Confusion with the equation $\int_0^{\infty}f(x)\sin(\alpha x)\,dx=-\int_0^{\infty}xf(x)\cos(\alpha x)\,dx$

Question

I’m given the question:

If:

$\displaystyle \int_0^{\infty}f(x)\sin(\alpha x)\,dx=-\displaystyle \int_0^{\infty}xf(x)\cos(\alpha x)\,dx$

and $f(1)=1$, then find $f$.

My attempt:

To me, if we add $\int_0^{\infty}xf(x)\cos(\alpha x)\,dx$ to both sides then we get:

$\displaystyle \int_0^{\infty}(f(x)\sin(\alpha x)+xf(x)\cos(\alpha x ))\,dx=0$

Now the above integral is the Fourier integral of $0$.

Therefore:

$f(x)=\dfrac{1}{\pi}\displaystyle \int_{-\infty}^{\infty}0\cdot\sin(\alpha x)\,d\alpha=0$

which is not in line with original assumptions of the question.

Where am I wrong?

Answer 1

Hint: Take $$I(\alpha) =\int_0^\infty f(x) \sin (\alpha x).$$ Your hypothesis is that $I'(\alpha)=-I(\alpha) $, so the solution is $I(\alpha)=Ce^{-\alpha}.$

Answer 2

It seems that there is no integrable function $f$ that is continuous at $1$ satisfying $f(1)=1$ and the integral equation in the OP for all $a$. To wit, set $J(a)=\int^\infty_0 f(x) \sin(xa)\,dx$. Define $g(x)=f(x)$ for $x\geq0$ and $g(x)=-f(-x)$ for $x<0$. Then $g$ is an odd function and $g\in L_1(\mathbb{R})$ iff $f\in L_1([0,\infty))$. Assuming integrability, $$\widehat{g}(a)=\int_\mathbb{R} e^{-i2\pi ax} g(x)\,dx= -i2 J(2\pi a) $$ Since $g$ is an odd function, $\widehat{g}(-a)=-\widehat{g}(a)$ for all $a\in\mathbb{R}$.

If in addition $ x f(x)\in L_1(0,\infty)$, then the conditions of the problem give $$ J'(a)=\partial_a\Big(\int^\infty_0 f(x)\sin (ax)\,dx\Big)=\int^\infty_0 x f(x)\cos(ax)\,dx=-J(a)$$

This is a first order linear ODE with solution $$ J(a)=J(0)e^{-a}$$ since $J(0)=0$, then $J(a)\equiv0$ and so, $\widehat{g}(a)\equiv0$. This means that $g(x)=0$ a.s., which in turn implies that $f(x)=0$ a.s. for $x\geq0$.

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Edit:

If the integrability assumptions are dropped and the integrals involved converge in the sense of improper Riemann integrals, It is possible to find for each $a$ in some interval, a function $f_a:(0,\infty)\rightarrow\mathbb{R}$ such that \begin{align}\int^\infty_0 f_a(x)\sin ax =-\int^\infty_0 x f_a(x)\cos ax \,dx\tag{*}\label{star}\end{align} I will impose sufficient conditions on $f_a$ as we move forward.

• Assuming that $f_a$ is differentiable on $(0,\infty)$, integration by parts on the right-hand-side of \eqref{star} yields \begin{align} -\int^\infty_0 x f_a(x)\cos ax \,dx&=-\frac1a\Big(\lim_{M\rightarrow\infty} M f_a(M)\sin aM- \lim_{m\rightarrow0} m f(m)\sin am\Big)\\ &\qquad\qquad + \frac1a\int^\infty_0(xf'_a(x)+f_a(x))\sin ax\,dx \end{align} Assume that \begin{align} \lim_{M\rightarrow\infty} M f_a(M)\sin aM=0= \lim_{m\rightarrow0} m f_a(m)\sin am \tag{**}\label{dstar} \end{align} By substituting back into \eqref{star} and matching integrands, we further assume that $f_a$ satisfies the first order linear ODE \begin{align} xf'_a(x)+(1-a) f_a(x)=0 \end{align} Solving this ODE with the condition $f_a(1)=1$ yields $$f_a(x)=x^{a-1}$$ This solution is compatible with \eqref{dstar} when $-1<a<0$. Thus, for $a\in (-1,0)$, there is a solution $f_a$ to \eqref{star} with $f_a(1)=1$.

• Had we integrated by parts the left-hand-side of \eqref{star}, we would have obtained \begin{align} \int^\infty_0 f_a(x)\sin ax \,dx &= -\frac1a\Big( \lim_{M\rightarrow\infty}f_a(M)\cos aM - \lim_{m\rightarrow0}f_a(m)\cos am\Big) \\ &\qquad\qquad + \frac1a\int^\infty_0 f'_a(x)\cos ax\,dx\tag{*'}\label{starp} \end{align} By assuming \begin{align} \lim_{M\rightarrow\infty}f_a(M)\cos aM = \lim_{m\rightarrow0}f_a(m)\cos am\in\mathbb{R}, \tag{**'}\label{dstarp} \end{align} substituting back into \eqref{star} and then matching integrands, we would have proposed the ODE $$f'_a(x) + a x f_a(x)=0$$ whose solution $$f_a(x)=e^{\frac{a}{2}} e^{-\frac{a}{2}x^2}$$ satisfies $f_a(1)=1$. However, no $f_a$ of this form is compatible with \eqref{dstarp}.

Answer 3

$\int_0^\infty f(x)\sin(\alpha x) dx=-\int xf(x)\cos(\alpha x) dx$, find $f(x)$.

$u=f(x). du=f'(x). dv=\cos(\alpha x). v=\frac{1}{\alpha} \cos(\alpha x)$

$C=\frac{f(x)}{\alpha}\sin(\alpha x)|_0^\infty-\int_0^\infty\frac{f'(x)\cos(\alpha x)}{\alpha}dx =-\int_0^\infty xf(x)\cos(\alpha x) dx$

$f'(x)=\alpha xf(x)\implies \ln f(x)=\alpha x^2/2+C_1\implies f(x)=C_2e^{(\alpha x^2/2)}$

$f(1)=1\implies C_2=e^{-\alpha /2}$