Tag Info

Hot answers tagged

17

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 ...


11

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 $$\int_{-\infty}^{\infty} du \, \frac{e^{i u k}}{\pi^2+4 u^2} = \frac14 \frac{\pi}{\pi/2} e^{-\pi |k|/2} $$ ...


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 ...


10

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 ...


9

One way is to use 'Feynman's trick', and differentiate under the integral. Note $$\int_0^\infty x^4e^{-\alpha x^2}=\int_0^\infty \frac{\partial^2}{\partial \alpha^2}e^{-\alpha x^2}=\frac{\partial^2}{\partial \alpha^2}\int_0^\infty e^{-\alpha x^2}$$ Calculate the inner definite integral first, then differentiate it twice with respect to $\alpha$.


9

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 ...


9

Outline of Method Perform the substitution $u = \ln x$. You will get $$ \int_{-\infty}^0 x^n e^x dx.$$ This is very nearly the Gamma function. Perform the substitution $u = -x$. You will get $$ (-1)^n \int_0^\infty x^n e^{-x} dx.$$ This now is the Gamma function. In particular, you have $$ (-1)^n \Gamma(n+1) = (-1)^n n!.$$ We might do a sanity check when ...


8

Notice, let $n$ be an integer then using property of definite integral, we have $$\int_{0}^{n}[x]dx=\int_{0}^{1}[x]dx+\int_{1}^{2}[x]dx+\int_{2}^{3}[x]dx+\ldots+\int_{n-1}^{n}[x]dx$$ $$=\int_{0}^{1}(0)dx+\int_{1}^{2}1dx+\int_{2}^{3}2dx+\ldots+\int_{n-1}^{n}(n-1)dx$$ $$=0+1+2+3+\ldots+(n-1)$$ $$\implies \int_{0}^{n}[x]dx=1+2+3+\ldots ...


8

Hint: $\lfloor x \rfloor = k$ for $x \in [k,k+1)$ for $k = 0,1,2,\ldots,n-1$. Hence $$\int_0^n \lfloor x \rfloor \; dx = \sum_{k=0}^{n-1} \int_{k}^{k+1} \lfloor x \rfloor \; dx = \sum_{k=0}^{n-1} \int_{k}^{k+1} k \; dx,$$ where the integrand is a constant with respect to $x$.


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.


7

Given $$ \int_0^1 \ln^n(x) dx $$ Let us write $$ I_n = \int_0^1 \ln^n(x) dx. $$ Integrate by parts $$ \begin{eqnarray} I_n &=& \int_0^1 \ln^n(x) dx\\ &=& \Bigg[ x \ln^n(x) \Bigg]_0^1 - n \int_0^1 \ln^{n-1}(x) dx\\ &=& \Bigg[ \exp(y) y^n\Bigg]_{-\infty}^0 - n \int_0^1 \ln^{n-1}(x) dx = - n I_{n-1}. \end{eqnarray} $$ ...


5

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\\ ...


5

By making the Euler substitution $\sqrt{x^{2}+x} = x+t$, we find $$ \begin{align} \int_{0}^{1} \frac{\log(x)}{\sqrt{x^{2}+x}} \, dx &= 2 \int_{0}^{\sqrt{2}-1} \frac{\log \left(\frac{t^{2}}{1-2t}\right)}{1-2t} \, dt \\ &= 4 \int_{0}^{\sqrt{2}-1} \frac{\log t}{1-2t} \, dt - 2\int_{0}^{\sqrt{2}-1} \frac{\log(1-2t)}{1-2t} \, dt. \end{align}$$ The first ...


5

Let: $$ f(s)=\int_{0}^{1}\frac{x^s}{\sqrt{1+x}}\,dx. $$ We have, by integration by parts: $$ f(s+1)+f(s) = \int_{0}^{1} x^s\sqrt{1+x}\,dx = \frac{1}{s+1}\left(\sqrt{2}-\frac{1}{2}\,f(s+1)\right)$$ hence: $$ \left(2s+3\right)\, f(s+1)+(2s+2)\,f(s) = 2\sqrt{2},$$ $f(0)=2\sqrt{2}-1$ and $\lim_{s\to +\infty}f(s)=0$. We have: $$ f(s)=\sum_{n\geq 0}\frac{(-1)^n ...


5

Perform the integration in steps: $$ \begin{array}{rcl} I_8 &=& \displaystyle \int_0^1 \frac{x^8}{\sqrt{x^3 + 1}} dx = \displaystyle \int_0^1 x^6 \frac{x^2}{\sqrt{x^3 + 1}} dx\\ &=& \displaystyle \left[ \frac{2}{3} x^6 \sqrt{x^3+1} \right]_0^1 - \int_0^1 4 x^3 x^2 \sqrt{x^3+1} dx\\ &=& \displaystyle \left[ ...


5

Let $I = \displaystyle\int_{0}^{\pi}xf(\sin x)\,dx$. Substitute $x' = \pi - x$ to get $I = -\displaystyle\int_{\pi}^{0}(\pi-x')f(\sin (\pi - x'))\,dx' = \displaystyle\int_{0}^{\pi}(\pi-x')f(\sin x')\,dx'$. So, $2I = I+I = \displaystyle\int_{0}^{\pi}xf(\sin x)\,dx + \int_{0}^{\pi}(\pi-x)f(\sin x)\,dx$ Can you take it from here?


