New answers tagged definite-integral
0
The integral in the LHS is a partial case of formula 3.665.2 from Gradshteyn & Ryzhik. The integral in the RHS also is well-known (ibid 3.621.1 ). I don't find any motivation to reinvent the wheel.
1
This can be rewritten
$
\int_0^{\pi} \frac{1}{\sqrt{a^2 + b^2 + c^2 +1 - 2a \cos t - 2b \sin t}} dt
$
which has the form
$
A \int_0^{\pi} \frac{1}{\sqrt{1 + B \sin(t+\phi)}} dt
$
for appropriate constants $A$, $B$, and $\phi$. I just tried the indefinite integral
$
\int \frac{1}{\sqrt{1 + B \sin(t+\phi)}} dt
$
in Wolfram Alpha. It provided a closed ...
3
Note that $(a-\cos(t))^2 + (b-\sin(t))^2 + c^2 = a^2 + b^2 + c^2 + 1 - 2(a \cos(t) + b \sin(t))$.
Also note that $a \cos (t) + b \sin(t) = \sqrt{a^2 + b^2} \sin (t + F)$.
Let $d = a^2 + b^2 + c^2 + 1$ and $e = 2\sqrt{a^2 + b^2}$. Then, our integral becomes
$\displaystyle\int_0^\pi \frac{1}{\sqrt{d - e \sin(t + F)}}\mathrm dt$, where $d,e$, and $F$ are ...
-1
By using the formula http://upload.wikimedia.org/wikipedia/en/math/7/2/a/72a1058ad2087aec467af24bddcf9479.png, we have $\int_0^1\sqrt{\tan^{-1}x}~dx=\sum\limits_{n=1}^\infty\sum\limits_{m=1}^{2^n-1}\dfrac{(-1)^{m+1}}{2^n}\sqrt{\tan^{-1}\dfrac{m}{2^n}}$
5
Hint:
Fix $x=-x \implies dx=-dx$
$$I=-\int_{\pi}^{-\pi}\dfrac{\sin n(-x)}{(1+(\pi)^{-x})\sin (-x)} dx$$
$$I= \int_{-\pi}^{\pi}\dfrac{\sin n(-x)}{(1+(\pi)^{-x})\sin (-x)} dx$$
$$I=\int_{-\pi}^{\pi}\dfrac{\pi^x \sin n(x)}{(\pi^x+1)\sin x} dx$$
Add this to $I=\int_{-\pi}^{\pi}\dfrac{\sin nx}{(1+\pi^{x})\sin x}dx$
$$2I= \int_{-\pi}^{\pi} \dfrac{\sin ...
5
You can have a solution as an infinite sum
$$\int_0^1 \sqrt {\tan^{-1}x}\space dx = \frac{\sqrt {\pi }}{2}-\frac{1}{\sqrt {\pi }}\sum _{n=0}^{\infty }{\frac {\psi \left( 2\,n+1
,\frac{1}{2} \right) }{ 16^n\left( 4\,n+3 \right) \left( 2\,
n+1 \right) !}} \sim 0.6298233443,$$
where $\psi(x)$ is the digamma function.
0
Since the function $x\mapsto\sqrt{x}$ is nondecreasing, every Riemann sum associated to the subdivision $s=(x_i)_{0\leqslant i\leqslant n}$ is between $R^+(s)$ and $R^-(s)$ with
$$
R^-(s)=\sum_{i=1}^n(x_i-x_{i-1})\sqrt{x_{i-1}},\qquad R^+(s)=\sum_{i=1}^n(x_i-x_{i-1})\sqrt{x_{i}}.
$$
If the mesh of $s$ is $\delta(s)$,
$$
...
1
Do you mean something like this:
$f(x)= \sqrt{x}$,
$$\int_a^b \sqrt{x}=\lim_{n\to \infty} \dfrac{b-a}{n} \sum_1^n\sqrt{x_i}$$
2
Let us divide the region $a\le x\le b$ into $n$ sub-region as $a,ar,ar^2,\cdots ar^n=n$ such that $\delta_i=ar^i-ar^{i-1}$
If real $m\ne-1,$
$$\int_a^b x^mdx=\lim_{n\to\infty}\left(\sum_{1\le i\le n}(ar^i)^m\delta_m\right)$$
$$=\lim_{n\to\infty}\left(\sum_{1\le i\le n} (ar^i)^m(ar^i-ar^{i-1})\right)$$
$$=a^{m+1}\lim_{r\to1}(r-1)\sum_{1\le i\le n} ...
