Evaluate $\int_{0}^{\pi} x\sin\big(\frac{1}{x}\big)\cos x \,dx$

Question

I wonder if an integral of the form $$\int_{0}^{\pi} x\sin\Bigl(\frac{1}{x}\Bigr)\cos x \,dx$$ which can be further simplified to $$\int_{0}^{\pi} \frac{\sin (x^{-1})}{(x^{-1})}\,\cos x\, dx=\cos x\biggl(\int_{0}^{\pi} \frac{\sin (x^{-1})}{(x^{-1})}\,dx\biggr)\Bigg|_{0}^{\pi}+\int_{0}^{\pi}\sin x \bigg(\int \frac{\sin (x^{-1})}{(x^{-1})}dx\bigg)dx$$ will have an explicit form. One simplifies the integral $$\int_{0}^{\pi} \frac{\sin (x^{-1})}{(x^{-1})}dx=\int_{0}^{\pi} \int_{0}^{\infty}e^{-(x^{-1}t)}\sin (x^{-1})\,dx\,dt$$ where we know that $\int_{0}^{\infty}e^{-xt}dt=\frac{1}{x}$. We also know that $$\int_{0}^{\infty}e^{-(xt)}\sin (x)\,dx=\frac{1}{t^2+1}.$$

Any idea on how to compute $$\int_{0}^{\infty}e^{-(x^{-1}t)}\sin (x^{-1})\,dx?$$

Answer 1

With regard to the last definite integral, I do not think that we can compute it because of the singularities.

However, what is "amazing" is that the antiderivative does exist (given by a CAS) $$\int e^{-\frac{t}{x}} \sin \left(\frac{1}{x}\right)\,dx=-\frac{1}{2} i e^{-\frac{t+i}{x}} \left(e^{\frac{t+i}{x}} \left((t-i) \text{Ei}\left(-\frac{t-i}{x}\right)-(t+i) \text{Ei}\left(-\frac{t+i}{x}\right)\right)+\left(-1+e^{\frac{2 i}{x}}\right) x\right)$$

Answer 2

NOT THE SOLUTION:

DAMN, I just got to the of the solution as below and realised I had the bounds wrong the whole time - Thought I would leave in case you're interested!

Here we are addressing the integral:

\begin{equation} I = \int_0^\infty x \sin\left(\frac{1}{x}\right)\cos(x)\:dx \end{equation}

Here we will employ Feynman's Trick and introduce a new function with two variables:

\begin{equation} J(a,b) = \int_0^\infty x \sin\left(\frac{a}{x}\right)\cos(bx)\:dx \end{equation}

We see that $J(1,1) = I$. Now $\frac{\partial}{\partial b}\sin(bx) = x\cos(bx)$ and so,

\begin{equation} J(a,b) = \int_0^\infty x \sin\left(\frac{a}{x}\right)\cos(bx)\:dx = \int_0^\infty \sin\left(\frac{a}{x}\right)\cdot x\cos(bx)\:dx = \int_0^\infty \sin\left(\frac{a}{x}\right)\cdot\frac{\partial}{\partial b}\sin(bx)\:dx \end{equation}

