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 $t=0$.