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Asymptotic Expansion of Integrals via Integration by Parts - Leading Behavior

$$\int_1^\infty \frac{\cos(xt)}{t}\,dt=-\text{Ci}(x)$$ Expanded as series $$\int_1^\infty \frac{\cos(xt)}{t}\,dt=-\log (x)-\gamma+\sum _{n=1}^\infty (-1)^{n+1} \frac {x^{2n}}{ 2n\,(2n)! }$$
Claude Leibovici's user avatar
0 votes

Evaluating $\lim_{t\to\infty}\left(\left(\log\left(t^2+\frac1{t^2}\right)\right)^{-1}\int_1^{\pi t}\frac{\sin^25x}{x}dx\right)$

Asssuming that $k$ and $t$ are both positive integers $$\int_1^{\pi t}\frac{\sin ^2(k x)}{x}\,dx=\frac{1}{2} (\text{Ci}(2 k)-\text{Ci}(2 k \pi t)+\log (\pi t))$$ Expanded as series, the numerator is ...
Claude Leibovici's user avatar
0 votes

Cardinality of the set of arguments in integration

If one wants to take the union of all these finite sets and say that they are being evaluated "during integration", then the cardinality of such evaluations is indeed $\aleph_0$. This is ...
athanos lee's user avatar
4 votes

Evaluating $\lim_{t\to\infty}\left(\left(\log\left(t^2+\frac1{t^2}\right)\right)^{-1}\int_1^{\pi t}\frac{\sin^25x}{x}dx\right)$

Consider $\int_1^ {\pi t} \frac{\cos^2(5x)}{x} dx$. Then this integral has the same growth behavior as $\int_1 ^{\pi t} \frac{\sin^2(5x)}{x} dx$ (that is, their quotient converges to 1; you can show ...
Pengin's user avatar
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1 vote
Accepted

Asymptotic expansion of $\int_0^{2\pi} \frac{\cos(nt)}{1+t^2} dt$

Start with $$\frac 1{1+t^2}=\frac 1{(t+i)(t-i)}=\frac i 2 \left(\frac{1}{t+i}-\frac{1}{t-i}\right)$$ Use obvious changes of variable, expand the trigonometir functions to obtain the antiderivative in ...
Claude Leibovici's user avatar
0 votes

evaluate $\int_{|z-e|=2} \frac{1}{(z-1)\log z}dz$

Here, $\mathrm{Log}z$ is the principal branch of the logarithm, defined in $\mathbb C\setminus\{x:x\le 0\}$. The function according to Cauchy integral formula: $$ f(z)=\frac{z-1}{{\mathrm{Log}}\,z} $$ ...
Yiorgos S. Smyrlis's user avatar
-1 votes

How do I integrate this? Problem(error) with integration

$$f'(x) = \frac{-4x+1}{2\times\sqrt{-4x^2+2x}}$$ Putting this derivative value here in the integral, $$\int_0^\frac{1}{2} \sqrt{1+(f'(x))^2} dx=\int_0^\frac{1}{2} \sqrt{1+\frac{(1-4x)^2}{4(2x-4x^2)}} ...
Gwen's user avatar
  • 1,280
1 vote

How to integrate $\int_0^1\frac{\ln^3(1+x)\,\ln^3x}x\mathrm{d}x$

Incomplete solution: Integrate by parts followed by writing $\frac{1}{1+x}=\frac{1}{x}-\frac{1}{x(1+x)}$, and then using the two generating functions: $$\ln^2(1-x)=2\sum_{n=1}^\infty\frac{H_{n-1}}{n}x^...
Ali Shadhar's user avatar
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1 vote

Integral representation of $\tan(z)$

Making the story short, using the Gaussian hypergeometric function $$I=\frac 2 \pi \int_0^p \frac{t^{a}-1}{t^2-1}\,dt=\frac{2 \tanh ^{-1}(p)}{\pi }-\frac{2 \,p^{a+1} }{(a+1)\pi }\, _2F_1\left(1,\...
Claude Leibovici's user avatar
2 votes

