Show that $\int^{\infty}_{0} x^{-1} \sin x dx = \frac\pi2$ by integrating $z^{-1}e^{iz}$ around a closed contour $\Gamma$ consisting of two portions of the real axis, from -$R$ to -$\epsilon$ and from $\epsilon$ to $R$ (with $R > \epsilon > 0$) and two connecting semi-circular arcs in the upper half-plane, of respective radii $\epsilon$ and $R$. […]

$$I = \int {e^{3x} – e^x \over e^{4x} + e^{2x} + 1} dx$$ Substituting for $e^x$, $$I = \int {u^2 – 1 \over u^4 + u^2 + 1} du = \int { u^4 + u^2 + 1 + – 2 – u^4 \over u^4 + u^2 + 1} du = u – \int {u^4 + […]

We can state that, with $n$ integer, $$\int_1^n \log x \ \mathrm{dx} \leq \sum_{m = 1}^n \log m$$ because the second is the area of $n$ rectangles with unity base, while the first is “just” the area under the function. 1) How can it analitically or geometrically be proved? 2) Can this be stated in […]

I’m not sure how to handle limits and integral and I would like some help with the following one: let $f:[0,\infty)\rightarrow \Bbb{R}$ be a continuous and bounded function, show that $$\lim_{h\to \infty}h\int_{0}^\infty{{ {e}^{-hx}f(x)} dx}=f(0)$$ I tried many things from The fundamental theorem of calculus and define $F$ such that $F’=f$ and use integration by parts […]

Help me please with integral: $$\int \frac{2x-\sqrt{4x^{2}-x+1}}{x-1}\;dx$$ I must solve it without using Euler substitution. Thanks!

How do I calculate the following indefinite integral? $$\int\frac{1}{x^3+x+1}dx$$ Approach: $x^3+x+1=(x-a)(x^2+ax+c)$ where $a:$ real solution of the equation $a^3+a+1=0$ $c:$ real solution of the equation $c^3-c^2+1=0$ Then $$\int\frac{1}{x^3+x+1}dx=\int\frac{1}{(x-a)(x^2+ax+c)}dx=\int\frac{A}{(x-a)}dx+\int\frac{Bx+C}{(x^2+ax+c)}dx$$

This is my problem,I tried to use change of variable, but no result so far. Can anyone help me? $$\int_0^1 {\frac{{\arcsin x}}{x}dx} $$

How to solve the following integral: $$\int_{-\infty }^{\infty }\exp \left ( i\left ( ax^3+bx^2 \right ) \right )dx$$ Standard CAS seem to get it totally wrong, see: http://www.walkingrandomly.com/?p=5031 So what is the right ansatz and solution? EDIT There seems to be a problem with the way this question is posed… which I quite frankly don’t […]

I would like to show that for $A(x) = \int_{0}^{x}\frac{1}{1+t^2}dt$, we have $A\left(\frac{2x}{1-x^2}\right)=2A(x)$, for all $|x|<1$. My idea is to start with either $2\int_0^x\frac{1}{1+t^2}dt$ or $\int_0^{2x/(1-x^2)}\frac{1}{1+t^2}dt$, and try to transform one into the other by change of variables. (It would make more sense for the moment if we did not do any trigonometric substitutions, since […]

$$I_1=\int_1^{\infty}\exp\left(-\left(\frac{x(2n-x)}{b}\right)^2\right)\mathrm dx,$$ I set $$t=\frac{x(2n-x)}{b},$$ and, solving for $x$ and $dt$ I got $$I_1=\frac{b}{2 n} \int_1^{\infty} e^{-t^2}\left(1-\frac{bt}{n^2}\right)^{\frac12}\mathrm dt.$$ I then expand $$\left(1-\frac{bt}{n^2}\right)^{\frac12} \approx 1+\frac{bt}{2n^2}$$ in the first two terms of Binomial series, and obtain something like $$I_1 \approx \frac{b}{2n}\left(\frac{\sqrt{\pi}(1-\mathrm{erf}(1))}{2}+\frac{b}{4n^2e}\right).$$ I am not 100% sure about this derivation (although the result is sensible), especially about the […]

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