# Linked Questions

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### Two sums for $\pi$

I was looking for a simple way to evaluate the integral $\int_0^\infty \frac{\sin x}{x}dx$ ( a belated look at this question). There are symmetries to be exploited, for one thing. So I had an idea, ...
2answers
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### Difficulties understanding a proof of $\int_0^{\infty} \frac{\sin(x)}{x} \, dx = \frac{\pi}{2}$

I got a homework and I've trying to do this problem about 2 days, but I "lost my fight". So I turn to you. I have to prove that $$\int _0^\infty \frac{\sin (x)}{x} \, dx = \frac{\pi}{2}.$$ I can't use ...
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How can I show that $\int_0^\infty \frac{\sin(1/x)}{x}\ dx$ converges? I have that $\sin(x)\leq x$ for $x\geq 0$ so then $\sin(1/x)\leq 1/x$ for $x\geq 0$. It follows then that $\int_1^\infty \frac{\... 3answers 135 views ### How to bound$\int_{0}^{a}{\frac{1-\cos x}{x^2}}$? I was trying to prove $$\left|\int_{0}^{a}{\frac{1-\cos{x}}{x^2}}dx-\frac{\pi}{2}\right|\leq \frac{3}{a}$$ or$\leq \frac{2}{a}$. My work: I would like to use Fubini's theorem to prove it. I ... 2answers 319 views ### Prove that:$ \int_{0}^{\infty} \frac{2 x \sin x+\cos 2x-1}{2 x^2} = 0$How would you prove that? $$\int_{0}^{\infty} \frac{2 x \sin x+\cos 2x-1}{2 x^2} dx= 0$$ I'm looking for a solution at high school level if possible. Thanks. 2answers 637 views ### Integrate:$\int_0^\infty \frac{\sin^2(x)}{x^2}dx$[duplicate] I am trying to integrate$\displaystyle \int_0^\infty \frac{\sin^2(x)}{x^2} dx$by method of contour. I am considering the following contour but I am not being able to. Also I am not sure if it's ... 3answers 231 views ### Using Laplace Transforms to solve$\int_{0}^{\infty}\frac{\sin(x)\sin(x/3)}{x(x/3)}\:dx$So, I've come across the following integral (and it's expansion) many times and in my study so far, Complex Residues have been used to evaluate it. I was hoping to find an alternative approach using ... 2answers 432 views ### How the calculate$\int_0^{+\infty} \frac{\sin^2 x}{x^2} \,\mathrm{d} x$? [duplicate] Just as the title say, consider the integral: $$I=\int_0^{+\infty} \frac{\sin^2 x}{x^2} \,\mathrm{d} x=\frac{1}{2}\int_{-\infty}^{+\infty} \frac{\sin^2 x}{x^2} \,\mathrm{d} x,$$ how to apply the ... 2answers 534 views ### uniform bound for sine integral function Prove that for any$0<a<b$, $$\left|\int_a^b\frac{\sin x}{x}\,dx\right|\le4$$ Here is my approach. I used integration by parts to prove that LHS is bounded by$3$when$a\ge 1$. I will be done ... 3answers 213 views ### Prove that$\int\limits_1^{\infty} \frac{\cos(x)}{x} \, \mathrm{d}x$converges Prove the convergence of $$\int\limits_1^{\infty} \frac{\cos(x)}{x} \, \mathrm{d}x$$ First I thought the integral does not converge because $$\int\limits_1^{\infty} -\frac{1}{x} \,\mathrm{d}x \... 1answer 205 views ### How may be prove that \int_{-\infty}^{+\infty}\sin(\cosh x)\cos(\sinh x)\mathrm dx={\pi\over 2}? Given this integral$$\int_{-\infty}^{+\infty}\sin(\cosh x)\cos(\sinh x)\mathrm dx={\pi\over 2}\tag1$$My try: Recall$$2\sin(A)\cos(B)=\sin(A-B)+\sin(A+B)$$(1) becomes$$\frac12\int_{-\... 1answer 265 views ### How to show that$\int_{0}^{\infty}{\sin(x)\sin(2x)\sin(4x)\cdots\sin(2^kx)\over x^{k+1}}\mathrm dx=2^{0.5(k^2-k-2)}\pi?$$$\int_{0}^{\infty}{\sin(x)\sin(2x)\sin(4x)\cdots\sin(2^kx)\over x^{k+1}}\mathrm dx=2^{0.5(k^2-k-2)}\pi\tag1$$$k\ge0$Experimental using wolfram integrator, let me to conclude the closed form for$(...

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