3 votes
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Using Feynman's technique to find $\int_0^{\infty}\frac{\cos{x}}{x^2+1} dx$

The trick for the problem is to take the first derivative integral and multiply the integrand by $\frac{x}{x}$: $$\int_{0}^{\infty} \frac{-x^2}{x^2+1}\cdot\frac{\sin ax}{x}\:dx = -\int_0^\infty\frac{\...
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  • 25.3k
3 votes
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Indication: $\displaystyle \lim_{x \to \infty}{\int^{x}_{1}{\ln(f(t))dt}}$

$$\int^{x}_{1}\ln(4t^3+1)\textrm{d}t \ge \int^{x}_{1}\ln(4t^3)\textrm{d}t =\int^{x}_{1}\ln(4)+\ln (t^3)\textrm{d}t \ge \int^{x}_{1}\ln(4)\textrm dt= (\ln 4)(x-1)$$
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  • 74.9k
3 votes

$I(x) = -\int_0^1 \frac{1}{z}\ln\left(\frac{1-x z + \sqrt{1-2 x z+ z^2}}{2}\right)\,dz$

Here's an approach using differentiation under the integral sign: $$\begin{align}I'(x) &= \int_{0}^{1}\frac{\frac{1}{\sqrt{-2xz+z^2+1}}+1}{\sqrt{-2xz+z^2+1}-xz+1}\, dz\\ &=\left.\frac{(x+1)\ln\...
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  • 2,967
2 votes

Evaluate the line integral $\int_{L} \frac{-y \,d x+x \,d y}{x^{2}+y^{2}}$ for a line segment $L$

The "change in argument" is, informally, just the total net angle that the path winds around the origin anticlockwise (a negative change in argument indicates a net clockwise winding). We ...
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2 votes

$\iint_Sz^2d\sigma$ where $S$ is an area of the cone $z=\sqrt{x^2+y^2}$ between planes $z=0$ and $z=1.$

The factor $\sqrt2$ arises from the area element of the given cone $$ dS = \sqrt{1+( z_x’)^2+ ( z_y ’)^2} \ dxdy= \sqrt2\ dxdy $$
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  • 67.3k
2 votes
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$\iint_Sz^2d\sigma$ where $S$ is an area of the cone $z=\sqrt{x^2+y^2}$ between planes $z=0$ and $z=1.$

The problem in your approach is the fact $\displaystyle \iint_{S}f(x,y,z)\, {\rm d}\sigma$ is a surface integral. You don't have $\displaystyle \iint_{D}f(x,y)\, {\rm d}A$ that is a double integral in ...
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2 votes

How to Evaluate $\int_1^4 (\frac{1}{2t}+i)^2 dt$

Since $$ \left(\frac1{2t}+i\right)^2=\frac1{4t^2}-1+\frac it, $$ you have $$ \begin{align*} \int_1^4\left(\frac1{2t}+i\right)^2\,dt&=\int_1^4\frac1{4t^2}-1\,dt+\left(\int_1^4\frac1t\,dt\right)i\\ &...
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2 votes

How to Evaluate $\int_1^4 (\frac{1}{2t}+i)^2 dt$

We have \begin{align*}\int \limits _1^4\left (\frac{1}{2t}+i\right )^2\,dt & =\int \limits _1^4\left (\frac{i}{t}+\frac{1}{4t^2}-1\right )\,dt \\ & =i\int \limits _1^4\frac{1}{t}\,dt+\frac{1}{...
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2 votes

Find $\displaystyle \int \dfrac{dx}{(x^2 + a^2)^3} $

Alternatively, integrate by parts to avoid the trig substitution \begin{align} I_3=&\int \frac{1}{(x^2 + a^2)^3 }dx =\int \frac{1}{4a^2x^3}d\left(\frac{x^4}{ (x^2 + a^2)^2}\right) = \frac{x}{4a^2 (...
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  • 67.3k
1 vote

Gaussian integral in 3rd dimensions

Is there no actual method to define $\frac13!$ ? The Gamma function is the definition of non-integer factorials. Explicitly $$ \frac13! : = \int_{0}^{\infty} x^{\frac13}e^{-x}\, \mathrm{d}x \tag{1} $$...
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  • 4,990
1 vote
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Confused what I did wrong for $\int_{0}^{\infty} \frac{1}{1 + x^4} \, dx$

You need to think of the first integral $$\sqrt{i}\int_{0}^{\infty} \frac{1}{1 - u^4} \, du$$ as being an integral of a function defined on the complex plane, and the path of integration is along the ...
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