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1

Here are closed forms for the first and second integrals respectively $$ I_1 = { \frac {\sqrt {2\pi }}{b\sqrt {T}}}{{\rm e}^{-\,{\frac {{b}^{2}}{2T}}}} $$ $$I_2= \, \pi- {{\rm erf}\left( {\frac {b}{\sqrt {2T}}}\right)}.$$ You can test them numerically.


1

it is a good habit to remove irrelevant complications before getting down to detail. here you may observe that the average value of the function $y=\tan(2x)$ over the interval $[0,\frac{\pi}{8}]$ is the same as the average value of $y=\tan x$ over the interval $[0,\frac{\pi}{4}]$. so, using the substitution $w=\tan x$, we obtain: $$ \text{average} = ...


1

$$ \int \tan(2x)\,dx = \frac 1 2\int\frac{1}{\cos(2x)}\Big(\ \underbrace{2\sin(2x)\,dx}_{du}\ \Big) = \frac 1 2 \int \frac 1 u \, du $$ where $u=\cos(2x)$, etc.


4

The general formula for the average value of a function $f(x)$ over $a \le x \le b$ is $$f_{\text{avg}} = \dfrac{1}{b-a}\displaystyle\int_a^bf(x)\,dx$$ Here, you want the avgerage value of the function $f(x) = \tan 2x$ over $0 \le x \le \dfrac{\pi}{8}$. Can you use the above formula to get the answer? EDIT: Since you are having trouble integrating $\tan ...


3

Hint : Try the substitution $$t=\cfrac{T}{u^2+1}$$ The first integral has the shape of the gaussian. The second one leads you to $$\alpha \int_{\mathbb{R}^+} \cfrac{1}{u^2+1} \exp(-\beta(u^2+1) ) \,du$$ This integral is a bit more tricky : Bruno Joyal has found a close form in this answer


1

We have $$ I = \int_{0}^{\pi/4}\frac{\log(1-2\sin^2\theta)}{\theta}\,d\theta = \int_{0}^{\pi/4}\frac{\log(\sin(2\theta))}{\pi/4-\theta}\,d\theta,\tag{1}$$ hence by considering the Taylor series of $\log(1-x)$ we end with a series of CosIntegral values, not so appealing. An interesting approach may be to represent both ...


0

To solve b), you want to find $E(x)=\int_{0}^{\infty}x(2e^{-2x}) dx$. To solve c), you want to solve $\int_{0}^{m}2e^{-2x} dx=\frac{1}{2}$ for $m$.


1

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1

Here's yet another way: \begin{align} & \int \cos^2\theta\,d\theta = \int(\cos\theta) \Big(\cos\theta\,d\theta\Big) = \int u\,dv = uv-\int v\,du \\[10pt] = {} & -\cos\theta\sin\theta -\int(\sin\theta)\, \Big( -\sin\theta\, d\theta\Big) \\[10pt] = {} & -\cos\theta\sin\theta + \int \sin^2\theta\,d\theta = -\cos^2\theta + ...


1

If you look at the plot of $y=\cos^2x$, you will notice that the curve is perfectly symmetric around $y=\frac12$ and has period $\pi$, as confirmed by $cos^2x=\frac12\cos2x+\frac12$. For this reason, the average value over a period is $\frac12$, and the area under the curve is $\frac\pi2$.


3

$$ \int_{-\pi/2}^{\pi/2} \cos^2\theta\,d\theta + \int_{-\pi/2}^{\pi/2} \sin^2\theta\,d\theta = \int_{-\pi/2}^{\pi/2} 1\,d\theta=\pi. $$ If you can show the two integrals are equal, then they each have to be $\pi/2$. But they have to be equal since the graph of $\sin^2$ has the same size and shape as that of $\cos^2$ and the interval from $-\pi/2$ to $\pi/2$ ...


0

Some hints: First, it's convenient to shift coordinates so that your integrand is of the form $\sqrt{x^2+y^2}$. This minimizes the labor of integration. The cost is that the region of integration is shifted off the origin; how do we manage that? The obvious way of writing a circle of radius $R$ in polar coordinates is $(x,y)=(R\cos\phi,R \sin\phi)$ (i.e. ...


7

Note that $$\int_{- \pi /2}^{\pi / 2} \cos^2 \theta d \theta= \int_{- \pi /2}^{\pi / 2} \sin^2 \theta d \theta \ $$ and that $$\pi = \int_{- \pi /2}^{\pi / 2} 1 d \theta = \int_{- \pi /2}^{\pi / 2} (\cos^2 \theta +\sin^2 \theta) d \theta = 2 \int_{- \pi /2}^{\pi / 2} \cos^2 \theta d \theta$$ hence $$\int_{- \pi /2}^{\pi / 2} \cos^2 \theta d \theta= ...


