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19

First note that by substituting $x\mapsto kx$, we get $$ \int_0^\infty\log\left(\frac{x^2+2kx\cos(a)+k^2}{x^2+2kx\cos(b)+k^2}\right)\frac{\mathrm{d}x}{x} =\int_0^\infty\log\left(\frac{x^2+2x\cos(a)+1}{x^2+2x\cos(b)+1}\right)\frac{\mathrm{d}x}{x} $$ Let $u=\frac{x+\cos(a)}{\sin(a)}$. Then $$ \begin{align} ...


16

Consider following parametric integral $$I(\alpha)=\int_{0}^{\infty}\sin{x}\arctan\left({\dfrac{\alpha}{x}}\right)\,\mathrm dx$$ We have $I(0)=0$ and $I(1)$ yields required Integral. Differentiating wrt $\alpha$, we get $$I'(\alpha)=\int_{0}^{\infty}\frac{{x}\sin{x}}{x^2+{\alpha^2}}\,\mathrm dx=\frac{\pi}{2}e^{-\alpha}$$ $I'(\alpha)$Integrating wrt ...


10

Here is another way to evaluate the integral. Notice that $$\int_0^1 \frac{x}{x^2+y^2}\mathrm dy=\arctan\left(\frac{1}{x}\right)$$ We also have $$\int_{0}^{\infty}\frac{{x}\sin{x}}{x^2+{y^2}}\,\mathrm dx=\frac{\pi}{2}e^{-y}$$ Hence \begin{align} \int_{0}^{\infty}\sin{x}\arctan\left({\frac{1}{x}}\right)\,\mathrm dx&=\int_{0}^{\infty}\sin{x}\int_0^1 ...


10

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10

\begin{align} \int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}&=\int_0^\infty\frac{1}{x}\int_a^b \frac{\mathrm d}{\mathrm dy}\ln \left(x^2+2kx\cos y+k^2\right) \;\mathrm dy\,\mathrm dx\\[10pt] &=-\int_0^\infty\frac{1}{x}\int_a^b \frac{2kx\sin y}{x^2+2kx\cos y+k^2} \;\mathrm dy\,\mathrm dx\\[10pt] ...


9

A Generalisation: \begin{align} \int^\infty_0\frac{x}{(x^2+w^2)(1+e^{2\pi x})}{\rm d}x\tag1 =&\int^\infty_0\frac{xe^{-x}}{(x^2+4\pi^2w^2)(1+e^{-x})}{\rm d}x\\ \tag2 =&\int^\infty_0\frac{x}{x^2+4\pi^2w^2}\left(\sum^\infty_{n=1}(-1)^{n-1}e^{-nx}\right){\rm d}x\\ \tag3 ...


9

Here is an elementary way to evaluate the integral without involving any special functions or advance formulas. Notice that $$\int_0^\infty e^{-y}\sin(xy)\;\mathrm dy=\frac{x}{1+x^2}$$ Hence we have \begin{align} \int_0^\infty \frac{x}{\left(1+x^2\right)\left(e^{2\pi x}+1\right)}\,\mathrm dx&=\int_0^\infty\int_0^\infty \frac{e^{-y}\sin(xy)}{e^{2\pi ...


8

With an integration by parts, we have $$ \int_{0}^{\infty}\sin{x}\arctan{\dfrac{1}{x}}\,\mathrm dx=\left.-\cos{x}\arctan{\dfrac{1}{x}}\right|_0^{\infty}-\int_0^\infty\frac{\cos\;x}{1+x^2}\mathrm{d}x=\frac{\pi}{2}-\int_0^\infty\frac{\cos\;x}{1+x^2}\mathrm{d}x$$ then we use the standard integral $$\int_0^\infty\frac{\cos\;x}{1+x^2}\mathrm{d}x ...


8

$1 - \dfrac{\sin^2(2x)}{2} = \dfrac{1+\cos^2(2x)}{2}$, and $\sin x\cos x = \dfrac{\sin (2x)}{2} \Rightarrow \displaystyle \int \dfrac{\sin x\cos x}{\cos^4x+\sin^4x}dx = \displaystyle \int -\dfrac{1}{2}\dfrac{d(\cos(2x))}{1+\cos^2(2x)}dx = -\dfrac{1}{2}\arctan(\cos (2x)) + C$


6

Here is a complex-analytic method: Notice that $$ \int_{0}^{\infty} \log\left( \frac{x^{2} + 2x\cos b + 1}{x^{2} + 2x\cos a + 1} \right) \frac{dx}{x} = 2 \int_{0}^{\infty} \Re \frac{\log(1 + e^{ib}x) - \log(1 + e^{ia}x)}{x} \, dx. \tag{1} $$ Let $R$ be a positive large number. Then the function $z \mapsto \log(1+z)/z$ is analytic on $\Bbb{C} \setminus ...


