# Tag Info

5

We suppose $0 < a < b$. For $\epsilon > 0$, you can find $\delta > 0$ such that for $\vert x \vert \le \delta$ you have $\vert f(x)-f(0) \vert \le \epsilon$ (by continuity of $f$ at $0$). Then for $\vert c \vert \max(a,b) \le \delta$, we have $[ca,cb] \subset [0,\delta]$. Hence: $$\left\vert \int_{ca}^{cb} \frac{f(x)}{x} dx - \int_{ca}^{cb} ... 4 Let y=\pi(x+1). Then  \sin{\pi x} = \sin{(y-\pi)} = -\sin{y} , dx/(x+1) = dy/y  and the integral is$$ -\int_{\pi}^{2\pi} \frac{\sin{\pi y}}{y} \, dy = -\int_{0}^{2\pi} \frac{\sin{\pi y}}{y} \, dy + \int_{0}^{\pi} \frac{\sin{\pi y}}{y} \, dy = \operatorname{Si}(\pi) - \operatorname{Si}(2\pi) $$4 write \sin(x)^a=e^{a\log(\sin(x))} Then$$ I=\partial_a^2 \left. \int_0^\pi dx\sin^a(x) \right|_{a=0} $$This integral can be evaluated in terms of Euler's Beta function and yields after simplification$$ I=\sqrt{\pi }\partial_a^2 \left. \left(\frac{ \Gamma \left(\frac{a+1}{2}\right)}{\Gamma \left(\frac{a}{2}+1\right)}\right) \right|_{a=0} $$... 4 If \alpha \neq \beta, and we differentiate with respect to c, we get (\alpha-\beta)f(c) = 0, from which we get f(c) = 0, hence f = 0 on [a,b]. If \alpha = \beta, all we can conclude is that \alpha \int_a^b f(x) dx = 0, and so either \alpha = 0, or f has zero average over [a,b]. 3 I guess you need to assume \alpha \ne \beta. Then if one defines$$g(x)=\alpha \int_{a}^{x}f(x)dx+\beta \int_{x}^{b}f(x) dx$$then g(c)=0 for all c \in [a,b]. Thus g is a constant and by fundamental theorem of calculus g is differentiable. So g'(x)=0 for all x. Thus (\alpha-\beta)(f(x))=0 and we are done 3 By replacing x with \arctan t,$$ I = \int_{0}^{\pi/2}\frac{dx}{4\sin^2 x+5\cos^2 x} = \int_{0}^{+\infty}\frac{dt}{4t^2+5}=\color{red}{\frac{\pi}{4\sqrt{5}}}.$$3 From the equation x^2+y^2=r^2, you may express your area as the following integral$$ A=\int_0^r\sqrt{r^2-x^2}\:dx. $$Then substitute x=r\sin \theta, \theta=\arcsin (x/r), to get$$ \begin{align} A&=\int_0^{\pi/2}\sqrt{r^2-r^2\sin^2 \theta}\:r\cos \theta \:d\theta\\ &=r^2\int_0^{\pi/2}\sqrt{1-\sin^2 \theta}\:\cos\theta \:d\theta\\ ...

3

Hint. You may use $$\sin^2 \frac{x}2=\frac{1-\cos x}2$$ and $\sin u \geq0, \, u \in [0,\pi].$

3

Hint Substitute $$u := \frac{x}{c}, \quad du = \frac{dx}{c}.$$

3

I=$\int _{ 0 }^{ \pi }{ { e }^{ { \sin ^{ 2 } x } } } { \cos((2n+1)x) } dx$ Let $u=\pi -x$, $dx=-du$, $${ { e }^{ { \sin ^{ 2 } x } } }={ { e }^{ { \sin ^{ 2 }( {\pi -u} )} } }={ { e }^{ { \sin ^{ 2 } u } } }$$, ${ \cos((2n+1)x) }$= ${ \cos((2n+1)(\pi-u) }$= ${ \cos(2n\pi + (\pi - (2n+1)u)) }$= ${ \cos(\pi - (2n+1)u)) }$= $-{ \cos(2n+1)u)) }$ ...

