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I want to prove that

$$\int_{0}^{+\infty} \frac{\sin(\pi x)}{x} dx$$

is convergent;

I know that $\int \frac{\sin(\pi x)}{x}=\frac{1-\cos(\pi x)}{\pi x}+ \int \frac{1-\cos(\pi x)}{\pi x^2} dx$.

Jack D'Aurizio
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Giulia B.
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  • First consider $\int_0^R$. Try splitting integral as $\int_0^1 + \int_1^2 + \cdots +\int_{\lfloor R \rfloor}^R$. – Sungjin Kim Jul 03 '16 at 16:49

2 Answers2

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We have that $\sin(\pi x)$ has a bounded primitive and $\frac{1}{x}$ is decreasing towards zero as $x\to +\infty$, hence the integral is convergent by Dirichlet's test (integral version).

This question deals with the computation of that integral.

Community
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Jack D'Aurizio
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1

Hint. Integrating by parts as you did is Ok, one has for $M>0$, $$ \int_{0}^M \frac{\sin(\pi x)}{x} dx=\left.\frac{1-\cos(\pi x)}{\pi x}\right]_0^M+ \int_0^M \frac{1-\cos(\pi x)}{\pi x^2} dx $$ then letting $M \to \infty$ gives $$ \left|\int_{0}^\infty \frac{\sin(\pi x)}{x} dx\right|=\left|\lim_{M \to \infty}\int_{0}^M \frac{\sin(\pi x)}{x} dx\right|\le \int_0^1 \left|\frac{1-\cos(\pi x)}{\pi x^2}\right| dx+\int_1^\infty \frac{|2|}{\pi x^2} dx<\infty $$

Olivier Oloa
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