By using Leibniz's Integral Rule we find: \begin{equation} J(a,b) = \int_0^\infty \sin\left(\frac{a}{x}\right)\cdot\frac{\partial}{\partial b}\sin(bx)\:dx = \frac{\partial}{\partial b} \underbrace{\int_0^\infty \sin\left(\frac{a}{x}\right)\sin(bx)\:dx}_{H(a,b)} \end{equation} We now will address $H(a,b)$. To proceed we employ Fubini's Theorem and take the Laplace Transform with respect to $a$: \begin{align} \mathscr{L}_{a \rightarrow s} \left[H(a,b)\right] &= \int_0^\infty \mathscr{L}_{a \rightarrow s} \left[\sin\left(\frac{a}{x}\right)\right]\sin(bx)\:dx = \int_0^\infty \frac{\frac{1}{x}}{s^2 + \left(\frac{1}{x}\right)^2}\cdot \sin(bx)\:dx \\ &= \int_0^\infty \frac{x}{s^2x^2 + 1}\cdot \sin(bx)\:dx \end{align} We now use the same trick from before: $ \frac{\partial}{\partial b} -\cos(bx) = x\sin(bx) $. Thus, \begin{align} \mathscr{L}_{a \rightarrow s} \left[H(a,b)\right] &= -\frac{\partial}{\partial b} \int_0^\infty \frac{\cos(bx)}{s^2x^2 + 1}\:dx \end{align} Or \begin{align} H(a,b) &= -\frac{\partial}{\partial b}\left( \mathscr{L}_{s \rightarrow a}^{-1} \left[\int_0^\infty \frac{\cos(bx)}{s^2x^2 + 1}\:dx\right]\right) \end{align} Or, in terms of $J(a,b)$ \begin{align} J(a,b) &= \frac{\partial}{\partial b} H(a,b) = -\frac{\partial^2}{\partial b^2}\left( \mathscr{L}_{s \rightarrow a}^{-1} \left[\int_0^\infty \frac{\cos(bx)}{s^2x^2 + 1}\:dx\right]\right) \end{align} We now need only address the much simpler integral (below) and then taken it's inverse Laplace Transform: \begin{equation} K(a,b) = \int_0^\infty \frac{\cos(bx)}{s^2x^2 + 1} \end{equation} Here we again use Fibini's Theorem and take the Laplace Transform with respect to $b$: \begin{align} \mathscr{L}_{b \rightarrow w} \left[ K(a,b) \right] &= \int_0^\infty \frac{\mathscr{L}_{b \rightarrow w} \left[ \cos(bx) \right]}{s^2x^2 + 1} = \int_0^\infty \frac{1}{s^2x^2 + 1} \cdot \frac{w}{w^2 + x^2}\:dx \\ &= \frac{w}{s^2w^2 - 1} \int_0^\infty \left[\frac{s^2}{s^2 x^2 + 1} - \frac{1}{w^2 + x^2 } \right]\:dx = \frac{w}{s^2w^2 - 1} \left[s\arctan(sx) - \frac{1}{w}\arctan\left(\frac{x}{w}\right)\right]_0^\infty \\ &= \frac{w}{s^2w^2 - 1}\left[ s\cdot\frac{\pi}{2} - \frac{1}{w} \cdot \frac{\pi}{2} \right] = \frac{sw - 1}{s^2w^2 - 1} \cdot \frac{\pi}{2} = \frac{\pi}{2\left(sw + 1\right)} \end{align} We now take the Inverse Laplace Transform with respect to $w$ to resolve $K(s,b)$: \begin{align} K(s,b) &= \mathscr{L}_{w \rightarrow b}^{-1} \left[ \frac{\pi}{2\left(sw + 1\right)} \right] = \frac{\pi}{2s}e^{-\frac{b}{s}} \end{align} From here we can now address J(a,b) \begin{align} J(a,b) &= \frac{\partial}{\partial b} H(a,b) = -\frac{\partial^2}{\partial b^2}\left( \mathscr{L}_{s \rightarrow a}^{-1} \left[ K(s,b) \right] \right) = -\frac{\partial^2}{\partial b^2}\left( \mathscr{L}_{s \rightarrow a}^{-1} \left[\frac{\pi}{2s}e^{-\frac{b}{s}} \right] \right) = -\frac{\partial^2}{\partial b^2} J_{0}\left(2\sqrt{ab} \right) \\ &= -\frac{\pi}{4b}\left[t\left( J_{0}\left(2\sqrt{ab} \right) + J_{2}\left(2\sqrt{ab} \right) \right) - \frac{ J_{1}\left(2\sqrt{ab} \right)}{\sqrt{b}} \right] \end{align}

Where $J_{\alpha}(x)$ is the Modified Bessel Function. From here we now need only let $a,b = 1$ to solve $I$

\begin{equation} I = J(1,1) = -\frac{\pi}{4\cdot 1}\left[t\left( J_{0}\left(2\sqrt{1\cdot 1} \right) + J_{2}\left(2\sqrt{1\cdot 1} \right) \right) - \frac{ J_{1}\left(2\sqrt{1\cdot 1} \right)}{\sqrt{b}} \right] = -\frac{\pi}{4}\big[J_0(2) + J_2(2) - J_1(2)\big] \end{equation}

Thus, \begin{equation} \int_0^\infty x \sin\left(\frac{1}{x}\right)\cos(x)\:dx = -\frac{\pi}{4}\big[J_0(2) + J_2(2) - J_1(2)\big] \end{equation}

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In fact, we we can go further using this method and have:

\begin{align} J_p(a,b) = \int_0^\infty x^{2p + 1} \sin\left(\frac{a}{x}\right)\cos(bx)\:dx = (-1)^{p + 1} \frac{\partial^{2p + 2}}{\partial b^{2p + 2}} J_{0}\left(2\sqrt{ab} \right) \\ \end{align}