2
HINT:
If $$y_n=\int_0^1\frac{x^n}{x+5}dx$$
$$\int_0^1\frac{x^n}{x+5}=\int_0^1\frac{x^{n-1}(x+5-5)}{x+5}dx=\int_0^1x^{n-1}dx-5\int_0^1\frac{x^{n-1}}{x+5}dx$$ as $x+5\ne0$
Alternatively,
$$\text{So,}y_n+5y_{n-1}$$
$$=\int_0^1\frac{x^n}{x+5}dx+5\int_0^1\frac{x^{n-1}}{x+5}dx$$
$$=\int_0^1\frac{x^{n-1}(x+5)}{x+5}dx=\int_0^1x^{n-1}dx$$ as $x+5\ne0$
0
$$\int_{-\infty}^{+\infty} \quad x^k\quad \left(\frac{\sin(\pi a x)}{\pi ax}\right)^2\quad \exp(-bx^2)\,dx$$
can be solved using known integrals involving bessel function of the first kind.
Since
$$\left(\frac{\sin(\pi a x)}{\pi ax}\right)^2=\frac{1}{2ax}J_{\frac{1}{2}}(\pi ax)J_{\frac{1}{2}}(\pi ax)$$
can be written as
$$\int_{-\infty}^{+\infty} ...
7
I think the most ecological approach to the problem is as follows:
Denote $x=\sin\varphi$ and recall that $\arctan x=\frac{1}{2i}\ln\frac{1+ix}{1-ix}$. One then obtains the integral
...
1
Using the change of variables $ u=e^{-x} $, we have
$$\int_{0}^{\infty}\frac{x^3}{e^x-1}\ln(e^x - 1)\,dx = \int _{0}^{1}\!{\frac { \left( \ln \left( u \right) \right) ^{3}
\ln \left( 1-u \right) }{u-1
}}{du}- \int _{0}^{1}\!{\frac { \left( \ln \left( u \right) \right)^{4} }{u-1
}}{du}. $$
Now, just use the technique which has been used to find ...
0
Wolfram Alpha checked that this improper integral does not converge.
1
You made a mistake there somewhere: the value of $X(0)$ makes no sense. Here's what I get for the LT:
$$X(s) = \frac{\dot{x}_0 + (s+\delta) x_0}{s^2+\delta s+\omega_0^2} + \frac{\gamma s}{(s^2+\delta s+\omega_0^2)(s^2+\omega^2)} $$
where $x_0 = x(0)$ and $\dot{x}_0 = \dot{x}(0)$. As you point out, the ILT is given by
$$\frac{1}{i 2 \pi} \lim_{R \to ...
11
Mathematica says that the answer is
$$\pi^2\zeta(3)+12\zeta(5)$$
I will try to figure out how this can be proven.
Added: Let me compute the 2nd integral in Ron Gordon's answer:
\begin{align}\int_{0}^{\infty}\frac{x^3 e^{-x}}{1-e^{-x}}\ln(1-e^{-x})\,dx
&=-\frac32\int_0^{\infty}x^2\ln^2(1-e^{-x})\,dx=\\&=-\frac32\left[\frac{\partial^2}{\partial ...
8
How about pulling factors of $e^{-x}$ from both the denominator and log terms? Then you end up with two separate integrals:
$$\int_0^{\infty}dx \frac{x^4 \, e^{-x}}{1-e^{-x}} + \int_0^{\infty}dx \frac{x^3 \, e^{-x}}{1-e^{-x}} \log{(1-e^{-x})}$$
In both cases, you Taylor expand the denominator in $e^{-x}$. For the first integral, this results in
...
2
Hint: Note that when $r<1$ this is the real part of
$$\frac12 \int_{|z|=r} \log(1-z)^2\,\frac{dz}{iz}$$
for the usual branch of $\log$.
When $r>1$, you're going to need to make a branch cut.
2
Two hints:
The integrand is an even function of $\theta$ $\Rightarrow$ the integral can be written as $\frac12\int_0^{2\pi}$.
$1-2r\cos\theta+r^2=(1-re^{i\theta})(1-re^{-i\theta})$.
6
A closed form indeed exists for this integral:
$$\int_0^\infty\frac{\log\left(1+\frac{\pi^2}{4\,x}\right)}{e^{\sqrt{x}}-1}\mathrm dx=\pi^2\left(\log\frac{\pi\,A^9\sqrt{2}}{\Gamma(\frac{1}{4})^2}-\frac{9}{8}\right)+2\,\pi\,C,$$
where $A$ is the Glaisher-Kinkelin constant and $C$ is the Catalan constant.