4

From the hint, $\cos^2 x= \frac{1}{2}(\cos 2x+1)$ so our integral becomes $$\int_{-\pi/2}^{\pi/2} \frac{1}{2}(\cos 2x+1)dx= \left. \left (\frac{1}{4}\sin 2x+\frac{1}{2}x \right ) \right |_{-\pi/2}^{\pi/2}=\pi/2$$


3

Consider $$ \mathcal{I}(y,t)=\int_{-\infty}^{\infty}\frac{\cos xt}{x^2+y^2}\ dx=\frac{\pi e^{-yt}}{y}\quad;\quad\text{for}\ t>0.\tag1 $$ Differentiating $(1)$ with respect $t$ and $y$ yields \begin{align} \frac{\partial^2\mathcal{I}}{\partial y\partial t}=\int_{-\infty}^{\infty}\frac{2xy\sin xt}{(x^2+y^2)^2}\ dx&=\pi te^{-yt}\\ ...


2

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1

Hint: $$u^3+2u^2-1=(u+1)(u^2+u-1).$$ Completing the square $$ u^2+u-1= u^2+u\frac{1}{4}-\frac{1}{4}-1=(u+1/2)^2-5/4.$$ Now use partial fraction \begin{align} \frac{1}{(u+1)((u+1/2)^2-5/4)}=\frac{A}{u+1}+\frac{Bu+C}{( u+1/2)^2-5/4} \end{align} We find that $ A=-1, B=1, C=0$, so that \begin{align} I&=\int { \frac{-1}{u+1} du}+\int{ ...


2

As $$u^3+2u^2-1=(u+1)(u^2+u-1)$$ Write, $$\frac u{u^3+2u^2-1}=\frac A{u+1}+B\frac{\dfrac{d(u^2+u-1)}{du}}{u^2+u-1}+\frac C{u^2+u-1}$$ $$u=A(u^2+u-1)+(2Bu+B)(u+1)+C(u+1)$$ Set $\displaystyle u+1=0$ to find $A$ Now comparing the coefficients of $u^2$, $0=A+2B\iff B=?$ Again comparing the constants $0=-A+B+C\implies C=?$ Observe that the second integral ...


1

Hint: Obserse that $u^3 + 2u^2 -1$ has $u+1$ as a factor.


3

We can use the following way to solve. It is very simple. In fact \begin{eqnarray} I&=&\int_0^\infty \frac{\ln x}{1+x^4}dx=\int_0^1 \frac{\ln x}{1+x^4}dx+\int_1^\infty \frac{\ln x}{1+x^4}dx\\ &=&\int_0^1 \frac{\ln x}{1+x^4}dx-\int_0^1 \frac{x^2\ln x}{1+x^4}dx\tag{1}\\ &=&\int_0^1\sum_{n=0}^\infty(-1)^n(x^{4n}-x^{4n+2})\ln xdx\tag{2}\\ ...


0

Let $1_{C}$ denote the characteristic function of set $C$. And define $f\left(x,y\right)=\left(x^{2}+y^{2}\right)1_{C}\left(x,y\right)$ Then: $$\underset{C}{\iint}x^{2}+y^{2}dxdy=\iint f\left(x,y\right)dxdy$$ Now fix $y$ and have a look at function $g_{y}\left(x\right)=f\left(x,y\right)$. It is prescribed by $x\mapsto0$ if $x\neq y$ and $x\mapsto2y^{2}$ ...


5

If you use polar coordinates you will get the following $$ \int_{0}^{2\pi} \int_{0}^{1} r^2 r dr \ d\theta $$


8

Use polar coordinates, We know that $$r^2 = x^2 + y^2$$ So our double integral becomes $$\int_{0}^{2\pi} \int_0^{1}r^2\cdot rdrd\theta$$ Now solve. EDIT I see that your computation is correct, I am simply offering another alternative and more easier way to solve this double integral.


0

If $C$ is exactly as you gave it, then your solution is incorrect, since $C$ is just the diagonal of the square with sides from $-1$ to $1$ and the integral is zero. Integrating in $R^2$ over a line segment is the same as integrating in $R$ over a point. The integrals are a way to understand what area is. The area of a point in a line is zero, and the area ...


0

From the figure, $$ A = \int_{-1}^{x_1}[f(x) - g(x)]dx + \int_{x_1}^{(1 + \sqrt{5})/2}[g(x) - f(x)]dx $$ where $x_1$ is a root of equation $f(x) = g(x)$.


1

Almost right, you have made sign errors: To the right of $\frac{1-\sqrt{5}}{2}$ $x^2 < x+1$ so you should be integrating $x+1-x^2$. Similarly your integrand for the first term is the negative of what it should be. That is the reason you may be getting a peculiar negative answer...