Evaluating $\lim_{t\to\infty}\left(\left(\log\left(t^2+\frac1{t^2}\right)\right)^{-1}\int_1^{\pi t}\frac{\sin^25x}{x}dx\right)$

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Felix Marin's user avatar
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1 vote

Definite integral of $(\sin (x)/x)^{100}$

Consider the general case of $$I_n=\int_{-\pi}^{+\pi}\left(\frac{\sin (x)}{x}\right)^n \,dx$$ Expanded as series around $x=0$ $$\left(\frac{\sin (x)}{x}\right)^n=1-\frac{n }{6}x^2+\frac{n(5 n-2)}{360} ...
Claude Leibovici's user avatar
2 votes

Integral $\int \frac{dx}{(1-\frac{c}x{} ) (\frac{3c}{x}-1)^{\frac{1}{2}}} $

Making the problem more general $$I=\int \frac{1}{\left(1-\frac{c}{x}\right) \sqrt{\frac{kc}{x}-1}}\,dx$$ The integral is not bad at all using what @David Quinn commented $$\sqrt{\frac{c k}{x}-1}=t \...
Claude Leibovici's user avatar
3 votes

Question About Function Integrability - Proposition 2.3.10 from Measury Theory by Donald Cohn

If $f\chi_{A_t}$ is not integrable, then $\mu(A_t) \le \frac{1}{t}\int f\chi_{A_t}$ holds for the reason that the right-hand side is $+\infty$.
Alex Ortiz's user avatar
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0 votes

Request for crazy integrals

Bit late to this post, but I've personally been on the hunt for the most exotic integrals I could come up with, here are some of the best in my opinion. The 4th one includes 6 fundamental constants. $$...
Aiden McDonald's user avatar
0 votes

Irrationality of $\sqrt{2}$ invoking some fact of definite integrals

Following is an alternative presentation of Jack D'Aurizio's answer that clarifies its relationship with Kostya_I's answer to a related question that I posed on MathOverflow. Kostya_I explained how ...
Timothy Chow's user avatar
1 vote

A common technique in number theory to evaluate integration

Calling this "dyadic summation" or "dyadic integration" or "dyadic decomposition" would all be understood. I'll note that generically it is not necessary to add a log ...
davidlowryduda's user avatar
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3 votes

Error with Cauchy Integral Formula

Expanding in a series at $\theta = \frac{3\pi}{2}$ gives that the integrand is $$\frac{1}{2 (1 + \sin \theta)} = \frac{1}{\left(\theta - \frac{3 \pi}{2}\right)^2} + g(\theta),$$ for some integrable (...
Travis Willse's user avatar
6 votes
Accepted

How to approximate $\frac{(2n+1)}{2^n} \int_0^1 \left( 1-x^2+\sqrt{1+2x^2-3x^4}\right)^ndx$

Since $-\ln f(x) \sim x^4$ as $x\to 0$ and $f(x)$ has its absolute maximum of $1$ there, Laplace's method gives $$\int_0^1f(x)^n \mathrm{d} x\sim\tfrac{\Gamma(\tfrac{1}{4})}{4} n^{-1/4}\approx\tfrac{1....
K B Dave's user avatar
  • 7,995
3 votes

Integral $\int_0^{\frac{\pi}{2}}\frac{\log\left(\sin x\right)}{\cos^2x+y^2\sin^2x}{d}x=-\frac{\pi}{2}\frac{\log\left(1+y\right)}{y}$

Use the following result Two integrals: $$\int_{0}^{\infty}\frac{\ln u}{1+y^2u^2}{\rm d}u=\frac{-\pi}{2y}\ln y,\quad \frac12\int_{0}^{\infty}\frac{\ln(1+u^2)}{1+y^2u^2}{\rm d}u =\frac{\pi}{2y}\ln\frac{...
Riemann's user avatar
  • 7,335
5 votes