A more general result:
...
0
If $x=2\arctan t$ then $\cos x = \cos\left(2\arctan t\right)$. Use the fact that $\cos(2u)=\cos^2u-\sin^2u$. So you get
$$
\cos x = \cos^2\left(\arctan t \right) -\sin^2\left(\arctan t \right)
$$
If $\varphi=\arctan t$ then $\dfrac t1=t = \tan\varphi=\dfrac{\text{opposite}}{\text{adjacent}}$, so $\cos\varphi=\dfrac{\text{adjacent}}{\text{hypotenuse}}$. ...
0
Hint:
When $\tan \dfrac{x}{2}=t$ you have $ \cos x= \dfrac{1-t^2}{1+t^2}$
3
If $\;t = \tan\left(\frac 12 x\right)$, xo $\,x = 2\tan^{-1}t,\,$ what should $\dfrac 1{2 - \cos x}$ then be?
We need to replace the function (integrand) of $x$ to one expressed as a function of $t$.
What are the new limits for $\,t\,$ if $\;t = \tan\left(\frac 12 x\right)$?
When $x = 0,\;$ $t = \tan\left(\frac 02\right) = 0$. Okay. But, when $x = \pi/2$, ...
0
You are in the good way.
The integral can be expressed as
$$
I(\varphi)=\int_{0}^{\pi}f(\theta,\varphi)d\theta
$$
where
$$ f(\theta,\varphi)=\left[1+\left(\dfrac{2\sin\left(\dfrac{\theta+\varphi}{2}\right)\sin\left(\dfrac{\theta-\varphi}{2}\right)}{\sin\theta}\right)^{2}\right]^{-\frac{n}{2}}
$$
where $x=\cos\varphi$.
Now, ...
2
The Gegenbauer polynomials were built to solve problems like this.
Suppose $m\in\mathbb{N}$.
We have
$$\begin{eqnarray*}
\int_0^{\pi} d\theta \,
\frac{\sin^{n+2m} \theta}{(1-2x \cos \theta+x^2)^{n/2}}
&=& \int_0^\pi d\theta \, \sin^{n+2m} \theta
\sum_{k=0}^\infty C_k^{(n/2)}(\cos\theta) x^k \\
&=& \int_{-1}^1 d u \, ...
1
I have got the following answers with help of Wolfram Mathematica 8.0:
$\int_{-\infty}^{\infty} \left(\frac{sin(\pi a x)}{\pi a x}\right) e^{-b x^2} dx = \frac1{a^2 \cdot \pi^\frac{3}{2}} \cdot \left( -\sqrt{b} + \sqrt{b} \cdot e^{-\frac{a^2 \cdot \pi^2}{b}} + a \cdot \pi^\frac{3}{2} \cdot \operatorname{Erf}\left( \frac{\pi a}{\sqrt{b }} \right) \right)$ ...
1
The first thing I would try is to replace $\int_{0}^{\infty}$ by $\frac12\int_{-\infty}^{\infty}$, then to consider this as a complex integral and pull the integration contour to $i\infty$. Since the only singularity of the hypergeometric function $_2F_1(\ldots,z)$ on the main sheet is the branch point $z=1$, in the end we would have to integrate the jump of ...
3
First one:
The only non-zero moments correspond to even $k=2m$. In this case we have
$$I_k=\int_{-\infty}^{\infty}x^k\left(\frac{\sin(\pi a x)}{\pi ax}\right)^2\quad \exp(-bx^2)\,dx=\frac{(-1)^{m-1}}{\pi^2a^2}\frac{\partial^{m-1}}{\partial b^{m-1}}\int_{-\infty}^{\infty}\sin^2\pi a x\, e^{-bx^2}dx$$
But the last integral can be written as
\begin{align}
...
0
By using the following (which can be computed using standard integrals, integration by parts and this limit)
$$\int^\infty_{a+h}\frac{1}{(x-a)^{\alpha+1}}dx=\left.-\frac{1}{\alpha(x-a)^\alpha}\right|^\infty_{a+h}=\frac{1}{\alpha h^\alpha},$$
...
1
For what it's worth, WolframAlpha will show you the steps to evaluate the indefinite integral associated with this definite integral. It's then just a matter of plugging in the limits and subtracting.
Here are the steps returned by WolframAlpha via the Mathematica interface:
2
Whenever you have $p(x)f(x)$, where $p(x)$ is a polynomial, and $f(x)$ is a function like $\sin(x)$ or $e^x$, such that differentiating doesn't increase its "order," integration by parts is always a good technique. Differentiate the polynomial so it becomes "less complicated," and integrate the other function.
$$\int_{-L}^Lx\sin\left(\frac{n\pi ...
