# Tag Info

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Too long for a comment (a work in progress) $$\int \frac{\mathrm{e}^{\pm iax}}{\sqrt{bx^2+cx+d}}dx$$ Focusing on the denominator $$b\left(x^2+\frac{c}{b}x+\frac{d}{b}\right) =b\left[\left(x+\frac{c}{2b}\right)^2+\frac{d}{b}-\left(\frac{c}{2b}\right)^2\right]$$ lets change variables $$t = ... 1 I re-posted this on MO and got the desired answer. The answer to the question is contained in the same paper of G. N. Watson which is referred to in the question. The integral in the question comes up in the transformation formulas for the one of the several mock theta functions defined by Ramanujan. The series in equation (1) of the question is the value ... 1 I had a previous answer in which I think I made the same mistake as yours for the calculation of S_j, j \ge 0. What we want is the area of the region, so we can't say that:$$S_j = \int_{j \pi}^{(j+1)\pi} \sin(x) e^{-x} dx$$Because this assume that the curve of y = \sin(x) e^{-x} is above the x-axis, which is not true. To correct that:$$S_j = ...

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Some heuristics suggest $\displaystyle\int\limits_0^\infty\operatorname{Li}_2\left(e^{-\pi x}\right)\arctan(x)\operatorname{d}x=\frac{\pi^2}{18}-\frac{3\zeta\left(3\right)}{8}.$

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Extending a comment made earlier: As the integral was first defined, with $\phi(x) = e^{|\sin x|} \, \cos x$, an identity was presented as, in corrected form, \begin{align} \int_{0}^{\pi} \frac{\phi(x) \, dx}{1 + e^{\tan x}} = - \frac{1}{2} \, \int_{0}^{\pi} \tanh\left(\frac{\tan x}{2} \right) \, \phi(x) \, dx. \end{align} Wolfram Alpha provides a ...

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Approach 1: For the first integral \begin{align} 2\int^1_{0}\frac{{\rm d}x}{\pi^2+\ln^2\left(\frac{1-x}{1+x}\right)} &=-\frac{4}{\pi}\mathrm{Im}\int^1_0\frac{{\rm d}x}{(\ln{x}+\pi i)(1+x)^2}\tag1\\ &=-\frac{4}{\pi}\mathrm{Im}\int^1_{-1}\frac{{\rm d}x}{\ln{x}(1-x)^2}\tag2\\ ...

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Using @Jack D'Aurizio's idea, there is a simplification of his closed-form. $$\sum^\infty_{n=1}\frac{(-1)^nH_n}{(2n+1)^2} = 2\,\Im\left[\operatorname{Li}_3\left(\frac{1+i}2\right)\right]-\frac{\pi^3}{64}-\frac{\pi}{16}\ln^2 2 - G \ln 2,$$ where $G$ is Catalan's contant, and $\operatorname{Li_3}$ is the trilogarithm function.

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Although not as creative and simple as nospoon's answer, here's another approach for the first one. \begin{align} \int_{0}^{\infty} \frac{1}{\cosh^{s}(x)} \, dx &= 2^{s} \int_{0}^{\infty} \frac{1}{(e^{x}+e^{-x})^{s}} \\ &= 2^{s} \int_{1}^{\infty} \frac{1}{(u+u^{-1})^{s}} \, \frac{du}{u} \\ &= 2^{s} \int_{1}^{\infty} ... 1 Hint: Use the formular:\frac{3x^2-ax}{(x-2a)(x^2+a^2)}=\frac{A}{(x-2a)}+\frac{Bx+C}{(x^2+a^2)}$$to resolve \frac{3x^2-ax}{(x-2a)(x^2+a^2)} into partial fraction by finding the values of A, B and C. Hope you can do that? Then$$\int_0^a \frac{3x^2-ax}{(x-2a)(x^2+a^2)}dx = \int_0^a \frac{A}{(x-2a)}dx +\int_0^a \frac{Bx+C}{(x^2+a^2)}dx.$$2 Use partial fraction$$\frac{3x^2-ax}{(x-2a)(x^2+a^2)}=\frac{A}{(x-2a)}+\frac{Bx+C}{(x^2+a^2)}$$Now find A,B,C$$ A=2\\B=1\\C=a\\\frac{3x^2-ax}{(x-2a)(x^2+a^2)}=\frac{2}{(x-2a)}+\frac{x+a}{(x^2+a^2)}=\\\frac{3x^2-ax}{(x-2a)(x^2+a^2)}=\frac{2}{(x-2a)}+\frac{x}{(x^2+a^2)}+\frac{a}{(x^2+a^2)}$$So$$\int_0^a \frac{3x^2-ax}{(x-2a)(x^2+a^2)}dx=\\ \int_0^a ...

