Solving $\int_{-\infty}^{\infty} \frac{x\sin(3x)}{x^4+1}\,dx $ without complex integration

Question

I am looking for a way to solve :

$$\int_{-\infty}^{\infty} \frac{x\sin(3x)}{x^4+1}\,dx $$

without making use of complex integration.

What I tried was making use of integration by parts, but that didn't reach any conclusive result. (i.e. I integrated $\sin(3x)$ and differentiated the rest)

I can't see a clear starting point to solve this question. Any help appreciated.

This problem is posted by Vilakshan Gupta on Brilliant.

Answer 1

I only write the key step for central issue, let for a>0 (this makes problem easy to deal without abs function) $$ f(a) = \int_{0}^{\infty} \frac{\sin(ax)}{x(x^4+1)} \,\mathrm{d}x $$ then you have ODE $$ f^{(4)}(a) + f(a) = \int_{0}^{\infty} \frac{\sin(ax)}{x} \,\mathrm{d}x =\frac{\pi}{2} $$ with boundary value $f(0)=f^{(2)}(0)=0$, $f^{(1)}(0)=\pi/2\sqrt2$, $f^{(3)}(0)=-\pi/2\sqrt2$ solved as $$ f(a) = \frac{\pi}{2}\left(1-e^{-a/\sqrt2}\cos\left(\tfrac{a}{\sqrt2}\right)\right) $$ then you just need to find $-f^{(2)}(3)$ to obtain the final answer.

Answer 2

@Nanayajitzuki has given you a very nice solution to this problem using Leibniz' integral rule (or Feynman trick if you are a Physicist!) Really, this integral is ridiculously difficult without Complex Analysis. It's doable... but any real method is going to be highly non-trivial.

For another solution, we could again parameterize the integral and use the Laplace Transform. Technically, formally inverting the Laplace Transform at the end would require Complex Integration - however, we can figure out the inverses of many standard functions very easily by using properties of the Laplace Transform along with knowing the Laplace Transform of standard functions.

Define

$${I(t)=\int_{0}^{\infty}\frac{x\sin(3xt)}{x^4 + 1}dx}$$

So we have

$${\Rightarrow \int_{0}^{\infty}\int_{0}^{\infty}\frac{x\sin(3xt)}{x^4 + 1}e^{-st}dxdt=\mathcal{L}\{I(t)\}(s)}$$

Interchanging the integrals gives us

$${\Rightarrow \int_{0}^{\infty}\int_{0}^{\infty}\left(\frac{x}{x^4 + 1}\right)\sin(3xt)e^{-st}dtdx=\int_{0}^{\infty}\left(\frac{x}{x^4 + 1}\right)\left(\mathcal{L}\{\sin(3xt)\}\right)dx}$$

We can use the well known formula for the Laplace Transform of ${\sin(ax)}$:

$${\Rightarrow \mathcal{L}\{I(t)\}=\int_{0}^{\infty}\frac{x}{x^4 + 1}\left(\frac{3x}{9x^2 + s^2}\right)dx=\int_{0}^{\infty}\frac{3x^2}{(9x^2 + s^2)(x^4 + 1)}dx}$$

This is a ridiculous integral to evaluate (you can look at Wolfram alpha as to how big the anti-derivative answer actually is) - but again, completely doable using real methods. If you do evaluate it though, the end result is

$${\Rightarrow \frac{3\pi}{2\sqrt{2}\left(s^2 + 3\sqrt{2}s + 9\right)}=\mathcal{L}\{I(t)\}}$$

(Writing down the complete way of solving that integral would be huge - but essentially you need to do partial fractions and substitutions a bunch. It is possible to evaluate it using real methods though - simply because it's a monster integral, I won't write the steps for it here. But you can give it a go if you are feeling brave :P It shouldn't be hard - just tedious).

Now the last part is to invert the Laplace Transform! To do this, notice that

$${\mathcal{L}\{e^{-at}\sin(bt)\}=\frac{b}{(a+s)^2 + b^2}}$$

(you can also see it as part of a Inverse Laplace table - see: https://tutorial.math.lamar.edu/classes/de/laplace_table.aspx)

And notice that

$${\mathcal{L}\{I(t)\}=\frac{3\pi}{2\sqrt{2}}\left(\frac{\sqrt{2}}{3}\right)\left(\frac{\frac{3}{\sqrt{2}}}{\left(s+\frac{3\sqrt{2}}{2}\right)^2 + \left(\frac{3}{\sqrt{2}}\right)^2}\right)=\frac{\pi}{2}\left(\frac{\frac{3}{\sqrt{2}}}{\left(s+\frac{3\sqrt{2}}{2}\right)^2 + \left(\frac{3}{\sqrt{2}}\right)^2}\right)}$$

And so we get

$${I(t)=\frac{\pi}{2}e^{-\frac{3\sqrt{2}}{2}t}\sin\left(\frac{3}{\sqrt{2}}t\right)}$$

Giving us overall

$${\Rightarrow \int_{-\infty}^{\infty}\frac{x\sin(3x)}{x^4+1}dx=2I(1)=\pi e^{-\frac{3}{\sqrt{2}}}\sin\left(\frac{3}{\sqrt{2}}\right)}$$

Answer 3

Too long for a comment

The more general problem $$I=\int_{-\infty}^\infty \frac {x^m\,\sin(px)}{P_{2n}(x)} \,dx \qquad\text{where}\qquad m <2n\qquad p > 0$$ is not too bad if you feel conformatble with the manipulation of complex numbers. For sure, the condition is that $P_{2n}(x)$ has no real root ($P_{2n}(x)$ has $n$ pairs of complex conjugate roots).

So, let us write $$P_{2n}(x)=\sum_{k=0}^{2n} a_k\,x^k=a_{2n} \prod_{i=0}^{2n}(x-r_i)$$ Now partial fraction decomposition leads to $$I=\frac 1 {a_{2n}}\sum_{i=0}^{2n} b_i \int_{-\infty}^\infty \frac {\sin(px)}{x-r_i}\,dx$$ The antiderivative involves sine and cosine integrals but at the end $$J_r=\int_{-\infty}^\infty \frac {\sin(px)}{x-r}\,dx=\pi\, e^{i p r}$$

Edit

For the specific case of $$I=\int_{-\infty}^\infty \frac {x\,\sin(px)}{x^4+1} \,dx \qquad\text{where}\qquad p > 0$$ we then have $$\frac {x}{x^4+1}=\frac i 4\left(\frac{1}{ x+\frac{1-i}{\sqrt{2}}}+\frac{1}{ x-\frac{1-i}{\sqrt{2}}}-\frac{1}{ x+\frac{1+i}{\sqrt{2}}}+\frac{1}{ x-\frac{1+i}{\sqrt{2}}}\right)$$ $$I=\frac i 4\left(-4 i \pi e^{-\frac{p}{\sqrt{2}}} \sin \left(\frac{p}{\sqrt{2}}\right)\right)=\pi \, e^{-\frac{p}{\sqrt{2}}}\, \sin \left(\frac{p}{\sqrt{2}}\right)$$