3

Yes you can evaluate it using spherical coordinates, but it is a long and tedious calculation. The final result is fairly simple though $$\color{red}{\int\limits_{0\leq x_i,~\sum_{i=1}^D x_i^{a_i}\leq 1}\sqrt{1-\sum_{i=1}^Dx_i^{a_i}}{\rm d}V = \frac{\sqrt{\pi}}{2}\frac{\prod_{i=1}^{D}\Gamma\left(a_i^{-1}\right)a_i^{-1}}{\Gamma\left(\frac{3}{2} + ... 2 You can neglect the \ln{} and the x^3 terms, as they are odd over a symmetric interval. Thus we have$$2 \int_0^1 dx \, e^{-x^4} (1-4 x^4) $$Integrate by parts to get$$4 \int_0^1 dx \, x^4 e^{-x^4} = - \int_0^1 d(e^{- x^4}) x = [x e^{-x^4}]_1^0 + \int_0^1 dx \, e^{-x^4}$$Thus the integral is$$2 \int_0^1 dx \, e^{-x^4} - \frac{2}{e} - 2 ...

2

Notice that $x\mapsto\ln (x+\sqrt{x^2+1})$ is an odd function since \begin{align} \ln [(-x)+\sqrt{(-x)^2+1}]&=\ln\left[\frac{\sqrt{x^2+1}-x}{1}\cdot\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}\right]\\ &=\ln\left[\frac{1}{\sqrt{x^2+1}+x}\right]\\ &=-\ln (x+\sqrt{x^2+1}) \end{align} Also $x\mapsto x^3$ is odd, then we have \begin{align} \int_{-1}^1 ...

2

Hint: You can make $x=t+\frac{\pi}{2}$, so \begin{align} \int_0^{\pi}{e^{\sin^2 x}\cos((2n+1)x) dx}&=\int_{-\pi/2}^{\pi/2}{e^{1-\cos^2 (t+\frac{\pi}{2})}\cos\left((2n+1)t+n\pi+\frac{\pi}{2}\right) dt}\tag{1} \end{align} Also, if $n\in\mathbb{Z}$, we have \begin{align} \cos\left((2n+1)t+n\pi+\frac{\pi}{2}\right)&=\cos [(2n+1)t]\cos\left(n\pi ...

2

by using the Taylor series $$\frac{\sin \pi x}{1+x}=\sum_{n=1}^{\infty }(-1)^n\frac{(\pi)^{(2n-1)}}{(2n-1)!}(x+1)^{2n-2}$$

2

This response will only address the $n=4$ case, $$I_{4}:=\int_{[0,1]^{4}}\frac{\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z\,\mathrm{d}w}{\left(1+x\right)\left(1+y\right)\left(1+z\right)\left(1+w\right)\left(1+xyzw\right)}.\tag{1}$$ According to WolframAlpha, the multiple integral $(1)$ above has the approximate numerical value $I_{4}\approx0.223076.$ Starting ...

2

Hint $$\int_{-1}^1\sqrt{1-x^2}dx=\int_{\arcsin(-1)}^{\arcsin(1)}\sqrt{1-\sin^2(x)}\cos(x)dx=...$$

2

\begin{align*} \int_{-1}^{1}\sqrt{1-x^2}\,dx &= 2\int_{0}^{1}\sqrt{1-x^2}\,dx \\&= 2\int_{0}^{\pi/2}\cos^2\theta\,d\theta\\&=\int_{0}^{\pi/2}(\cos(2\theta)+1)\,d\theta\\&=\int_{0}^{\pi/2}1\,d\theta=\color{red}{\frac{\pi}{2}}\end{align*} as expected. We exploited the parity of the function $\sqrt{1-x^2}$, the substitution $x=\sin\theta$, ...

2

If $1 + \cos 2\theta = 2 \cos^2 \theta$, then $$1 - \cos 2\theta = 2 \sin^2 \theta,$$ because by adding these two equations together, you would get $$2 = 2 \cos^2 \theta + 2 \sin^2 \theta,$$ where upon dividing by $2$ you recover the circular identity $$1 = \cos^2 \theta + \sin^2 \theta.$$ Consequently, $$\frac{1 - \cos 2\theta}{2} = \sin^2 \theta,$$ and ...

