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Properties of the Dirac Delta Function 📂Functions

Properties of the Dirac Delta Function

Properties

δ(−x)=δ(x) \begin{equation} \delta (-x) =\delta (x) \end{equation}

δ(kx)=1∣k∣δ(x) \begin{equation} \delta (kx)= \frac{1}{|k|} \delta (x) \end{equation}

Proof

Proof of (1)(1)

Substituting ∫−∞∞f(x)δ(−x)dx\int_{-\infty }^ { \infty } f(x) \delta (-x) dx with −x≡y-x \equiv y gets us x=−yx=-y and dx=−dydx=-dy,

∫−∞∞f(x)δ(−x)dx= −∫∞−∞f(−y)δ(y)dy= ∫−∞∞f(−y)δ(y)dy= f(0)= ∫−∞∞f(x)δ(x)dx \begin{align*} \int_{-\infty } ^{ \infty } f(x) \delta (-x) dx =&\ -\int_{ \infty }^{-\infty} f(-y) \delta (y) dy \\ =&\ \int_{-\infty } ^{\infty } f(-y) \delta (y) dy \\ =&\ f(0) \\ =&\ \int_{-\infty } ^{\infty } f(x) \delta (x) dx \end{align*}

∫−∞∞f(x)δ(−x)dx=∫−∞∞f(x)δ(x)dx \int_{-\infty }^ { \infty } f(x) {\color{blue}\delta (-x)} dx = \int_{-\infty } ^{\infty } f(x) {\color{blue}\delta (x)} dx

Therefore,

δ(−x)=δ(x) \delta (-x) = \delta (x)

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Proof of (2)(2)

Substituting ∫−∞∞f(x)δ(kx)dx\int_{-\infty }^ { \infty } f(x) \delta (kx) dx with kx≡ykx \equiv y gets us x=1kyx=\frac{1}{k}y and dx=1ky dx=\frac{1}{k}y,

∫−∞∞f(x)δ(kx)dx=1k∫−∞∞f(1ky)δ(y)dy \int_{-\infty }^ { \infty } f(x) \delta (kx) dx = \frac{1}{k} \int_{-\infty }^ { \infty } f(\frac{1}{k}y) \delta (y) dy

However, this result is under the condition k>0k>0. When it’s k<0k<0, the integration interval must be considered.

When it’s k<0k<0 and substituting with kx≡ykx \equiv y, it gets us x=1kyx=\frac{1}{k}y, dx=1kydx=\frac{1}{k}y and hence (x→∞ , y→−∞)(x \rightarrow \infty \ ,\ y \rightarrow -\infty), (x→−∞ , y→∞)( x \rightarrow -\infty \ ,\ y \rightarrow \infty),

∫−∞∞f(x)δ(kx)dx= 1k∫∞−∞f(1ky)δ(y)dy= (1−k)(−∫∞−∞f(1ky)δ(y)dy)= 1∣k∣∫−∞∞f(1ky)δ(y)dy \begin{align*} \int_{-\infty } ^{ \infty } f(x) \delta (kx) dx =&\ \frac{1}{k} \int_{\infty }^ { -\infty } f(\frac{1}{k}y) \delta (y) dy \\ =&\ \left( \frac{1}{-k} \right) \left(-\int_{\infty }^ { -\infty } f(\frac{1}{k}y) \delta (y) dy \right) \\ =&\ \frac{1}{|k|}\int_{-\infty }^ { \infty } f(\frac{1}{k}y) \delta (y) dy \end{align*}

Therefore, considering for all real numbers kk,

∫−∞∞f(x)δ(kx)dx= 1∣k∣∫−∞∞f(1ky)δ(y)dy= 1∣k∣f(0)= 1∣k∣∫−∞∞f(x)δ(x)dx= ∫−∞∞f(x)1∣k∣δ(x)dx \begin{align*} \int_{-\infty }^ { \infty } f(x) \delta (kx) dx =&\ \frac{1}{|k|}\int_{-\infty }^ { \infty } f(\frac{1}{k}y) \delta (y) dy \\ =&\ \frac{1}{|k|} f(0) \\ =&\ \frac{1}{|k|}\int_{-\infty }^ { \infty } f(x) \delta (x) dx \\ =&\ \int_{-\infty }^ { \infty } f(x) \frac{1}{|k|}\delta (x) dx \end{align*}

∫−∞∞f(x)δ(kx)dx=∫−∞∞f(x)1∣k∣δ(x)dx \int_{-\infty }^ { \infty } f(x) {\color{blue} \delta (kx)} dx=\int_{-\infty }^ { \infty } f(x) {\color{blue} \frac{1}{|k|}\delta (x) } dx

∴δ(kx)=1∣k∣δ(x) \therefore \delta (kx) = \frac{1}{|k|}\delta (x)

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