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I think Math-fun's second approach based on changing the order of integration is a good strategy. Appropriate use of substitutions and trig identities along the way clean up a lot of the resulting "mess": \begin{align} \mathcal{I} ... 5 I will follow @user15302's idea. In this answer, I showed that \int_{0}^{\infty} \frac{dx}{(a + \cosh x)^{s}} \, dx = \frac{1}{(a+1)^{s}} \int_{0}^{1} \frac{v^{s-1}}{\sqrt{(1-v)(1-bv)}} \, dv, $$where b = \frac{a-1}{a+1}. Now let I denote the Vladimir's integral and set s = \frac{1}{4} and a = 7. Then we have b = \frac{3}{4} and$$ I = ...

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Extending @nospoon's idea, we notice that $$a + \cosh 2x = (a+1)\cosh^2 x (1 - b \tanh^2 x), \qquad b =\frac{a-1}{a+1}.$$ If $a > -1$, then $b < 1$ and using the substitution $u = \tanh^2 x$ we get $$\int_{0}^{\infty} \frac{dx}{(a + \cosh x)^{s}} \, dx = \frac{1}{(a+1)^{s}} \int_{0}^{1} \frac{(1 - u)^{s-1}}{(1 - b u)^s\sqrt{u}} \, du.$$ Making ...

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For the first one, \begin{align} \int_0^{\infty} (\operatorname{sech}x)^{2s}dx \\ &= \int_0^{\infty} (\operatorname{sech}^2x)^{s-1}\operatorname{sech}^2x\, dx \\ &= \int_0^{\infty} (1-\tanh^2x)^{s-1}\,\mathrm{d}(\tanh x)\\ &= \int_0^1 (1-x^2)^{s-1} \mathrm{d}x\\ &= \frac12 \int_0^1 (1-x)^{s-1} x^{-\frac12} \mathrm{d}x\\ &= \frac12 ... 0 This is a partial answer to the second question. Mathematica could evaluate\int_0^\infty\frac{dx}{\sqrt[4]{a+\cosh x}},$$in term of the following Appell function:$$ F_1\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{5}{4},\sqrt{a^2-1}-a,\frac{1}{\sqrt{a^2-1}-a}\right). $$For a=0 and a=1 there is a closed-form of this Appell function, so we get ... 2 Note that for 1\ge x>1/2,$$xe^{\frac1n \left(\left(\frac{1-x}{x}\right)^n-\frac12\left(\frac{1-x}{x}\right)^{2n}\right)}<\left(x^n+(1-x)^n\right)^{1/n}=x\left(1+\left(\frac{1-x}{x}\right)^n\right)^{1/n}<xe^{\frac1n \left(\frac{1-x}{x}\right)^n}$$where we used the estimates x-\frac12 x^2<\log (1+x)<x. Thus, by the Squeeze Theorem ... 0 I=\lim_{n\to\infty}\int_0^1(x^n+(1-x)^n)^{1/n}\ dx=\lim_{n\to\infty}\int_0^1x(1+((1-x)/x)^n)^{1/n}dx. Now \lim_{n\to\infty}(1+(1-x)/x)ⁿ)^{1/n}=1. This can be verified by taking log. Assuming uniform convergence we get the result as \int_0^1xdx=1/2. 10 Hint: It looks like there may be a mistake in your proposed simplification step, but at any rate, you can break the limit up into \lim_{n \to \infty} \int_0^{1/2} f_n dx + \int_{1/2}^1 f_n dx where f_n = (x^n + (1 - x)^n)^{1/n}. You should be able to show that in each piece of the broken up integral, either x^n is hugely dominating or (1-x)^n is ... 0 As f is a bijective decreasing function.Therefore,using basic definition of decreasing functions i.e.ifx_1<x_2\Rightarrow f(x_1)>f(x_2).Therefore,f(2)=5,f(4)=3 \int\limits_{2}^{4}f(t)dt-\int\limits_{3}^{5}f^{-1}(t)dt=\int\limits_{2}^{4}f(t)dt+\int\limits_{5}^{3}f^{-1}(t)dt=4\times 3-2\times 5=2 ... 7 Let f(x)=\sqrt{1+x^3}. \\ Easily show that f^{-1}(x+1)=\sqrt[3]{x^2+2x}. You are asked to find$$\int_0^2 f(x)dx +\int_0^2 f^{-1}(x+1)dx \\ =\int_0^2f(x)dx+\int_1^3 f^{-1}(x)dx \\ =\int_0^2f(x)dx+\int_{f(0)}^{f(2)}f^{-1}(x)dx$$. Draw a picture. 10 Hint. The function  x \mapsto f(x):=\sqrt{1+x^3} is strictly increasing on [0,2], then you may use the following property:$$ \int_a^bf(x)dx+\int_{f(a)}^{f(b)}f^{-1}(x)dx=b{f(b)}-a{f(a)} \tag1 $$(here  x \mapsto f^{-1}(x+1)=\sqrt[3]{(x+1)^2-1}=\sqrt[3]{x^2+2x},\quad f^{-1}(0+1)=0,\,f^{-1}(2+1)=2) obtaining ... 2 The first graph represents the domain you have before applying Fubini's theorem, while the second is the same domain but viewed from a different perspective. 1 Your mistake is in this:$$\int_0^1 \int_{2y}^2 xy^2 \ dx \ dy$$Limits of integration in yours is 0\ \text{to} \ 2y, it should be 2y \ \text{to} \ 2 as given above. 1 The error is in the bounds on your second integral. If you sketch the region of integration described in your first integral, it is the right triangle with hypotenuse y = \frac{x}{2}; one leg the interval [0,2] on the x-axis, and one leg the vertical line segment connecting (2,0) and (2,1). When you change the bounds by switching dy and dx, ... 4 Using the substitution \displaystyle \theta=\frac{1}{2}\arctan\left(\frac{x}{2}\right) so \displaystyle \text{d}\theta=\frac{1}{x^2+4} \text{d}x and \displaystyle x=2\tan(2\theta) we get$$\int_{0}^{\infty}\frac{\ln(x)}{x^2+4}\text{d}x=\int_{0}^{\frac{\pi}{4}}\ln\left(2\tan(2\theta)\right)\text{d}\theta= \\ ...