5

Another way to look at it (that probably isn't simpler) : Let $u=x^2$, then the integral becomes: $$\frac{1}{2} \int_0^\infty u^{3/2}e^{-\alpha u}du$$ which is the Laplace transform of $u^{3/2}$, times ${1}/{2}$. We know that the Laplace transform of $u^n$ is $n!*\alpha^{-n-1}$ or generally $\Gamma(n+1)*\alpha^{-n-1}$. Here, $n=3/2$ so $$\Gamma(3/2 ...


4

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 ...


4

Integrating by parts, we have $$\begin{align} I_n&=\int_0^1\frac{x^n}{(1+x^3)^{1/2}}dx\\\\ &=\left.\frac23 x^{n-2}(x^3+1)\right|_0^1-\frac23(n-2)\int_0^1x^{n-3}(x^3+1)^{1/2}dx \tag 1\\\\ &=\frac232^{1/2}-\frac23(n-2)\int_0^1\frac{x^{n-3}(x^3+1)}{(x^3+1)^{1/2}}\,dx \\\\ &=\frac232^{1/2}-\frac23 (n-2)I_n-\frac23(n-2)I_{n-3}\\\\ ...


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= \\ ...


4

It is easier to evaluate the integral in cylindrical coordinates. We have $$\int_0^b \int_0^{2\pi} \int_0^a 5\rho^2\cos^2\phi \rho \,d\rho\, d\phi\, dz=5\pi b\frac{a^4}{4} \tag 1$$ where we used $\cos^2\phi =\frac{1+\cos 2\phi}{2}$ and $\int_0^{2\pi}\cos 2\phi d\phi=0$ to evaluate the integral in $(1)$ In Cartesian coordinates, we can evaluate the volume ...


4

Hint: Put $-\alpha x^2=u \Rightarrow -2\alpha x \mathrm dx=\mathrm du$,then $x\mathrm dx=\frac {-1}{2\alpha}\mathrm du$. you will have, $$\displaystyle\int xe^{-\alpha x^2}\,dx=\frac {-1}{2\alpha}\displaystyle\int e^u \mathrm du$$


3

After making the substitution $u = \text{arctanh}(x)$, we could use the Laplace transform $$\int_{0}^{\infty} \cos(ax) \, e^{-bx} \, dx = \frac{b}{a^{2}+b^{2}} \, , \, \text{Re} (b) >0 $$ and then switch the order of integration. Specifically, $$ \begin{align} \int_{-\infty}^{\infty} \frac{\text{sech}^{2}(u)}{4u^{2} + \pi^{2}} \, du &= ...


3

\begin{align*}I_n&=\int_0^1 \dfrac{x^n}{\sqrt{x^3+1}}dx=\int_0^1 \dfrac{x^{n-3}(x^3+1-1)}{\sqrt{x^3+1}}dx = \int_0^1 x^{n-3}\sqrt{x^3+1}dx - \int_0^1 \dfrac{x^{n-3}}{\sqrt{x^3+1}}dx\\ &= \int_0^1 x^{n-3}\sqrt{x^3+1}dx-I_{n-3}\end{align*}This integral is handled with integration by parts: $$\int_0^1 ...


3

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 ...


3

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 = ...


3

We have $$\begin{align} I&=\int_0^{\infty}e^{-(x^2+1/x^2)}dx \tag 1\\\\ &=\int_0^{\infty}\frac{1}{x^2}e^{-(x^2+1/x^2)}dx \tag 2 \end{align}$$ where in going from $(1)$ to $(2)$ we enforced the substitution $x\to 1/x$. We also are given $$J=\int_0^{\infty}x^2e^{-(x^2+1/x^2)}dx $$ Thus, forming the difference $J-I$ we find $$\begin{align} ...


3

Let \begin{align}\tag{1} I = \int_{0}^{\infty} e^{- \left( u^{2} + \frac{1}{u^{2}} \right) } \, du. \end{align} Now let $u = 1/x$ to obtain \begin{align}\tag{2} I = \int_{0}^{\infty} e^{- \left( x^{2} + \frac{1}{x^{2}} \right)} \, \frac{dx}{x^{2}}. \end{align} Adding (1) and (2) leads to \begin{align} 2 I = \int_{0}^{\infty} e^{- \left( u^{2} + ...


3

I tried 2 ways to find a closed form, though unsuccessful up to this point. 1st trial. Let $I$ denote the integral, and write $$ I = 8 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \int_{0}^{\infty} \frac{e^{-x}(1 - e^{-(2n+1)x})}{x (1 + e^{-x})^{2}} \, dx. $$ In order to evaluate the integral inside the summation, we introduce new functions $I(s)$ and ...


3

Here, you'll want to integrate by part to find the Gauss integral : $$I = \int_0^{\infty} x^4 e^{-\alpha x^2} dx = \int_0^{\infty} x^3 \times x e^{-\alpha x^2} dx$$ Now, you let $u = x^3$ and $v' = x e^{-\alpha x^2} $, and this gives you $$I = \left [ -x^3 \frac{e^{-\alpha x^2}}{2\alpha} \right]_0^{\infty}+\int_0^{\infty} \frac{3}{2\alpha} x^2 e^{-\alpha ...



Only top voted, non community-wiki answers of a minimum length are eligible