2
If $n=0,$ this is simple, so suppose not. Use the substitution $$u=\frac{n\pi}lx,$$ so that $$x=\frac{l}{n\pi}u,$$ and so $$\int_{-l}^lx\sin\left(\frac{n\pi}lx\right)\,dx=\frac{l^2}{n^2\pi^2}\int_{-n\pi}^{n\pi}u\sin u\,du,$$ then integrate by parts.
2
So, you can start out by writing the denominator like this: $ y^4+y^2(C-2)+1=(y^2-\frac{2-C-\sqrt{C(C-4)}}{2})(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})$. This is simply the solution of a quadratic equation (ugly, but it will do). So, now, we want to simplify this expression somehow, like by writing:
$ ...
0
Ron:
I am not trying to be pretentious, and I realize this is an old post, but if I may offer a little input on the evaluation of your zeta sum. If I can assume you're still interested at this point.
I believe it was Choi who has done research on these series.
Using the Barnes G function, many series involving zeta have been evaluated.
Here is a ...
2
Integrate by parts letting $u = x, dv = \sin(ax) dx$, and so $du = dx, v = -\cos (ax)/a$ and you have
$$\int x \sin(ax) dx = \int u \cdot dv = uv - \int v \cdot du
= \frac{-x\cos(ax)}{a} + \frac{1}{a} \int \cos(ax) dx,$$
which should be much easier.
4
Let $f : \Bbb{R} \to \Bbb{C}$ be an integrable even function. Then we claim that
$$ \int_{0}^{\infty} f\left( y - \frac{1}{y} \right) \, dy = \int_{0}^{\infty} f(x) \, dx. \tag{1} $$
Indeed, let $I$ denote the LHS of $(1)$. Then by the substitution $y \mapsto y^{-1}$ we have
$$ I = \int_{0}^{\infty} f\left( \frac{1}{y} - y \right) \, \frac{dy}{y^2} = ...
5
Note that
$$\frac{y^2}{y^4-(2-C)y^2+1}=\frac{1}{\alpha_+-\alpha_-}\left(\frac{\alpha_+}{y^2+\alpha_+}-\frac{\alpha_-}{y^2+\alpha_-}\right),$$
where $-\alpha_{\pm}$ denote two roots of the equation $z^2-(2-C)z+1=0$. For the initial integral to converge, $\alpha_{\pm}$ should be either both positive, either conjugate to each other. Consider for simplicity the ...
7
$$\int_0^{\infty}\frac{x^3}{(x^4+1)(e^x-1)}\mathrm dx=-\frac{\pi}{\sqrt8}-\log\sqrt{2\pi}-\frac{1}{2}\Re\ \psi\left(\frac{\sqrt[4]{-1}}{2\pi}\right),$$
where $\Re\ \psi(z)$ denotes the real part of the digamma function, $\psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}$.
Solution: Use the approach from sos440's answer.
To calculate the infinite sum and simplify the ...
0
I have found how to handle with this problem. The key point is to use the following integral representation of $\frac{1}{\sqrt{x-y}}$:
$\frac{1}{\sqrt{x-y}} = \frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\frac{da}{\sqrt{\pi}} e^{-a(x-y)}$.
Using Mathematica, we can carry out the integral over $y$ and $a$ (in this order), obtaining a complex result in terms of ...
1
Often, this type of integral is workable using the substitution
$$
z=\tan(x/2)\quad\text{and}\quad\mathrm{d}x=\frac{\mathrm{2\,d}z}{1+z^2}\\
\sin(x)=\frac{2z}{1+z^2}\quad\text{and}\quad\cos(x)=\frac{1-z^2}{1+z^2}
$$
A couple of identities that simplify this integral are
$$
\sin^2(x)=\frac{1-\cos(2x)}{2}\quad\text{and}\quad\cos^2(x)=\frac{1+\cos(2x)}{2}
$$
...
5
Note that this is a definite integral, and hence the result should be a number and not a function.
First, note that the inner integral is quite tough to compute. It's not obvious what an anti-derivative of $\cos(y^4)$ should be. Therefore, it is sometimes convenient to change the order of integration, i.e. in this exercise integrate with respect to $x$ ...
3
Actually $$\int_0^{\pi/2}\!\dfrac{2a}{a^2 +b^2 cot^2 x}\, dx $$ is also the way. Then $u = \cot x $ and $du = - \csc^2x = -(1+u^2)$ which will lead you to partial fractions.