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Sub $x=\tanh{u}$, $dx = \operatorname{sech^2}{u} \, du$. Then the integral is $$\int_{-\infty}^{\infty} du \, \frac{\operatorname{sech^2}{u}}{\pi^2+4 u^2}$$ Now, use Parseval. The Fourier transforms of the pieces of the integrand are $$\int_{-\infty}^{\infty} du \, \frac{e^{i u k}}{\pi^2+4 u^2} = \frac14 \frac{\pi}{\pi/2} e^{-\pi |k|/2}$$ ...

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We will go through a sequence of integrals, and, remarkably, we will see that at each step an integrand will have a continuous closed-form antiderivative in terms of elementary functions, dilogarithms and trilogarithms, so evaluation of an integral is then just a matter of calculating values (or limits) at end-points and taking a difference. I used ...

3

$I = \int_{-\infty}^\infty \text dz\ \frac{1}{z-i\varepsilon}e^{- a z^2+i b z} = \int_{-\infty}^\infty \text dz\ \frac{1}{z-i\varepsilon}e^{- a z^2+i b (z - i\varepsilon +i\varepsilon)} = e^{-b\varepsilon}\int_{-\infty}^\infty \text dz\ \frac{1}{z-i\varepsilon}e^{- a z^2+i b (z - i\varepsilon)}$, that is $$e^{b\varepsilon}I = \int_{-\infty}^\infty \text dz\ ... 1 No, there is no point trying to render this into Cartesian form: it is better to stick to parametric form Hint...if you can find the right t values for the limits, you only need$$\int y\frac{dx}{dt}dt$$0 div\vec F=\bigg(\frac{\partial(4x)}{\partial x}+\frac{\partial (-y^2)}{\partial y}+\frac{\partial (z^2)}{\partial z}\bigg)=\bigg(4-2y+2z\bigg)$$\iint_{S}\vec F\cdot \hat n \;ds=\iint_{V}div\vec F \;dv=\iiint_{V}\bigg(4-2y+2z\bigg) \;dxdydzx=\rho \cos \varphi,\; y=\rho \sin \varphi, \; z=z0\leq z \leq3, \; 0\leq \rho \leq 2, \; ...