1

Theorem: if $f$ is a bounded function and has a finite amount of discontinuities on a compact interval $[a,b]$, then $f$ is Riemann integrable on $[a,b]$, Check out this answer, $f$ bounded on $[a,b]$ with one or finite discontinuities implies $f$ Riemann-integrable.


1

$$F(x)=\int f(x) \:\mathrm{d}x= \begin{cases} \int 1 \:\mathrm{d}x & x\geq1\\ \int 0 \:\mathrm{d}x & x<1 \end{cases} $$ $$\therefore\:F(x)=\begin{cases} x+C_1 & x\geq1\\ 0+C_2 & x<1 \end{cases}$$ $$\int_0^1 f(x) \:\mathrm{d}x=\left[ 0\right]_0^1=0$$


2

(This is more a comment than answer, but I couldn't get MathJax to properly show it in comments) Here is a nice identity (equation (21) of this paper with $x=-1/7$): $$_2F_1 \left(a,a+\frac{1}{2};\frac{4a+5}{6};-\frac{1}{7}\right)=\left(\frac{7}{4}\right)^a {_2}F_1 \left(\frac{a}{3},\frac{a+1}{3};\frac{4a+5}{6};-27\right)$$ It's an example of a cubic ...


4

Too long for a comment and this is not the answer too, but following @David H's comment. Perhaps your brother means $$\int_{\color{red}{\large\frac12}}^1\frac{\ln (2x-1)}{\sqrt[\large 6]{x(1-x)(1-2x)^4}}\ dx.\tag1$$ If so, then setting $u=2x-1$ to $(1)$ yields $$ \frac1{\sqrt[\large 3]{4}}\int_{0}^1\frac{\ln u}{\sqrt[\large 6]{(1+u)(1-u)u^4}}\ ...


4

I highly doubt that one can evaluate that integral by hand, however the integral does have a closed form found by Mathematica to be: $$\frac{2 \left(4 (-1)^{5/6} \, _4F_3\left(1,1,1,\frac{7}{6};2,2,2;\frac{1}{9}\right)+8 (-1)^{5/6} \log (3) \, _3F_2\left(1,1,\frac{7}{6};2,2;\frac{1}{9}\right)+432 (-1)^{5/6} \log ^2(2)-189 (-1)^{5/6} \log ^2(3)+648 (-1)^{5/6} ...


9

Here is an approach without using contour integration (Cauchy theorem). I've found the following general result. Theorem 1. Let $a$ be any real number. Then $$ \begin{align} \displaystyle \int_{0}^{\pi} \frac{x}{x^2+\ln^2(2 e^{a} \sin x)}\mathrm{d}x & = \, \frac{2 \pi^2}{\pi^2+4a^2},\tag1\\\\ \int_{0}^{\pi} \frac{\ln(2 e^{a} \sin ...


1

I don't think a closed form exists after computing that integral numerically in Mathematica, and looking up in the Inverse Symbolic Calculator. It is approximately equal to: $-0.10617124113817\ldots$ If you need more digits just ask.


3

The integral converges only when $a$ is an odd multiple of $\pi/2$, and does not seem to vanish even for those $a$. For large $\left|z\right|$ it is known that $$ J_0(z) = \sqrt{\frac2{\pi\left|z\right|}} \bigl(\cos (\left|z\right|-\frac\pi4) + O(1/\left|z\right|) \bigr). $$ Therefore we have for large $x$ (either $x>0$ and $x<0$): $$ ...


1

HINT: Denote the integral by $ I $. Since $$ x^a-x^{-a}=e^{a\ln(x)}-e^{-a\ln(x)}=2\sinh(a\ln x) .$$ Substituting $\ln x= u $, we get $$ I=\int_{-\infty}^0 \frac{ 2\sinh au}{e^u-1 }e^udu $$ To change the limits of the integral to more convenient one, again substitue $ u=-t$ we have \begin{align} I&=\int_0^{\infty}\frac {2\sinh at}{e^{-t }-1} e^{-t} ...


8

I also tried expanding it with a series, but it didn't help. It should have helped. If we write $$\frac{x^{-a}-x^a}{1-x} = (x^{-a}-x^a)\sum_{n=0}^\infty x^n,$$ by the monotone convergence theorem, we have $$\begin{align} \int_0^1 \frac{x^{-a}-x^a}{1-x}\,dx &= \sum_{n=0}^\infty \int_0^1 (x^{-a}-x^a)x^n\,dx\\ &= \sum_{n=0}^\infty ...


0

To illustrate intuitively why this doesn't and shouldn't work, think of function on the interval $[0,L]$ rotated around the x-axis. It should be clear that the area of the surface of revolution should depend on the length of the curve in that interval. So $y = c$, for some constant $c$, has a length of L, and the surface generated is the side of a right ...


