Why do I get two different values for an integral?

[Mr Davis 97]

Suppose ##t \ge 0##. Let ##\displaystyle I(t) = \int_{-\infty}^{\infty}\frac{x \sin (tx)}{x^2+1}~\text{dx}##. Call this form 1.

Note that we can also write the integral as
$$
\begin{align*}
I(t) &= \int_{-\infty}^{\infty}\frac{x \sin (tx)}{x^2+1}~\text{dx} \\
&= \int_{-\infty}^{\infty}\frac{(x^2+1-1) \sin (tx)}{x(x^2+1)}~\text{dx} \\
&= \int_{-\infty}^{\infty}\frac{\sin (tx)}{x}~\text{dx} - \int_{-\infty}^{\infty}\frac{\sin (tx)}{x(x^2+1)}~\text{dx} \\
&= \pi - \int_{-\infty}^{\infty}\frac{\sin (tx)}{x(x^2+1)}~\text{dx} \\
\end{align*}
$$
Call this form 2.

Now, in form 1, it seems that ##I(0) = 0##, while in form 2 it seems that ##I(0) = \pi##. Why am I getting two different values, and what am I doing wrong?

 

[stevendaryl]

Suppose ##t \ge 0##. Let ##\displaystyle I(t) = \int_{-\infty}^{\infty}\frac{x \sin (tx)}{x^2+1}~\text{dx}##. Call this form 1.

Note that we can also write the integral as
$$
\begin{align*}
I(t) &= \int_{-\infty}^{\infty}\frac{x \sin (tx)}{x^2+1}~\text{dx} \\
&= \int_{-\infty}^{\infty}\frac{(x^2+1-1) \sin (tx)}{x(x^2+1)}~\text{dx} \\
&= \int_{-\infty}^{\infty}\frac{\sin (tx)}{x}~\text{dx} - \int_{-\infty}^{\infty}\frac{\sin (tx)}{x(x^2+1)}~\text{dx} \\
&= \pi - \int_{-\infty}^{\infty}\frac{\sin (tx)}{x(x^2+1)}~\text{dx} \\
\end{align*}
$$
Call this form 2.

Now, in form 1, it seems that ##I(0) = 0##, while in form 2 it seems that ##I(0) = \pi##. Why am I getting two different values, and what am I doing wrong?

The culprit is the integral ##\int_{-\infty}^{+\infty} \dfrac{sin(tx)}{x} dx##. Call that integral ##F(t)##. If ##t > 0##, then you can perform a variable change ##x \rightarrow tx## to transform the integral into: ##\int_{-\infty}^{+\infty} \dfrac{sin(x)}{x} dx = \pi##. If ##t < 0##, then since ##sin(tx) = - sin(-tx)##, it should be clear that ##F(t) = -\pi##. When ##t=0##, ##F(t) = 0##.

 

[fresh_42]

Your initial integral is well-defined, but then you extended it by ##\frac{x}{x}## and integrated over the now singularity at ##x=0##.

 

[Fred Wright]

I suggest that you extend the integral to the complex plane and use the residue theorem. I get:$$ I(t)= i\pi sinh(t) $$
I don't see how to solve this integral w/o complex analysis.

 

[mathman]

Suppose ##t \ge 0##. Let ##\displaystyle I(t) = \int_{-\infty}^{\infty}\frac{x \sin (tx)}{x^2+1}~\text{dx}##. Call this form 1.

Note that we can also write the integral as
$$
\begin{align*}
I(t) &= \int_{-\infty}^{\infty}\frac{x \sin (tx)}{x^2+1}~\text{dx} \\
&= \int_{-\infty}^{\infty}\frac{(x^2+1-1) \sin (tx)}{x(x^2+1)}~\text{dx} \\
&= \int_{-\infty}^{\infty}\frac{\sin (tx)}{x}~\text{dx} - \int_{-\infty}^{\infty}\frac{\sin (tx)}{x(x^2+1)}~\text{dx} \\
&= \pi - \int_{-\infty}^{\infty}\frac{\sin (tx)}{x(x^2+1)}~\text{dx} \\
\end{align*}
$$
Call this form 2.

Now, in form 1, it seems that ##I(0) = 0##, while in form 2 it seems that ##I(0) = \pi##. Why am I getting two different values, and what am I doing wrong?

Why is ##I(0)=0## in form 1? The integrand is an even function of ##x##.

 

[stevendaryl]

Why is ##I(0)=0## in form 1? The integrand is an even function of ##x##.

Because ##sin(tx) = 0## when ##t=0##.