# Tag Info

8

Let $I(a)$ be the integral $$I(a)\equiv \int_0^{\infty}\log\left(1+\frac{a^2}{x^2}\right)dx$$ Taking a derivative with respect to $a$ reveals that \begin{align} I'(a)&=\frac{d}{da}\int_0^{\infty}\log\left(1+\frac{a^2}{x^2}\right)dx\\\\ &=2a\int_0^{\infty}\frac{dx}{x^2+a^2}\\\\ &=\pi \end{align} Integrating shows that $I(a)=\pi a +C$. ...

8

First notice that \begin{align} &\frac{1}{2} \int_{0}^{\infty} \log (4 \sin^{2} x) \Big(1 - x \, \text{arccot}(x) \Big) \, dx \\ &= \frac{\log(4)}{2} \int_{0}^{\infty} \Big( 1- x \, \text{arccot}(x) \Big) \, dx + \frac{1}{2} \int_{0}^{\infty} \log (\sin^{2} x) \Big(1 - x \, \text{arccot}(x) \Big) \, dx \\ &= \frac{\pi \log(2)}{4} + ... 7 The last series Special Addendum in the OP looks quite interesting! (just an observation, not an answer) We have by symmetry of the series:S = \sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{(-1)^k }{k2^k n 2^n (k+n)} = \sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{(-1)^n}{k2^k n 2^n (k+n)}$$Hence,$$\begin{align}\sum _{k=1}^{\infty } \sum ...

6

Using integration by parts: $$\int_0^\infty \ln\left(1+\frac{a^2}{x^2}\right)\,dx\,\,\begin{bmatrix}u=\ln\left(1+\frac{a^2}{x^2}\right)& du=\frac{\frac{-2a^2}{x}}{x^2+a^2}\,dx \\ dv=dx& v=x\end{bmatrix}\\=\underbrace{\left.x\cdot \ln\left(1+\frac{a^2}{x^2}\right)\right|_0^{\infty}}_{L}+\underbrace{\int_0^{\infty}\frac{2a^2}{x^2+a^2}\,dx}_{I}$$ ...

6

If one assumes that $f$ is sufficiently well-behaved (and in particular, differentiable), one can solve this by differentiating under the integral sign (this is sometimes called Feynman's trick). Define $$I(a) := \int_0^{\infty} \frac{f(x) - f(a x)}{x} \,dx.$$ Differentiating with respect to $a$ (and justifying passing the derivative sign through the ...

5

There are possible singularities at the origin and at infinity; by definition the integral is the limit of $\int_\delta^A$ as $\delta\to0$ and $A\to\infty$. You have $$\begin{eqnarray*} \int_\delta^A\frac{f(t)}{t}-\int_\delta^A\frac{f(2t)}{t}&=\int_\delta^A\frac{f(t)}{t}-\int_{2\delta}^{2A}\frac{f(t)}t ... 5$$\int_{0}^{+\infty}\log\left(1+\frac{a^2}{x^2}\right)\,dx = a\int_{0}^{+\infty}\log\left(1+\frac{1}{x^2}\right)\,dx $$then by setting x=\tan\theta we have:$$ \int_{0}^{+\infty}\log\left(1+\frac{1}{x^2}\right)\,dx = -2\int_{0}^{\pi/2}\frac{\log\sin\theta}{\cos^2\theta}\,d\theta=2\int_{0}^{\pi/2}\cot(\theta)\tan(\theta)\,d\theta=\color{red}{\pi},$$where ... 4 The sum$$\frac{1}{m+1} \sum_{j=0}^{n-1} \gamma^{m+1}(t_{j+1}) - \gamma^{m+1}(t_{j})$$telescopes since$$(\gamma^{m+1}(t_{1}) - \gamma^{m+1}(t_{0})) + (\gamma^{m+1}(t_{2}) - \gamma^{m+1}(1)) + \cdots + (\gamma^{m+1}(t_{n}) - \gamma^{m+1}(t_{n-1}))=-\gamma^{m+1}(t_0) + (\gamma^{m+1}(t_1) - \gamma^{m+1}(t_1)) + \cdots + (\gamma^{m+1}(t_{n-1}) - ...