3

It could be a way to use the digamma function $\psi$ which verify $$ \psi(a+1)=-\gamma+\int_0^1 \frac{1-x^a}{1-x} \mathrm{d}x, \quad |a|<1,$$ since $$\psi(a)-\psi(1-a)=-\pi \cot (\pi a),$$ $$\psi(1+a)-\psi(a)=\frac{1}{a}.$$ But this is not elementary. I doubt that you have to prove all from the beginning.


6

Since no answers have been posted, I'll expand on my comment above. There is a general formula that states $$\sum_{n=1}^{\infty}\frac{H_{n}^{(r)}}{n^{q}}=\zeta(r)\zeta(q)-\frac{(-1)^{r-1}}{(r-1)!}\int_{0}^{1}\frac{\text{Li}_{q}(x) \log^{r-1}(x) }{1-x}dx $$ where $$ H_{n}^{(r)} = \sum_{k=1}^{n} \frac{1}{k^{r}} .$$ A proof can be found here. Making the ...


2

Using the identity $$\int_0^{1} \left[\frac{1}{\log(s)}+\frac1{1-s}\right] \, ds= \gamma,$$ we have $$\int_0^k \left(\int_1^2 \frac{(s+1)^{n-1}+s-1}{s} \, dn\right) \, ds \\ = \int_0^k \left(\int_1^2 \frac{(s+1)^{n-1}}{s} \, dn\right) \, ds +\int_0^k \left( 1- \frac1{s}\right) \, ds\\=\int_0^k \frac{1}{\log(s+1)} \, ds +\int_0^k \left( 1- ...


1

Here is an idea. I have not made all the computations, and I hope that I does not have made computation errors. If you make the change of variable $2t=f\pi$, then your integral becomes $$C\int_0^{+\infty}\frac{(\sin(2t\tau))^4\sin(t)^2}{t^3\cos(t)^2} dt=C I$$ where $C$ is a computable constant. Now as $\tau$ is an integer, there exists a polynomial ...


15

$$I:=\int_{0}^{\infty}\frac{\ln{(x)}}{\sqrt{x}\,\sqrt{x+1}\,\sqrt{2x+1}}\mathrm{d}x.$$ After first multiplying and dividing the integrand by 2, substitute $x=\frac{t}{2}$: ...


3

Write the sin-function as the difference of two exponentials and obtain the corresponding integrals by contour integration. This gives the result: \begin{eqnarray*} I &=&\int_{-\infty }^{+\infty }dx\frac{1}{x-i}\sin x=\int_{-\infty }^{+\infty }dx\frac{1}{x-i}\frac{1}{2i}\left\{ e^{ix}-e^{-ix}\right\} \\ \int_{-\infty }^{+\infty ...


2

The general gaussian integration formula is $$\int_{-\infty}^{\infty}\ldots\int_{-\infty}^{\infty} e^{-\frac12 x^T Ax}dx_1\ldots dx_n=\frac{(2\pi)^{\frac{n}{2}}}{\sqrt{\operatorname{det}A}}.$$ In your case, $n=2$ and $A=\left(\begin{array}{cc} 1 & -1/2 \\ -1/2 & 1\end{array}\right)$.


11

Recall: $$x^2-xy+y^2=(x-\frac{1}{2}y)^2+\frac{3}{4}y^2$$ With that you get: $$\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}(x^2-xy+y^2)}dxdy=\\\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}\left((x-\frac{1}{2}y)^2+\frac{3}{4}y^2\right)}dxdy=\\ \int_{-\infty}^\infty\int_{-\infty}^\infty ...


1

The mistake is simple--$\mathrm{Ci}$ has a branch cut across the negative real axis, so $\mathrm{Ci}(-\infty - i)$ should indeed evaluate to $-i \pi$ rather than $i \pi$.


5

The main idea is that, near the largest point of the integrand (which occurs at $x=0$), we have $$ \frac{1}{\log x} + \frac{1}{1-x} \approx \frac{1}{\log x} + 1. $$ So we split the integral up as $$ \int_0^1 \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx = \left[\int_0^{1/e} + \int_{1/e}^1 \right] \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx. ...


1

Ah, I guess the problem is that $f$ blows up on the imaginary axis, so $z^{v-1/2}f(z)$ does not go to 0 uniformly as $z$ goes to infinity.


4

Proposition. $$ \int_{0}^{\pi/2} x^3 \ln^3(2 \cos x)\:{\mathrm{d}}x = \frac{45}{512} \zeta(7)-\frac{3\pi^2}{16} \zeta(5)-\frac{5\pi^4}{64} \zeta(3)-\frac{9}{4}\zeta(\bar{5},1,1) $$ where $\zeta(\bar{p},1,1)$ is the colored MZV (Multi Zeta Values) function of depth 3 and weight $p+2$ given by $$ \zeta(\bar{p},1,1) : = \sum_{n=1}^{\infty} ...



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