Integral $\int_0^{\frac{\pi}{2}}\frac{\log\left(\sin x\right)}{\cos^2x+y^2\sin^2x}{d}x=-\frac{\pi}{2}\frac{\log\left(1+y\right)}{y}$

\begin{align} &\int_0^{\frac{\pi}{2}}\frac{\ln\left(\sin x\right)}{\cos^2x+y^2\sin^2x}{d}x\\ =& -\frac12\int_0^{\frac{\pi}{2}}\frac{\ln(1+\cot^2x)}{\cos^2x+y^2\sin^2x}{d}x\\ = & -\int_0^{\...
Quanto's user avatar
  • 98k
5 votes
Accepted

Error with Cauchy Integral Formula

Let $I=\int\frac{1}{2+2\sin{\theta}} d\theta$ Now, let $x=\tan{\frac{\theta}{2}}$. Then $\sin{\theta}=\frac{2}{1+x^2}$ and $d\theta=\frac{2}{1+x^2}dx$ After some algebraic manipulation this gives $I=\...
Red Five's user avatar
  • 1,724
0 votes

Computing Cauchy-Principal value with residue theorem on the positive line?

$$\frac{y}{(y-b)\sqrt{1+a^2 y^2} }=\frac{1}{\sqrt{1+a^2 y^2}}+\frac{b}{(y-b)\sqrt{1+a^2 y^2} }$$ $$I=\int \frac{b}{(y-b)\sqrt{1+a^2 y^2} }\,dy$$ $$y=\frac{\sinh (t)}{a} \quad \implies \quad I=\int \...
Claude Leibovici's user avatar
1 vote

Integral of two Hermite polynomials, one shifted, and exponentials

Once you use the relation $H_{n}(x+y)=\sum_{r=0}^{n} nC_r\,H_{r}(x)(2y)^{(n-r)}$ we can map it into the Associated Laguerre polynomial and we can get the result, $$I_{nm}(a)=e^{-a^2/4}\times 2^n\sqrt{\...
Sijo Joseph's user avatar
1 vote

The vertices of a hexagon are random points on a unit circle; $a,b,c$ are the lengths of three random sides. Conjecture: $P(ab<c)=\frac35$.

Choosing the sides to compare at random isn't necessary, since every permutation of side lengths is equally likely; we may as well fix one vertex at $(1,0)$ and let $a,b,c$ be the lengths of the first ...
mjqxxxx's user avatar
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2 votes
Accepted

Applying Integration by parts to the Dirac Delta Sifting Property

If we define $H(x)$ as the step function so that $H'(x) = \delta(x)$ in some weak sense so that $H(x) = 0$ when $x < 0$ and assume that $f$ and its derivatives decay sufficiently fast, then ...
whpowell96's user avatar
  • 5,632
7 votes
Accepted

Integral $ \int \frac{1}{\sqrt[3]{(x-1)^7 (x + 1)^2}} \mathrm{d} x $

Substitute $x=\frac{1-t}{1+t}$ \begin{align} &\int \frac{1}{\sqrt[3]{(x-1)^7 (x + 1)^2}} {d} x \\ =& -\frac14\int (t^{-7/3}+ t^{-4/3})dt = \frac34 t^{-1/3}+\frac3{16}t^{-4/3}+C \end{align}
Quanto's user avatar
  • 98k
8 votes

What is the arc-length parameterization of an ellipse?

So, the standard parameterization of the ellipse $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$ is given by $$ r(t) = (a \cos t , b \sin t) \quad t \in [0,2\pi) $$ The arc-length parameterization is ...
PrincessEev's user avatar
  • 44.1k
8 votes
Accepted

What is the arc-length parameterization of an ellipse?

