MHB Definite Integral ∫xe^(ax)cos(x)dx

[MarkFL]

Evaluate the following:

$$I=\int_0^{\infty} xe^{ax}\cos(x)\,dx$$ where $a<0$

 

[lfdahl]

My attempt:

Spoiler

My result:

\[\int_{0}^{\infty}xe^{ax}\cos xdx = \frac{a^2-1}{(a^2+1)^2}, \;\;\; a < 0.\]

The solution is obtained by integration by parts several times:

\[I = \int_{0}^{\infty}xe^{ax}\cos xdx = \underbrace{\left [ xe^{ax}\sin x \right ]_0^\infty}_{0} - \int_{0}^{\infty}e^{ax}(1+ax)\sin xdx \\\\ = \underbrace{\int_{0}^{\infty}e^{ax}(-\sin x)dx}_{=I_1}+\underbrace{a\int_{0}^{\infty}xe^{ax}(-\sin x)dx}_{=I_2}\]

Solve the first of the two integrals, $I_1$, in the sum:

\[I_1 = \int_{0}^{\infty}e^{ax}(-\sin x)dx = \underbrace{\left [ e^{ax}\cos x \right ]_0^\infty}_{=-1}-a\int_{0}^{\infty}e^{ax} \cos xdx \\\\ =-1-a\left ( \underbrace{\left [ e^{ax}\sin x \right ]_0^\infty}_{0}-a\int_{0}^{\infty}e^{ax}\sin xdx \right ) = -1 -a^2\int_{0}^{\infty}e^{ax}(-\sin x)dx = -1 -a^2I_1\\\\ \Rightarrow (a^2+1)I_1 = -1 \Rightarrow I_1 = -\frac{1}{a^2+1}\]

The second integral, $I_2$:

\[I_2 = a\int_{0}^{\infty}xe^{ax}(-\sin x)dx = a\left (\underbrace{\left [ xe^{ax}\cos x \right ]_0^\infty}_0- \int_{0}^{\infty}e^{ax}(1+ax)\cos xdx \right ) \\\\ = a\left ( -\int_{0}^{\infty}e^{ax}\cos xdx -a\int_{0}^{\infty}xe^{ax}\cos xdx\right ) \\\\ =-a\left ( \underbrace{\left [ e^{ax}\sin x \right ]_0^\infty}_0+a\int_{0}^{\infty}e^{ax}(-\sin x)dx \right )-a^2I \\\\ = -a^2I_1-a^2I\]Finally, we get:

\[I = I_1 + I_2 = -\frac{1}{a^2+1} +\frac{a^2}{a^2+1} -a^2I \\\\ \Rightarrow I = \frac{a^2-1}{(a^2+1)^2}\]

 

[Fantini]

My attempt:

Spoiler

Let $$I(a) = \int_0^{\infty} e^{ax} \cos(x) \, dx$$ for $$a < 0$$. Then

$$\begin{aligned} \dfrac{d}{da} I(a) & = \dfrac{d}{da} \int_0^{\infty} e^{ax} \cos(x) \, dx \\ & = \int_0^{\infty} \dfrac{\partial}{\partial a} (e^{ax} \cos(x) ) \, dx \\ & = \int_0^{\infty} x e^{ax} \cos(x) \, dx. \end{aligned}$$

Thus, if we compute $$I(a)$$ in terms of $$a$$ the result follows by differentiation. Let $$A = \int_0^{\infty} e^{ax} \cdot e^{ix} \, dx$$. We have

$$\begin{aligned} A & = \int_0^{\infty} e^{(a+i)x} \, dx \\ & = \dfrac{e^{(a+i)x}}{a+i} \bigg\vert_0^{\infty} \\ & = - \dfrac{1}{a+i} \\ & = - \dfrac{a}{a^2+1} + \dfrac{i}{a^2+1}. \end{aligned}$$

Since $$e^{ix} = \cos(x) + i \sin(x)$$ we obtain

$$I(a) = \operatorname{Re} \left( \int_0^{\infty} e^{ax} \cdot e^{ix} \, dx \right) = - \dfrac{a}{a^2+1}.$$

Thus

$$\dfrac{d}{da} I(a) = \int_0^{\infty} x e^{ax} \cos(x) \, dx = \dfrac{d}{da} \left( - \dfrac{a}{a^2+1} \right) = \dfrac{a^2 -1}{(a^2+1)^2}.$$

 

[Theia]

My attempt:

Spoiler

Let $$I = \int_0^{\infty}xe^{ax}\cos x dx = - \int_0^{\infty} -x \cdot e^{ax}\cos x dx$$.

Now the last part is the Laplace transform of cosine:

$$\int_0^{\infty}e^{ax}\cos x dx = \mathcal{L}\{\cos x\} = \frac{a}{a² + 1}$$.

Using the properties of Laplace transform (function times variable) we further obtain

$$I = - \frac{d}{da} \frac{a}{a² + 1} = \frac{a^2 - 1}{(a^2 + 1)^2}$$.

 

[MarkFL]

Thanks everyone for all the varied methods! (Yes)

Here's the method I chose:

Spoiler

We are given to evaluate:

$$I=\int_0^{\infty} xe^{ax}\cos(x)\,dx$$ where $a<0$

I would write:

$$v(x)=\int xe^{ax}\cos(x)\,dx$$

So that we have the first-order linear inhomogeneous ODE:

$$\d{v}{x}=xe^{ax}\cos(x)$$

Since the homogeneous solution $v_h$ is a constant, we need only find the particular solution, which will have the form:

$$v_p(x)=e^{ax}\left((Ax+B)\cos(x)+(Cx+D)\sin(x)\right)$$

Differentiating w.r.t $x$, and substituting into our ODE, we obtain:

$$\d{v_p}{x}=e^{ax}\left(((aA+C)x+aB+A+D)\cos(x)+((aC-A)x+aD-B+C)\sin(x)\right)=e^{ax}\left((1x+0)\cos(x)+(0x+0)\sin(x)\right)$$

Equating coefficients gives rise to the system:

$$aA+C=1$$

$$aB+A+D=0$$

$$aC-A=0$$

$$aD-B+C=0$$

Solving this system, we obtain:

$$(A,B,C,D)=\left(\frac{a}{a^2+1},\frac{1-a^2}{\left(a^2+1\right)^2},\frac{1}{a^2+1},-\frac{2a}{\left(a^2+1\right)^2}\right)$$

And so, our particular solution may be written:

$$v_p(x)=\frac{e^{ax}}{\left(a^2+1\right)^2}\left((a\left(a^2+1\right)x+1-a^2)\cos(x)+(\left(a^2+1\right)x-2a)\sin(x)\right)$$

Hence, the definite integral in question may be written:

$$I=\frac{1}{\left(a^2+1\right)^2}\lim_{t\to\infty}\left(\left(e^{at}\left((a\left(a^2+1\right)t+1-a^2)\cos(t)+(\left(a^2+1\right)t-2a)\sin(t)\right)\right)-\left(e^{a(0)}\left((a\left(a^2+1\right)(0)+1-a^2)\cos(0)+(\left(a^2+1\right)(0)-2a)\sin(0)\right)\right)\right)$$

$$I=\frac{1}{\left(a^2+1\right)^2}\lim_{t\to\infty}\left(\left(e^{at}\left((a\left(a^2+1\right)t+1-a^2)\cos(t)+(\left(a^2+1\right)t-2a)\sin(t)\right)\right)-\left((1-a^2)\right)\right)=\frac{a^2-1}{\left(a^2+1\right)^2}$$