6

By clever substitutions your professor probably meant Euler Substitutions. $$\begin{align} I &=\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx\tag{1}\\ &=\frac12\int_0^1 \frac{t}{2(2-t)(1+t) + 3\sqrt{(2-t)(1+t)}}\,\mathrm dt\tag{2}\\ &=\frac13\int_{\sqrt2}^{1/\sqrt2} \frac{u^2-2}{(1+u)^2(u^2+1)}\,\mathrm du\tag{3}\\ ...


6

Hint: make use of the Binet's second formula http://mathworld.wolfram.com/BinetsLogGammaFormulas.html. $$\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}+1)} dx$$ $$=\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}-1)} dx-2\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{4\pi x}-1)} dx$$ $$=\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}-1)} ...


6

Just for references, I remark that the following proposition was proved in my answer: Proposition. If $0 < r < 1$ and $r < s$, then $$ I(r, s) := \int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}} \log \left( \frac{1 + 2rsx + (r^{2} + s^{2} - 1)x^{2}}{1 - 2rsx + (r^{2} + s^{2} - 1)x^{2}} \right) \, dx = 4\pi \arcsin r. \tag{*} $$ Recently, I ...


5

Use integration by parts to get $$ \displaystyle\int_0^1 \frac {\ln x}{1 + x^2} \, \mathrm{d}x = \left[ \arctan(x) \ln(x) \right]_0^1 - \displaystyle\int_0^1 \frac {\arctan(x)}{x} \, \mathrm{d}x = - \displaystyle\int_0^1 \frac {\arctan(x)}{x} \, \mathrm{d}x. $$Now, we use the Taylor expansion of $\arctan(x)$ to get $$ -\int_0^1 ...


5

One of the first observations we can make is that you're integrating a function over an interval that is symmetric around $0$ (i.e. it can be written as $[-c,c]$). I have a trick for you, whenever you find similar definite integrals start by writing the function you're integrating as a sum of an even function and an odd function. In fact, it can be proven ...


5

Continuing from where you left: $$I_n=3n\int_0^1 v^3(1-v^3)^{n-1}\,dv=3n\left(\int_0^1 (1-v^3)^{n-1}\,dv-\int_0^1(1-v^3)^{n}\,dv\right)$$ $$\Rightarrow I_n=3n\left(I_{n-1}-I_n\right) \Rightarrow I_n=\frac{3n}{3n+1}I_{n-1}$$


5

Let $x=\sin y$, and note that $(2-x^2)(1+x^2)$ simplifies to $2+\sin^2y\cos^2y$. Then integral becomes $$I=\int_0^{\pi/2}\frac{\sin^3y\cos y}{2(2+\sin^2y\cos^2y)+3\sqrt{2+\sin^2y\cos^2y}}dy.$$ Now set $y=z-\frac{\pi}{2}$, to observe that $$I=I_0=\int_0^{\pi/2}\frac{\cos^3y\sin y}{2(2+\sin^2y\cos^2y)+3\sqrt{2+\sin^2y\cos^2y}}dy.$$ With $\cos^3y\sin y+\cos ...


5

Let us first substitute $y = x^{2} + 1$. Then \begin{align*} I &:= \int_{0}^{1} \frac{x^{3}}{2(2-x^{2})(1+x^{2}) + 3\sqrt{(2-x^{2})(1+x^{2})}} \, dx \\ &= \frac{1}{2} \int_{1}^{2} \frac{y - 1}{2y(3-y) + 3\sqrt{y(3-y)}} \, dy \tag{1} \end{align*} Using the substitution $y \mapsto 3-y$, it follows that $$ I = \frac{1}{2} \int_{1}^{2} ...


5

Another approach is to use the Fourier series $$\sum_{k=1}^{\infty}\frac{x^{k} \cos(ka)}{k} = - \frac{1}{2} \log \left(x^{2} - 2 x \cos(a) +1 \right) \ , \ |x| <1 $$ which can be derived from the Maclaurin series of $\log(1-z)$ by replacing $z$ with $xe^{ia}$ and equating the real parts on both sides. $$ \begin{align} ...