4

We start with: $\displaystyle \int_0^1 \frac{t^2}{x^2+t^2}\,dt = 1-x\tan^{-1}\frac{1}{x}$ Then, \displaystyle \begin{align} \int_0^{\infty} \log (2\sin x)\left(1-x\tan^{-1}\frac{1}{x}\right)\,dx &= \int_0^{\infty}\int_0^1 \frac{t^2\log (2\sin x)}{t^2+x^2}\,dt\,dx\\&= -\sum\limits_{n=1}^{\infty} \int_0^1 t^2\int_0^{\infty} \frac{1}{n}\frac{\cos ... 3 Note that\frac{1}{\sqrt x}=x^{-\frac 12}$and that $$\int x^\alpha dx=\frac{1}{\alpha+1}x^{\alpha+1}+C$$ for$\alpha\not=-1$. 3 One may show that this integral is equivalent to the integral in this problem as follows: First, rewrite $$\frac12 \log{\left ( \frac{1+x}{1-x} \right )} = \tanh^{-1}{x}$$ Then note that $$\tan^{-1}{\left ( \frac1{x} \right )} = \frac{\pi}{2} - \tan^{-1}{x}$$ The integrand is then equal to $$\left [\frac{2}{\pi} \tan^{-1}{\left (\frac{2}{\pi} \left ... 3 Using the double angle identities to express \cos 2x in terms of \cos x and \sin x, we get:$$\frac{1-\cos 2x}{1+\cos 2x} = \frac{1- (1 - 2 \sin^2 x)}{1 + (2 \cos^2 x - 1)} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x$$3 &=&4\int_{0}^{\pi/4}\frac{dx}{\sin(2x)+\frac{\sqrt{3}}{2}}\\&=&4\int_{0}^{\pi/4}\frac{dy}{\cos(2y)+\frac{\sqrt{3}}{2}}\\&=&4\int_{0}^{1}\frac{dt}{2+\left(\frac{\sqrt{3}}{2}-1\right)(1+t^2)}\\&=&8\int_{0}^{1}\frac{dt}{\left(2+\sqrt{3}\right)-\left(2-\sqrt{3}\right)t^2}\\&=&8\,\text{arctanh}(2-\sqrt{3})=\color{red}{2\log ... 3$$\int_{0}^{\frac{\pi}{2}}\frac{1}{\cos (x-\frac{\pi}{3})\cdot\cos (x-\frac{\pi}{6})}\mathrm{d}x=\int_0^{\pi/2}\frac{4}{\sqrt{3}+2\sin(2x)}dx$$and the change of variables...$$=\lim_{M\rightarrow\infty}\int_{\sqrt 3}^{\sqrt{3}+M}\frac{4\sqrt 3}{u(2\sqrt{3}+u)}du=\cdots=\lim_{M\rightarrow\infty}\left[2\ln(3)+2\ln\left(\frac{M+\sqrt ... 3 I couldn't resist as if your limits were same for each variable say varying$a$to$b$, for any$a$and$b, the following trick I saw somewhere, is applicable, but in your case, there is no short cut, I guess. I=\int_{a}^b\int_{a}^b \int_{a}^b \int_{a}^b \frac{x_1-x_2+x_3{-}x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4 = \int_{a}^b\int_{a}^b \int_{a}^b ... 2 Hint: Use the fact that \dfrac{1-\cos2x}2=\sin^2x, in conjunction with the integral formulas for Bessel functions. 2 For \int e^{\sin t}~dt , \int e^{\sin t}~dt =\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n}t}{(2n)!}dt+\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}t}{(2n+1)!}dt =\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\sin^{2n}t}{(2n)!}\right)dt+\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}t}{(2n+1)!}dt For n is any natural number, ... 2 The book is wrong. Plain and simple. They transposed the last two digits, which is a simple typo. Originally, I was going to write the following, and I include it here for numerical reference. You need to do all the rounding right at the end. Since we have calculators today, which usually carry 12-16 digits of precision, round-off error is the least of our ... 2 Let the integral in question be \begin{align}\tag{1} I_{a} = \int_{0}^{\infty} e^{-st} \, \frac{\ln(1+a t)}{1 + t} \, dt. \end{align} For the case of a=1 the following is obtained. By utilizing \begin{align} \int_{0}^{\infty} e^{-s t} \, \ln^{2} t \, dt = \frac{1}{s} \, ( \zeta(2) + (\gamma + \ln s)^{2} ) \end{align} for the change t \to t+1 leads to ... 2 Integrating by parts and using the known series results, we get that\int_0^1 \frac{\text{Li}_2 \left(-\frac{1}{1-z}\right)-\text{Li}_2 \left(-\frac{1}{1+z}\right)}{z}dz=\int_0^1 \frac{\log (z+2) \log (z)}{z+1}dz+\underbrace{\int_0^1\frac{\log (2-z) \log (z)}{1-z} dz}_{\large \sum _{k=1}^{\infty } \frac{(-1)^k H_k}{k^2}=-5/8 \zeta ... 2 Your solution is OK. My (clumsier) solution, just to reinforce yours, is as follows We can approach the integral $$\int_a^bf'(x)f''(x)\ dx$$ via integration by part. The mnemonic is that\int u'v=uv-\int uv'. $Let$u'=f''$and let$v=f'$. Then$u=f'$and$v'=f''\$. So $$\int_a^b f'(x)f''(x) dx=\left[(f'(x))^2 \right]_a^b-\int_a^bf'(x)f''(x)\ dx.$$ ...

