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## Hot answers tagged definite-integrals

17

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

11

Recall: $$x^2-xy+y^2=(x-\frac{1}{2}y)^2+\frac{3}{4}y^2$$ With that you get: $$\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}(x^2-xy+y^2)}dxdy=\\\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}\left((x-\frac{1}{2}y)^2+\frac{3}{4}y^2\right)}dxdy=\\ \int_{-\infty}^\infty\int_{-\infty}^\infty ... 10 We will prove that$$I=-\frac{\pi^4}{2880}.$$Indeed, let$$ J=\int_0^{\pi/2}\log^2(\sin x)\log(\cos x)\tan x \,dx $$It is easy to see that$$\eqalign{J&=\int_0^{\pi/4}\log^2(\sin x)\log(\cos x)\tan x \,dx+ \int_{\pi/4}^{\pi/2}\log^2(\sin x)\log(\cos x)\tan x \,dx\cr &=\int_0^{\pi/4}\log^2(\sin x)\log(\cos x)\tan x \,dx+ \int_{0}^{\pi/4}\log^2(\cos ...

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 ... 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 ...

8

With the substitution $x^2=t$, $$I=\frac{1}{2}\int_0^{\infty} \frac{\sin^3(\pi t)\cos(4t)}{t^3}\,dt$$ Since, $$\sin(3x)=3\sin x-4\sin^3 x \Rightarrow \sin^3x=\frac{3\sin x-\sin(3x)}{4}$$ $$\Rightarrow \sin^3(\pi t)=\frac{3\sin(\pi t)-\sin(3\pi t)}{4}$$ i.e \begin{aligned} I &=\frac{1}{8}\int_0^{\infty} \frac{(3\sin(\pi t)-\sin(3\pi ... 8 Use polar coordinates, We know thatr^2 = x^2 + y^2$$So our double integral becomes$$\int_{0}^{2\pi} \int_0^{1}r^2\cdot rdrd\theta$$Now solve. EDIT I see that your computation is correct, I am simply offering another alternative and more easier way to solve this double integral. 7 Hint : Try the substitution$$t=\cfrac{T}{u^2+1}$$The first integral has the shape of the gaussian. The second one leads you to$$I(\beta) = \alpha \int_{\mathbb{R}^+} \cfrac{1}{u^2+1} \exp(-\beta(u^2+1) ) \,du$$Considering$$\begin{cases} I'(\beta)=\alpha \int_{\mathbb{R}^+} \exp(-\beta(u^2+1) ) \,du=\alpha\exp(-\beta)\cfrac{\sqrt{\pi}}{2\sqrt{\beta}} ...

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= ... 6 I am afraid that what you did is wrong : you are not integrating a polynomial expression. Just for your curiosity,$$\frac{d}{dt}\Big(\frac{e^{t+t^2}}{t^2/2+t^3/3}\Big)=\frac{6 e^{t+t^2} (t+2) \left(4 t^2-3\right)}{t^3 (2 t+3)^2} \neq e^{t+t^2}$$To compute$$I=\int e^{t}.e^{t^2} dt=\int e^{t^2+t} dt$$first complete the square for the exponent and perform a ... 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 ... 6 If you use polar coordinates you will get the following$$ \int_{0}^{2\pi} \int_{0}^{1} r^2 r dr \ d\theta $$5 We have \lim_{y\to-\infty}\phi(y)=0 and \lim_{y\to\infty}\phi(y)=1. Moreover, the function \phi is continuous. Thus, by the Intermediate Value Theorem, the function \phi is surjective. 5 We have:$$ J=\int_{0}^{+\infty}\frac{x^p}{1+x^2}\arctan(x)dx,$$but since \arctan x=\Im\log(1+xi)=-\Im\log(1-xi),$$ J=\frac{1}{2}\Im\int_{0}^{+\infty}\frac{x^p}{1+xi}\log(1+xi)\,dx-\frac{1}{2}\Im\int_{0}^{+\infty}\frac{x^p}{1-xi}\log(1-xi)\,dx$$hence:$$ ...

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 $$\int_0^1 \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx = \left[\int_0^{1/e} + \int_{1/e}^1 \right] \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx. ... 5 Use the substitution: x = \tan\theta. The integral is then equal to:$$I= \int_{0}^{\pi/4} \ln(1+\tan\theta) \ d\theta (*)$$Also,we know the property:$$\int_{0}^{b} f(x) \ dx = \int_{0}^{b} f(b-x) \ dx$$so we have$$I = \int_{0}^{\pi/4} \ln\biggl(1+\tan\Bigl(\frac{\pi}{4}-\theta\Bigr)\biggr) \ d\theta = \int_{0}^{\pi/4} ...

4


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