I have to calculate this dirac integral: $I=\int_{-1}^{0}\delta(4t+1)dt$

Question

How can I evaluate the following integral?

$$ I=\int_{-1}^{0}\delta(4t+1) \, dt $$

Here is my working out so far:

\begin{align*} I &=\int_{-1}^{0} \delta(4t+1) \, dt \\ &=\int \delta \cdot (4t+1) \, dt \\ &=\int \delta \cdot 4\cdot \left(t+\tfrac{1}{4}\right) \, dt \end{align*}

Noting $u=t+\frac{1}{4}$, $du = dt$, we have:

\begin{align*} 4\int \delta\cdot u\cdot du &= 4\left[\left(\delta \cdot \frac{u^{2}}{2}\right) - \left(\delta\cdot\frac{u^{2}}{2}\right)\right] \\ &= 4\left[\left(\delta \cdot \frac{0^{2}}{2}\right) - \left(\delta\cdot\frac{-1^{2}}{2}\right)\right] \\ \end{align*}

Answer 1

Well, it is not hard to show that:

$$\int\delta\left(\text{a}x+\text{b}\right)\space\text{d}x=\frac{\theta\left(\text{a}x+\text{b}\right)}{\text{a}}+\text{C}\tag1$$

Where $\theta(\cdot)$ is the Heaviside step function.

So, in your case you will get:

$$\int_{-1}^0\delta\left(4x+1\right)\space\text{d}x=\frac{\theta\left(4\cdot0+1\right)}{4}-\frac{\theta\left(4\cdot(-1)+1\right)}{4}=$$ $$\frac{\theta\left(1\right)}{4}-\frac{\theta\left(-3\right)}{4}=\frac{1}{4}-\frac{0}{4}=\frac{1}{4}\tag2$$

Answer 2

Hint: $$\int_{-\infty}^{+\infty}f(x)\delta(x)dx = f(0)$$ Use the change of variables and select appropriate $f(x)$.