2

This is not an answer but it is too long for a comment. As I wrote in comment, there is something wrong somewhere since $$\int\frac{\sqrt{\tan x}}{\sin x}dx=-2 \sqrt{\cos (x)} \, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};\cos ^2(x)\right)$$ and $$I=\int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx=\frac{\Gamma \left(\frac{1}{4}\right) \Gamma ... 2 The answer is a partial YES and a partial NO. If you split the Riemann improper integral into two pieces:$$\int_0^\infty \frac{f(x)}{x} dx \stackrel{def}{=} \lim_{\Lambda \to \infty, \lambda \to 0} \int_\lambda^\Lambda \frac{f(x)}{x} dx = \lim_{\Lambda\to\infty} \int_1^\Lambda \frac{f(x)}{x} dx + \lim_{\lambda\to 0} \int_\lambda^1 \frac{f(x)}{x}dx $$The ... 2 Other example. Take f(x)=e^{-x}. 2 No. Consider f=1_{[0,1]}, that is, f(x)=1 when x\in[0,1], and 0 otherwise. 2 Write \dfrac{1}{(x^2-x+1)^2} as a Taylor series \sum_{j=0}^\infty a_j x^j. The coefficients can be written as$$ a_j = \dfrac{j+1}{3} b_j + \dfrac{2}{3} c_j $$where b_j repeats 1,2,1,-1,-2,-1,\ldots for j = 0,1,2,\ldots while c_j repeats 1,1,0,-1,-1,0,\ldots, both with period 6. Now$$\int_0^1 \log^2(x)\; x^k\; dx = \dfrac{2}{(k+1)^3}$$so ... 2 f(t,t)+\int\limits_{0}^{t} f_{t}(x,t)dx=f(t,t)+\int\limits_{0}^{t} 1 dx =2t+t=3t, which agrees! 2 For the integral from 0 to 1 to exist, we need, as was pointed out in the OP, \alpha\gt -1. We now deal with the integral from 1 to \infty. Rewrite the integrand as \frac{x^\alpha}{e^x}e^x\cos(e^x), and integrate by parts, letting u=\frac{x^\alpha}{e^x} and dv=e^x\cos(e^x)\,dx. Then du=e^{-x}\left(\alpha x^{\alpha-1}-x^\alpha\right)\,dx, ... 2 We must have \alpha>-1, otherwise we have a non-integrable singularity in a right neighbourhood of the origin. Given that, we have:$$ \int_{0}^{M}x^\alpha \cos(e^x)\,dx = \int_{1}^{e^M}\frac{\log^\alpha(t)\cos t}{t}\,dt $$that is not absolutely converging, but is conditionally converging by Dirichlet's test, integral version, since \cos t has a ... 2 Here is my evaluation of the quadratic case. The starting point for the derivation is the algebraic identity$$6ab^2=\left(a+b\right)^3+\left(a-b\right)^3-2a^3.$$Then,$$\begin{align} \mathcal{I}_{2} &=\int_{0}^{1}\frac{\ln{\left(\frac{1}{t}\right)}\ln^{2}{\left(t+2\right)}}{t+1}\,\mathrm{d}t\\ ...

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