5

Here's a little help: at first, it may seem to integral is not treatable using Cauchy, since the problem lies at $1/\sin z$ which has a zero at $z=0$. Now consider $f(z)= \dfrac{z}{\sin z}$. This is now analytic and zero free in $|z|\leqslant 1/2$, and the same holds for $g(z)=f(z)(z-1)^{-1}$. Your integral now has the form ...


5

I finally figured it out due to a hint by Sameer Kailasa. $$\int_{0}^{\frac{\pi}{2}}{\frac{1}{1+\tan^\sqrt{2}(x)}} \ dx$$ $$u=\frac{\pi}{2}-x \implies du=-dx$$ $$= -\int_{\frac{\pi}{2}}^{0}{\frac{1}{1+\cot^\sqrt{2}(x)}} \ dx = \int_{0}^{\frac{\pi}{2}}{\frac{\tan^\sqrt{2}(x)}{\tan^\sqrt{2}(x)+1}} \ dx $$ ...


5

The case $a^2=b^2$ being simple, let's just consider, by symmetry, the case $a>b>0$. Observe that $$ \partial _x \left(\frac{1}{a^2 \cos^2x+b^2 \sin^2 x}\right)=2(a^2-b^2)\frac{\cos x\sin x}{(a^2 \cos^2x+b^2 \sin^2 x)^2} $$ then, integrating by parts, you may write $$ \begin{align} I(a,b)&=\int_0^{\pi/2}\frac{x\cos x\sin x}{(a^2 \cos^2x+b^2 \sin^2 ...


4

Denote the first integral by $I$ and the second by $J$. Then, $$\begin{aligned} J=&\int_0^{\pi/4} x\left(\frac{\pi}{2}-\arctan\sqrt{\frac{\cos 2x}{2\sin^2 x}}\right)\,dx \\ =&\frac{\pi^3}{64}-I \,\,\,\,\,\,\,(1) \end{aligned}$$ $I$ can be simiplified to: $$I=\int_0^{\pi/4} x\arccos(\sqrt{2}\sin x)\,dx=\left(\frac{x^2\arccos(\sqrt{2}\sin ...


4

\begin{align} \int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\cos^4x}\mathrm dx&=\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\left(1-\sin^2x\right)^2}\mathrm dx\\[7pt] &=\int_0^{\pi/2}\frac{\sin x\cos x}{2\sin^4x-2\sin^2x+1}\mathrm dx\\[7pt] &=\frac14\int_0^1\frac{\mathrm dt}{t^2-t+\frac12}\qquad\color{blue}{\implies}\qquad t=\sin^2x\\[7pt] ...


4

Here is one line proof $$\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\cos^4x}dx=\int_0^{\pi/2}\frac{\tan(x) (\tan(x))'}{\tan^4(x)+1}\ dx=\int_0^{\infty} \frac{x}{x^4+1}\ dx=\left[\frac{\arctan(x^2)}{2}\right]_0^{\infty}=\frac{\pi}{4}$$ Q.E.D.


4

$\sqrt[3]{2x^2-x^3} = x\sqrt[3]{\dfrac{2}{x}-1} \to u = \sqrt[3]{\dfrac{2}{x} - 1} \to u^3 = \dfrac{2}{x} - 1 \to x = \dfrac{2}{u^3+1}$. Can you take it from here?


4

Main Idea In short, reverse the order of integration. Then it works very simply.


4

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4

let $-I = \int_0^1{\ln x \over 1 + x^2} \ dx.$ we will need $$a_k = \int_0^\infty xe^{-kx}\ dx = {1 \over k^2} \int_0^\infty xe^{-x} \ dx = {1 \over k^2}$$ $I = \int_0^1{\ln x \over 1 + x^2} \ dx,$ by a change of variable $ u = -\ln x, x = e^{-u}, $ the integral $\begin{eqnarray} I &=& \int_0^\infty{ ue^{-u} \over 1 + e^{-2u}} \ du \\ ...


4

All you need is that $f$ is continuous with $f(1) = 1$. Note that for any $\delta \in (0,1)$, $$\int_0^{1-\delta} y x^y\; dx = \dfrac{y}{y+1} (1-\delta)^{y+1} \to 0$$ while $$ \int_0^1 y x^y\; dx = \dfrac{y}{y+1}\to 1$$ Take $\delta$ so that $|f(x) - f(1)| < \epsilon$ for $1-\delta < x \le 1$, and use the fact that $f$ is bounded...



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