No. You have stumbled on a well known difficult problem. In integral calculus, an elliptic integral is one of a number of related functions defined as the value of certain integrals, which were first ...
Ethan Bolker's user avatar
  • 95.6k
0 votes

Non-Uniform Partition Integrals

In the Riemann integral $dx$ is just a hint to represent the variable we're integrating with respect to. In fact, in most Analysis texts it is omitted and the notation is $\int_a^b f$. $f(dx)$ is not ...
Sgg8's user avatar
  • 1,362
0 votes

Prove $\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}$

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Felix Marin's user avatar
  • 89.7k
3 votes
Accepted

Evaluating the sum $\sum_{n=1}^{\infty}{\frac{(-1)^n n}{n^2+1}}$

To evaluate the sum decompose the summand into partial fractions – complexly: $$\frac i2\left(\sum_{n=1}^\infty\frac{(-1)^n}{1+in}-\sum_{n=1}^\infty\frac{(-1)^n}{1-in}\right)$$ Now use (6) on ...
Parcly Taxel's user avatar
3 votes
Accepted

Closed form of the integral $\int_{0}^{1} \log^n \left (\frac {1-x}{1+x}\right )dx$

Note that we have for $n>1$ $$\begin{align} i(n)&=\int_0^1 \log^n\left(\frac{1-x}{1+x}\right)\,dx\tag1\\\\ &=2\int_0^1 \frac{\log^n(x)}{(1+x)^2}\,dx\tag2\\\\ &=2\left.\frac{d^n}{dt^n}\...
Mark Viola's user avatar
  • 180k
10 votes

Closed form of the integral $\int_{0}^{1} \log^n \left (\frac {1-x}{1+x}\right )dx$

Substitute $t= \frac {1-x}{1+x}$ \begin{align}&\int_{0}^{1} \ln^n\frac {1-x}{1+x}\;dx\\ =&\ 2\int_0^1 \frac{\ln^n t}{(1+t)^2}dt =2\int_0^1 {\ln^n t}\ d\left(\frac t{1+t}\right)\\ \overset{ibp}=...
Quanto's user avatar
  • 98k
1 vote

How can we approximate $\sum_{j=0}^n{\sum_{k=0}^j{c^j k^{1/2}}}$ by integrals?

As I commented, the simplest for of Euler-MacLaurin summation gives a monster (I refuse to type it - too long). Trying for $c=\frac 12$ and $n=10^4$ gives $\color{red}{3.2931}33$
Claude Leibovici's user avatar
4 votes

differential equations, substitution suggested by the equation

You got rightfully stuck on that integral because you can't integrate it, it can only be solved numericallly. There must have been a typo in the question. My guess is that the question should have ...
Omar Dennaoui's user avatar
3 votes

What goes wrong with Stokes theorem if a surface is not orientable?

Since this has gotten bumped: The problem (as noted in the comments) is not with Stokes' theorem, but defining a flux integral (or the integral of a differential $2$-form) on a non-orientable surface ...
Andrew D. Hwang's user avatar
0 votes

Example of an integrable function $f$ such that $|\int(1+x^2)f(x)\,dx|<|\int f(x)\,dx|$

For $h_1(x)=1_{[0,1]}(x)$ and $h_2(x)={1\over 2}1_{(1,3]}(x)$ let $$f(x)=(1+x^2)^{-1}[h_1(x)-h_2(x)]$$ Then $$\int (1+x^2)f(x)\,dx = \int h_1(x)\,dx- \int h_2(x)\,dx =0$$ $$\int f(x)\,dx =\int h_1(...
Ryszard Szwarc's user avatar
5 votes

What is minimum of the integral function $I(x)= \int_0^\infty \frac{1}{(1+t^x)^x} \,dt$

$$I(x)= \int_0^\infty \frac{1}{(1+t^x)^x} \,dt=\frac{\Gamma \left(1+\frac{1}{x}\right) \Gamma \left(x-\frac{1}{x}\right)}{\Gamma(x)}$$ Take logarithms and differentiate $$\frac{I'(x)}{I(x)}=-\frac{x^2 ...
Claude Leibovici's user avatar
0 votes

Can one use asymptotic behaviour to deduce the induced distribution from a function?

