19
votes
3answers
3k views

Integral $\int_0^1\frac{\log(x)\log(1+x)}{\sqrt{1-x}}\,dx$

I'm trying to evaluate this definite integral: $$\int_0^1\frac{\log(x) \log(1+x)}{\sqrt{1-x}} dx$$ It's clear that the result can be expressed in terms of derivatives of a hypergeometric function with ...
15
votes
3answers
665 views

Evaluating the integral $\int_0^{\frac{\pi}{2}}\log\left(\frac{1+a\cos(x)}{1-a\cos(x)}\right)\frac{1}{\cos(x)}dx$

How can I evaluate the following integral? $$ \int_0^{\pi/2} \log\left(\frac{1 + a\cos\left(x\right)}{1 - a\cos\left(x\right)}\right)\, \frac{1}{\cos\left(x\right)}\,{\rm d}x\,, \qquad\left\vert\,a\...
17
votes
2answers
763 views

Integral $\int_0^\infty F(z)\,F\left(z\,\sqrt2\right)\frac{e^{-z^2}}{z^2}dz$ involving Dawson's integrals

I need you help with evaluating this integral: $$I=\int_0^\infty F(z)\,F\left(z\,\sqrt2\right)\frac{e^{-z^2}}{z^2}dz,\tag1$$ where $F(x)$ represents Dawson's integral: $$F(x)=e^{-x^2}\int_0^x e^{y^2}...
9
votes
4answers
1k views

What's the deal with integration?

So at uni we learned tricks and techniques for integration until cows came home. But to what end? Any/All definite integrals can be evaluated using numerical methods. Most integrals in application can ...
15
votes
4answers
522 views

Evaluate $\int_1^\infty \frac {dx}{x^3+1}$

I would like some help with the following integral. I would like to find a contour line to evaluate $$\int_1^\infty \frac {dx}{x^3+1}$$ So one can see that on any circumference it goes to $0$, but ...
15
votes
2answers
765 views

Integral $\int_1^{\sqrt{2}}\frac{1}{x}\ln\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)dx$

Calculate the following integral: \begin{equation} \int_1^{\sqrt{2}}\frac{1}{x}\ln\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)dx \end{equation} I am having trouble to calculate the integral. I tried ...
7
votes
4answers
2k views

How prove this $I=\int_{0}^{\infty}\frac{1}{x}\ln{\left(\frac{1+x}{1-x}\right)^2}dx=\pi^2$

Prove this $$I=\int_{0}^{\infty}\dfrac{1}{x}\ln{\left(\dfrac{1+x}{1-x}\right)^2}dx=\pi^2$$ My try: let $$I=\int_{0}^{\infty}\dfrac{2\ln{(1+x)}}{x}-\dfrac{2\ln{|(1-x)|}}{x}dx$$
30
votes
2answers
798 views

Does $\int_{-1}^1\frac{\arctan x}{\text{arctanh}\,x}\,\mathrm{d}x$ have a closed form?

$$\newcommand{\arctanh}{~\mathrm{arctanh}~}\newcommand{\sech}{~\mathrm{sech}~}$$ $$I=\int_{-1}^1\frac{\arctan x}{\arctanh x}\,\mathrm{d}x$$ Mathematica gives an approximate result of $I=1....

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