# Integral #5

#### Random Variable

##### Well-known member
MHB Math Helper
The typical way to evaluate $\displaystyle \int_{0}^{\infty} \frac{\cos mx}{a^{2}+x^{2}} \ dx$ is by contour integration.

In a recent thread I evaluated that integral using the Laplace transform.

My challenge question is to use the fact that $$\frac{1}{a^{2}+x^{2}} = 2 \int_{0}^{\infty} t e^{-(a^{2}+x^{2}) t^{2}} \ dt$$ to show that

$$\int_{0}^{\infty} \frac{\cos mx}{a^{2}+x^{2}} \ dx = \frac{\pi}{2a} e^{-am} .$$

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#### Random Variable

##### Well-known member
MHB Math Helper
I'm going to break the problem into three parts.

1) Show that $\displaystyle \int_{0}^{\infty} e^{-t^{2} x^{2}} \cos (mx) \ dx = \frac{\sqrt{\pi}}{2t} e^{-m^{2}/(4t^{2})}$.

2) Show that $\displaystyle \int_{0}^{\infty} e^{-[a^{2}t^{2} + m^{2}/(4t^{2})]} \ dt = \frac{\sqrt{\pi}}{2a} e^{-am}$.

3) Evaluate $\displaystyle \int_{0}^{\infty} \frac{\cos mx}{a^{2}+x^{2}} \ dx$ using the fact that $\displaystyle \int_{0}^{\infty} \frac{\cos mx}{a^{2}+x^{2}} \ dx= 2 \int_{0}^{\infty} \int_{0}^{\infty} \cos(mx) t e^{-(a^{2}+x^{2}) t^{2}} \ dt \ dx$.

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#### Pranav

##### Well-known member
I'm going to break the problem into three parts.

1) Show that $\displaystyle \int_{0}^{\infty} e^{-t^{2} x^{2}} \cos (mx) \ dx = \frac{\sqrt{\pi}}{2t} e^{-m^{2}/(4t^{2})}$.

2) Show that $\displaystyle \int_{0}^{\infty} e^{-[a^{2}t^{2} + m^{2}/(4t^{2})]} \ dt = \frac{\sqrt{\pi}}{2a} e^{-am}$.

3) Evaluate $\displaystyle \int_{0}^{\infty} \frac{\cos mx}{a^{2}+x^{2}} \ dx$ using the fact that $\displaystyle \int_{0}^{\infty} \frac{\cos mx}{a^{2}+x^{2}} \ dx= 2 \int_{0}^{\infty} \int_{0}^{\infty} \cos(mx) t e^{-(a^{2}+x^{2}) t^{2}} \ dt \ dx$.
I have figured out the first two parts, still working on third.

Woops, third one is trivial now.

Problem 1.
Let
$$I(m)=\int_0^{\infty} e^{-t^2x^2}\cos(mx)\,\,dx$$
Differentiate wrt $m$,
$$I'(m)=-\int_0^{\infty} xe^{-t^2x^2}\sin(mx)\,\,dx$$
Integrate by parts and using the fact that $\int xe^{-t^2x^2}dx=-e^{-t^2x^2}/(2t^2)$, we get:
$$I'(m)=-\frac{mI(m)}{2t^2}$$
Solving the differential equation,
$$\ln I(m)=-\frac{m^2}{4t^2}+C$$
Since $I(0)=\frac{\sqrt{\pi}}{2t}$, we have $C=\ln(I(0))$, hence
$$\ln\left(\frac{I(m)}{I(0)}\right)=-\frac{m^2}{4t^2}$$
$$\Rightarrow I(m)=I(0)e^{-m^2/(4t^2)}=\frac{\sqrt{\pi}}{2t}e^{-m^2/(4t^2)}$$

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Problem 2.
Let
$$I(m)=\int_0^{\infty} e^{-(a^2t^2+m^2/(4t^2))}dt=e^{-am}\int_0^{\infty} e^{-(at-m/t)^2}\,dt$$

Differentiate wrt m to get
$$I'(m)=-ae^{-am}\int_0^{\infty} e^{-(at-m/t)^2}\,dt+2e^{-am}\int_0^{\infty} e^{-(at-m/t)^2}\left(a-\frac{m}{t^2}\right)\,dt$$
I rewrite $a-m/t^2$ as $2a-a-m/t^2$ and split the integral as follows:
$$I'(m)=-aI(m)+4ae^{-am}\int_0^{\infty} e^{-(at-m/t)^2}\,dt-2e^{-am}\int_0^{\infty} e^{-(at-m/t)^2}\left(a+\frac{m}{t^2}\right)\,dt$$
$$I'(m)=3aI-2e^{-am}\int_0^{\infty} e^{-(at-m/t)^2}\left(a+\frac{m}{t^2}\right)\,dt$$
Use the substitution $at-m/t=u$ in the integral.
$$\Rightarrow I'(m)=3aI(m)-2e^{-am}\int_{-\infty}^{\infty} e^{-u^2}\,du$$
Since
$$\int_{-\infty}^{\infty} e^{-u^2}\,du=\sqrt{\pi}$$
hence, we have the linear differential equation,
$$I'(m)-3aI(m)=-2e^{-am}\sqrt{\pi}$$
The integrating factor is $e^{-3am}$, hence,
$$I(m)\cdot e^{-3am}=\frac{\sqrt{\pi}e^{-4am}}{2a}+C$$
Since $I(0)=\frac{\sqrt{\pi}}{2a}$, we have $C=0$, hence
$$I(m)=\frac{\sqrt{\pi}}{2a}e^{-am}$$

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Problem 3.
Rewrite the integral as
$$2\int_0^{\infty}te^{-a^2t^2}\int_0^{\infty} e^{-x^2t^2}\cos(mx)\,dx\,dt$$
From Problem 1, we can write:
$$2\int_0^{\infty}te^{-a^2t^2}\cdot\frac{\sqrt{\pi}}{2t}e^{-m^2/(4t^2)}\,dt=\sqrt{\pi}\int_0^{\infty}e^{-a^2t^2-m^2/(4t^2)}\,dt$$
From the result of problem 2, we have
$$\sqrt{\pi}\cdot \frac{\sqrt{\pi}}{2a}e^{-am}=\frac{\pi}{2a}e^{-am}$$

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