Yes, in a distributional sense means: "multiplied with any test function and then integrated". The terms you are manipulating are defined in that sense. Also they are not infinite when used ...
Jos Bergervoet's user avatar
9 votes

What is minimum of the integral function $I(x)= \int_0^\infty \frac{1}{(1+t^x)^x} \,dt$

Substituting $u = (1 + t^x)^{-1}$ transforms the integral to $$I(x) = \frac1x \int_0^1 u^{x - \frac1x - 1} (1 - u)^{\frac1x - 1} \,du = \frac1x \mathrm{B}\left(x - \frac1x, \frac1x\right) = \frac{\...
Travis Willse's user avatar
0 votes

Is there a closed form for $\int_0^\infty \frac{1-\cos(tx)}{e^t-1}dt$?

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Felix Marin's user avatar
  • 89.7k
1 vote

Asymptotics of $\int_{0}^{1} \frac{\tan^{-1}(x^n)}{\sqrt{1 - x^n}} \, dx $

Performing the change of integration variables from $x$ to $t$ via $x=t^{1/n}$ yields $$ I_n = \frac{1}{n}\int_0^1 {t^{1/n} \frac{{\arctan (t)}}{t}\frac{1}{{\sqrt {1 - t} }}{\rm d}t} . $$ We can ...
Gary's user avatar
  • 32.3k
2 votes

Show that $\int_{0}^{1} \frac{\tan^{-1}(x^2)}{\sqrt{1 - x^2}} \, dx = \frac{1}{2}\pi \tan^{-1}\left(\sqrt{\frac{1}{\sqrt{2}} - \frac{1}{2}}\right)$

Let’s start from $$I=\int_0^{\frac{\pi}{2}} \tan ^{-1}\left(\sin ^2 t\right) d t$$ with its parametrised integral $$\int_0^{\frac{\pi}{2}} \tan ^{-1}\left(a\sin ^2 t\right) d t$$ Differentiating w.r.t ...
Lai's user avatar
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1 vote
Accepted

Bound an integral by a function with an appropriate decay

We prove that $C = 2$ is valid. (1) $t > 1$ Let $$f(v) := (1 + t^v)\int_0^t \frac{1}{1 + s^v} \frac{1}{1 + t - s}\, \mathrm{d} s.$$ We have $$f'(v) = \int_0^t \left(\frac{t^v\ln t}{(1 + s^v)(1 + t -...
River Li's user avatar
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1 vote

Example of an integrable function $f$ such that $|\int(1+x^2)f(x)\,dx|<|\int f(x)\,dx|$

Let $f(x) = 1_{[0, 1)}(x) - 1_{[1, \frac{3}{2}]}(x)$. It is clearly in $L^1$. The LHS equals $\frac{5}{12}$ while the RHS equals $\frac{1}{2}$.
David Gao's user avatar
  • 5,095
2 votes

How to integrate $\int_{0}^{1}\frac{x^{100}}{1+x}dx$?

We have for $J$ the integral to be computed: $$ \begin{aligned} J &:= \int_{0}^{1}\frac{x^{100}}{x+1}\;dx = \int_{0}^{1}\underbrace{\frac{x^{100}+x}{x+1}} _{=x^{99}-x^{98}\pm\dots - x^4 + x^3 - x^...
dan_fulea's user avatar
  • 33.5k
0 votes

Seeking methods to solve: $\int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt$

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Felix Marin's user avatar
  • 89.7k
3 votes
Accepted

How to integrate $\int_{0}^{1}\frac{x^{100}}{1+x}dx$?

As suggested in the comments, since the interval of integration is between $0$ and $1$, $$\frac{1}{1+x}=\sum_{n\geq 0}(-x)^n<\infty$$ and so, the integrand becomes $$\frac{x^{100}}{1+x}=\sum_{n\geq ...
Conreu's user avatar
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0 votes

Volume of a Cone using Change of Variables

One way of doing it is; Let $u=x/a, v=y/b, w=z/c$. The Jacobian $\frac{d(x,y,z)}{d(u,v,w}$ for this transformation is : ${abc}$ So we are finding the volume bounded by : $u^2+v^2=w^2$ and multiplying ...
J.Dmaths's user avatar
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