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You must use the Gauss Theorem (divergent theorem), since it deals with flux: $$\nabla F = 4-4y+2z$$ $$\int_{V} \nabla F dV = \int\int \vec{F}\vec{n}dS$$ Your volume is define by the cylinder $x^2 + y^2 = 4$ and $0 \leq z \leq 3$.

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If $f(a)=c$ and $f(b)=d$, then \begin{align} \int_a^b f(x) \,\,dx+\int_c^d f^{-1}(y) \,\,dy &=\int_a^b f(x) \,\,dx+\int_a^b f^{-1}(f(x)) f'(x) \,\,dx\\\\ &=\int_a^b f(x) \,\,dx+\int_a^b x f'(x) \,\,dx\\\\ &=\int_a^b \left(f(x)+x f'(x)\right) \,\,dx\\\\ &=\int_a^b (xf(x))' \,\,dx\\\\ &=bf(b)-af(a)\\\\ &=bd-ac \end{align} Now, let ...

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The gaussian error function is defined through: $$\frac{2}{\sqrt{\pi}}\int_{0}^{t}e^{-x^2}\,dx = \text{Erf}(t).$$ In a similar way, the imaginary error function is defined through: $$\frac{2}{\sqrt{\pi}}\int_{0}^{t}e^{x^2}\,dx = \text{Erfi}(t).$$ It follows that: \int_{0}^{t} x^2\,e^{x^2}\,dx = \int_{0}^{t}\frac{x}{2}\left(2x\, e^{x^2}\right)\,dx = ... 0 There is no analytical closed form for this integral. It must be evaluated by some numerical procedure, like Gauss-Legendre, for example. 6 As an alternative, we may apply the substitution x\mapsto\sinh^2{x}. \begin{align} \int^1_0\frac{\ln{x}}{\sqrt{x(x+1)}}{\rm d}x &=4\int^{-\ln(\sqrt{2}-1)}_0\ln(\sinh{x})\ {\rm d}x\\ &=4\int^{-\ln(\sqrt{2}-1)}_0\left(\color{green}{\ln\left(1-e^{-2x}\right)}-\ln{2}+x\right)\ {\rm d}x\\ ... 0 while converting in polar co-ordinate you have directly taken dxdx=drd(theta), you have to use polar transformation Jacobin to do so, that is dxdx=rdr*d(theta), then integrate normally answer will be (ip/2). 0 Switching to polar coordinates, the Jacobian is given by  |J| where J = \dfrac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \dfrac{\partial x}{\partial r} & \dfrac{\partial y}{\partial r} \\ \dfrac{\partial x}{\partial \theta} & \dfrac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos\theta & \sin\theta \\ ...

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Multiply Jacobian in integrand $$\int_{\theta=\pi/4}^{\theta=3\pi/4}\bigg[\int_{r=0}^{r=2}\bigg(r^3\bigg)dr\bigg]d\theta$$

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\begin{align} I&=\int_{0}^{\infty}e^{-(x^2+\frac{1}{x^2})}dx\\ &=\int_{0}^{\infty}e^{-(x-\frac{1}{x})^2-2}dx\\ &=\frac{1}{e^2}\int_{0}^{\infty}e^{-(x-\frac{1}{x})^2}dx\\ &=\frac{1}{e^2}\int_{0}^{\infty}e^{-x^2}dx\\ &=\frac{\sqrt{\pi}}{2e^2} \end{align} where I have used this going from 3 to 4. Using the results from @Dr.V we have that ...

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A. Erdélyi et al., Higher Transcendental Functions Vol. I, section 3.6.2 formula (32) list the following Generalization of Neumann's formula (attributed to Gormley, 1934) $$Q_{\nu}^{\mu}(z) = \frac{1}{2}e^{i\mu\pi}(z^2-1)^{\mu/2}\int_{-1}^{1}(1-v^2)^{-\mu/2}\frac{P_{\nu}^{\mu}(v)}{z-v} dv$$ with the restrictions \$\nu+\mu = 0,1,\dots,\;\Re \nu > -1,\; z ...

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