 Let us consider a charge q smeared out over a thin line. We can describe this spread-out charge with a one-dimensional charge density row. Since it is one-dimensional, it depends only on one spatial coordinate x. The one-dimensional charge density is charge per length. That means we can calculate the charge q if we integrate charge density over this line on which the charge is smeared out. But what if we don't have a smeared-out charge q, but a charge concentrated in a single point, so if we have a point charge? Let's assume for the moment that q is located at the origin that is at x equals 0. What properties must the charge density have so that it describes a point charge? The charge density must fulfill two properties if it describes a single point charge. First property is, it must be 0 at every position x, except at the position where the point charge is located. That is at x equals 0. And second property is, the line integral over the charge density must of course yield the value q of the point charge if we integrate between two points a and b where the point charge is located. On the other hand, if we integrate over a region where the charge q is not present, then the integral should of course yield 0. Let's always include the charge q, for this we choose the integration limits from minus infinity to plus infinity. If we normalize the charge to the value q equals 1 and consider these two properties, then we write this density with a Greek delta and call it Dirac's delta function. Even if the name may suggest, the delta function is mathematically not a function, but another mathematical object which can be understood for example as a so-called distribution or as a Dirac measure. In this video, we are not interested in a mathematically clean definition. We just want to know how to work with the delta function. The delta function is illustrated with an arrow located at the position of the point charge. The height of the arrow is usually chosen to represent the value of the integral, in this case 1. Now let's consider an integral of the delta function together with another but well-behaved function f of x. Such an integral is very easy to calculate, because you know that the delta function is 0 everywhere except at the point x equals 0. Thus, the product is also 0 everywhere except at the point x equals 0. Only the function value f of 0 remains. Since f of 0 no longer depends on x, we can place it in front of the integral. The integral over the delta function is 1. So we know what the delta function does in the integral when it is multiplied by a function f of x. It picks the value of the function at the position x equals 0. We can, of course, move the charge to another location on the x-axis, for example to the positive location x equals 0. Then the delta function must be 0 everywhere except at this new position. We change the argument of delta to x minus x0, minus x0 because we have moved the delta function to the positive direction. The integral over the delta function is 1. We have only shifted the delta function to x0, therefore the result of the integral is the same as with the delta of x. What if the shifted delta function appears in the integral with another function f of x? Delta is 0 everywhere except at the point x0. To explain it graphically, the shifted delta in the integral picks the function value of f at the position where the delta function is currently located. You can prove this by substituting x minus x0 with y. The x minus x0 in the delta function becomes y. The x in the function becomes y plus x0. The derivative of y with respect to x is 1. Thus dx changes to dy. As we know, delta of y picks the value of the function at the point y equals 0. Therefore the integral yields the value f of x0. Next, let's look at what happens when we have minus x in the delta. Let's make a substitution, y is equal to minus x. Delta of minus x becomes delta of y. And f of x becomes f of minus y. The derivative of y with respect to x is minus 1. Thus we replace the dx with minus 1 times dy. The lower integration limit becomes plus infinity. The upper one becomes minus infinity. To make the integration again from negative to positive, we reverse the integration limits and thus get a minus sign in front of the integral. We already know what delta of y does in the integral. It picks the value of f at the point y equals 0. So this integral yields the value f of 0. But we also know that the integral of the delta function without the minus sign also yields f of 0. So it makes no difference whether we use delta of x or delta of minus x in the integral. The delta function is symmetric. What happens if we scale the x in the delta by a factor k? If k is negative, then we can write it as a negative magnitude and omit the minus sign in front of it because the delta function is symmetric. Let's make a substitution, y is equal to kx. This turns delta of kx to delta of y. And f of x becomes f of 1 over k times y. The derivative of y with respect to x is k. Thus we substitute dx with 1 over k times dy, placing 1 over k in front of the integral. As you know, this integral picks the value of f at the point y equals 0. Thus we get 1 over the magnitude of k times f of 0. Here you must note that k should not be 0, so that you do not divide by 0. Let's make a few examples. Let's look at this function in the integral with the delta function. The result of this integral is f of 0. Insert x equals 0, sign of 0 is 0, e to the power of 0 is 1, so the integral is equal to minus 2. This is how easy it is to calculate complicated integrals when a delta function occurs in the integrand. Here we integrate x squared with the delta of x minus 3. In this case, however, not from minus infinity to plus infinity, but from minus 1 to 1. We must therefore first ensure that the delta function is inside the integration limits. The delta function is here shifted to the position x equals 3 and is therefore outside the upper integration limit. The integral is thus 0. Here we integrate the cosine of x multiplied with delta of x plus pi from minus 6 to 0. The delta function is at x equals minus pi and is within the integration limits. So we have to evaluate cosine at the position minus pi. This yields the value of the integral minus 1. So far, we have considered only one-dimensional delta function. The charges over other point objects are usually located in a three-dimensional space. If the unit point charge is located at the origin, then we can describe its charge density with the product of three delta functions, delta of x times delta of y times delta of z. This product is non-zero only if the charge is at the position 0, 0, 0 at the origin. If we move the charge to the positive direction to the position x0, y0, z0, then we have to adjust the arguments of the delta functions as in the one-dimensional case. In order not to have to write three deltas, we combine them to one delta with a 3 on top and in the argument we write the position vector r and for the displacement we write, for example, the vector r0. The line integral with one delta function becomes a volume integral with three delta functions and yields one if we integrate over a volume v that includes the position of the three-dimensional delta function. So that's it. If you want to support me, then please become a member of the YouTube channel. With this in mind, bye and see you next time.