2
Essentially, we want to integrate
$$I = \int_0^{\pi/2} \dfrac{\sin^2(t)dt}{1+c \sin^2(t)}$$
where $c<1$.
We have
$$I = \sum_{k=0}^{\infty} (-c)^k\int_0^{\pi/2}\sin^{2k+2}(t)dt = \dfrac{\pi}8 \cdot\sum_{k=0}^{\infty}\left(\dfrac{-c}4 \right)^k \dbinom{2k+2}{k+1} = \color{red}{\dfrac{\pi}{2c} \left(\dfrac{\sqrt{1+c}-1}{\sqrt{1+c}}\right)}$$
$\dfrac1{1+r} = ...
0
$$\displaystyle \int_0^{\dfrac{\pi}{2}} \dfrac{2a \sin^2 x}{a^2\sin^2 x +b^2 cos^2 x} dx$$
$$\displaystyle 2a\int_0^{\dfrac{\pi}{2}} \dfrac{\csc^2x }{a^2\csc^2 x +b^2 \cot^2 x \csc^2 x} dx$$
$$\displaystyle 2a\int_0^{\dfrac{\pi}{2}} \dfrac{\csc^2x }{(\cot^2 x+1)(a^2+b^2\cot^2 x)} dx$$
Let $u=\cot x$
$$\displaystyle 2a\int_0^{\infty} \dfrac{du }{(u^2 ...
1
Your approach is fine, as long as you compare with $1/x^2$ when $|x|\gt1$ and then use the continuity of $\frac{x^6+6}{x^8+8}$ for $|x|\le1$. We can illustrate this with a slightly different comparison function.
First, note that
$$
\lim_{|x|\to\infty}\frac{(x^6+6)(x^2+1)}{x^8+8}=1\tag{1}
$$
Thus, $(1)$ says that there is an $m$ so that if $|x|\ge m$, then ...
15
My calculation shows that
\begin{align*}
\int_{0}^{\infty} \frac{x^3}{(x^4 + 1)(e^x - 1)} \, dx
&= \frac{\gamma}{2} - \log\sqrt{2\pi} + \frac{\pi}{4} \frac{\sin\frac{1}{\sqrt{2}}}{\cosh\frac{1}{\sqrt{2}} - \cos\frac{1}{\sqrt{2}}} \\
&\quad - \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n\left(1 + (2\pi n)^4\right)} \\
&\approx ...
0
You can use the fact that
$$(x^2+y^2)^k = \sum_{n=0}^k {k \choose n}x^{2n}y^{2(k-n)}$$
So
$$\mathrm{I}=\int_{\!-\infty}^{\!+\infty}\!\!\!\int_{\!-\infty}^{+\infty}dxdy(x^2\!\!+\!y^2)^k\! e^{-\frac{(x-x_0)^2+(y-y_0)^2}{a^2}}\!\!=\!\sum_{n=0}^k{k \choose n}\int_{-\infty}^{+\infty}\!\!\!x^{2n}e^{-\frac{(x-x_0)^2}{a^2}}dx ...
1
Suppose that the $x$-axis runs parallel to the long side, with the origin at a corner where the depth is $3$. Then the depth $z$ at the point with coordinates $(x,y)$ is given by $z=3+\frac{12x}{150}$ (here $0\le x\le 150$, $0\le y\le 50$).
If we want to find the volume using a triple integral, then we are integrating $dz \,dx\,dy$, where $z$ runs from ...
1
Let me first rewrite the statement in a bit more understandable form.
Under the restrictions you have on $S$ and $k$, you want to show that
$$\int_0^{\infty}\frac{\left[P_k^{(2,-4S-2)}(1+2x^2)\right]^2x^5dx}{(1+x^2)^{2+4S}}=\frac{(k+1)(k+2)}{2(4S-2k-1)(4S-k)(4S-k+1)}.\tag{1}$$
To compute the integral on the left, rewrite it as
\begin{align}
...
1
Integrate over a closed contour that is in the shape of a wedge in the 1st quadrant, having an angle of $\pi/5$. That is, consider
$$\oint_C dz \frac{z^2}{z^{10}+1}$$
where $C$ is that wedge contour. As mentioned above, the contour splits into 3 pieces:
$$\oint_C dz \frac{z^2}{z^{10}+1} = \int_0^R dx \frac{x^2}{x^{10}+1} + i R \int_0^{\pi/5}d\phi \, ...
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