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My teacher showed in class that starting with $\int_{-\infty}^{\infty} e^{inx}g(x)dx$ for an even function $g(x)$ we have $$\int_{0}^{\infty} \cos(nx)g(x)dx=\pi i\sum \operatorname{Res}\left( f(z)\right)$$

and for an odd function $g(x)$: $$\int_{0}^{\infty} \sin(nx)g(x)dx=\pi \sum \operatorname{Res} \left(f(z)\right)$$

with $n>0$, $f(z)=e^{inz}g(z)$ and $\lim\limits_{z\to\infty}g(z)=0$

However, when he tried to do an example, by mistake he forgot to add an $x$ in the numerator and left it as $$\int_{0}^{\infty} \frac{\sin(4x)}{x^2+4}dx$$

and added the $x$ later. While

$$\int_{0}^{\infty} \frac{x\sin(4x)}{x^2+4}dx$$

is not that hard to evaluate, many solutions are found here

I tried to do the first one when I got home, but the methods that I learned until now were unsuccessful. Can I get some help on how to evaluate $I=\int_{0}^{\infty} \frac{\sin(4x)}{x^2+4}dx$ please?

Dylan
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Zacky
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2 Answers2

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If we set $F(a)=\int_{0}^{+\infty}\frac{\sin(ax)}{x^2+1}\,dx$ we have $$ (\mathcal{L} F)(b) = \int_{0}^{+\infty}\int_{0}^{+\infty}\frac{\sin(ax) e^{-ab}}{x^2+1}\,dx \,da =\int_{0}^{+\infty}\frac{x\,dx}{(x^2+b^2)(x^2+1)}=\frac{\log b}{b^2-1}$$ hence $F(a)$ is given by the inverse Laplace transform of $\frac{\log b}{b^2}+\frac{\log b}{b^4}+\frac{\log b}{b^6}+\ldots$, i.e. by $$ \sum_{n\geq 0} \frac{a^{2n+1}}{(2n+1)!}\left(H_{2n+1}-\gamma-\log(a)\right) $$ and $$ \int_{0}^{+\infty}\frac{\sin x}{x^2+1}\,dx = \sum_{n\geq 0}\frac{H_{2n+1}-\gamma}{(2n+1)!} $$ is rather non-elementary (related to the exponential integral) but pretty simple to approximate numerically.

Jack D'Aurizio
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  • I see, and that series is equal to $\frac{Ei(1)-e^2 Ei(-1)}{2e}$ (according to wolfram-alpha) right? – Zacky Apr 05 '18 at 16:12
  • @Zacky: agreed. On the other hand I find the series representation a bit more intuitive. – Jack D'Aurizio Apr 05 '18 at 16:14
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    @Zacky: you may use the integral representation for the harmonic numbers, $$ H_n = \int_{0}^{1}\frac{z^n-1}{z-1},dz, $$ then switch $\sum_{n\geq 0}$ and $\int_{0}^{1}$ to recover an integral clearly related to $\text{Ei}$. Or just write $\sin x$ in terms of the complex exponential and apply a partial fraction decomposition to $\frac{1}{x^2+1}=\frac{1}{(x-i)(x+i)}$. – Jack D'Aurizio Apr 05 '18 at 16:19
  • Thanks, but I better stick with those which are taught in school. – Zacky Apr 05 '18 at 16:24
  • @Zacky: sometimes the things we teach/learn bring us pretty far from our starting point, but that is not a bad thing, on the contrary, sheer curiosity is the core of mathematical investigation. – Jack D'Aurizio Apr 05 '18 at 16:27
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Some hints: What is Fourier transform of $e^{-a|t|}$?

Do you know about connection between reverse Fourier transform of function and the function exact?

What about $\int_{0}^{\infty} e^{-at} \sin(bt)$?

Another attempt : you may consider $F(a) = \int_{0}^{\infty} \frac{\cos(ax)}{x^{2}+4}$. Then find a first derivation with respect to $a$. But you should prove uniform convergence of first integral on $E = [1,\infty)$.

openspace
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  • @Zacky no. You know that $F'(a) = \frac{\pi}{2e^{2a}}$ you can integrate the right part and you will have your $F(a)$ with constant , which easy to find with some substitution. – openspace Apr 05 '18 at 16:02
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    @Zacky that was my bad. By the same idea change $\cos(ax)$ by $\sin(ax)$ – openspace Apr 05 '18 at 16:18