2

Notice, the formula $$\cos x=1-2\sin^2 \frac{x}{2}\ dx$$ Hence, $$\int_{0}^{2\pi}\sqrt{\frac{1-\cos x}{2}}\ dx=2\int_{0}^{\pi}\sqrt{\frac{1-1+2\sin^2 \frac{x}{2}}{2}}\ dx$$ $$=2\int_{0}^{\pi}\sin\frac{x}{2} dx$$ $$=2(2)\left[-\cos\frac{x}{2}\right]_{0}^{\pi}=\color{red}{4}$$

2

Suppose we seek to evaluate $$\int_0^{\pi/2} \frac{1}{4\sin^2(x)+5\cos^2(x)} dx = \frac{1}{4} \int_0^{2\pi} \frac{1}{4\sin^2(x)+5\cos^2(x)} dx.$$ Introduce $z=\exp(ix)$ so that $dz=iz \; dx$ to get $$\frac{1}{4}\int_{|z|=1} \frac{1}{4(z-1/z)^2/(-4)+5(z+1/z)^2/4} \frac{dz}{iz} \\ = \int_{|z|=1} \frac{1}{-4(z-1/z)^2+5(z+1/z)^2} \frac{dz}{iz} \\ = \int_{|z|=1} ... 2 Let$$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\frac{1}{4\sin^2 x+5\cos^2 x}dx$$Now Divide both \bf{N_{r}} and \bf{D_{r}} by \cos^2 x\;, We get$$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\frac{\sec^2 x}{4\tan^2 x+5}dx\;,$$Now Put \displaystyle 2\tan x=t\;, Then \displaystyle 2\sec^2 xdx =dt\Rightarrow \sec^2 xdx = \frac{1}{2}dt and Changing ... 2 Use De Moivre's formula to compute the Fourier sine/cosine series of \sin(t)^n, then exploit:$$ \int_{0}^{+\infty}e^{-at}\sin(bt)\,dt = \frac{b}{a^2+b^2},\qquad \int_{0}^{+\infty}e^{-at}\cos(bt)\,dt = \frac{a}{a^2+b^2}.$$2 I give you some hints: Let$$ I_n=\int_0^{+\infty}e^{-at}\sin^n t\,dt=\mathcal L(\sin^n t)(a), $$i.e. I_n is the Laplace transform of \sin^n(t). Using the fact that$$ D^2(\sin^nt)=n(n-1)\sin^{n-2}t-n^2\sin^n t. $$we find that, for n>1, (using the rule for differentiation and Laplace transforms)$$ a^2I_n=n(n-1)I_{n-2}-n^2I_n, $$or$$ ...

2

Hint Let $$G(x)=\int_0^x e^{-u^2}\mathrm dx=\int_0^x g(u)\mathrm du.$$ You have that $$f(x)=G^2(x).$$ Therefore $f'(x)=2G(x)G'(x)$ and by definition, $G'(x)=g(x)$. I let you conclude.

2

It would be better to know what $b$ is in the original problem, since in full generality $$\int_{0}^{1}\sqrt{\frac{1+ax^2}{1+bx^2}}\,dx =\frac{1}{\sqrt{b}}\int_{0}^{\sqrt{b}}\frac{\sqrt{1+\frac{a}{b}x^2}}{\sqrt{1+x^2}}\,dx=\frac{1}{\sqrt{b}}\int_{0}^{\log(\sqrt{b}+\sqrt{b+1})}\sqrt{1+\frac{a}{b}\sinh^2 x}\,dx$$ is just an incomplete elliptic integral of ...

2

Hint: Differentiate your equation with respect to $c$ and use the fundamental theorem of calculus. For $\alpha -\beta \neq 0$, you will then see the claim to be true.

2

I know this isn't contour integration, but a proof of your expression using Feynman integration - let $$I(x) = \int_0^{\infty} \log(1+tx)t^{-p-1} \ \mathrm{d}t,$$ $$I'(x) = \int_0^{\infty} \frac{t^{-p}}{1+tx} \ \mathrm{d}t,$$ making the substitution $u=tx$, $$I'(x) = x^{p-1}\int_0^{\infty} \frac{u^{-p}}{1+u} \ \mathrm{d}u.$$ Recall the result that  ...

2

Like I mentioned in the comments, the mean value theorem for integrals shows that $f$ has at least one root. However, $f$ may have more than one root. There is a quadratic polynomial that satisfies all the conditions of the problem and has two roots in $(0,1)$, namely, $p(x) = 210x^2 - 192x + 27$.

2

The fundamental theorem of calculus: if $F(x)=\int_{a}^{x}f(t)dt$ then $F'(x_0)=f(x_0)$. We have $F(x)=\int_{6}^{x^3}\sin ^3 (5t)dt$. Let $\alpha = x^3$. then $F(\alpha ^ {\frac{1}{3}})=\int_{6}^{\alpha}\sin ^3 (5t)dt$ Now derive: $(F(\alpha ^ {\frac{1}{3}}) )'=\frac{1}{3}\alpha ^ {\frac{-2}{3}}F'(\alpha ^{\frac{1}{3}})$ That was from chain rule. But ...

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