# How Does a Definite Integral Yield an Imaginary Number?

• paulfr
In summary, an imaginary result of an integral is a complex number that contains both real and imaginary components. It differs from a real result in that it cannot be plotted on a traditional x-y axis and must be represented in the complex plane. While it may have physical significance in certain areas of science, such as quantum mechanics and electrical engineering, it must be interpreted by considering both components separately. It can also be converted into a real result by taking the absolute value of the complex number, but it is important to consider the context of the problem before doing so.
paulfr
Why and how does this definite integral result in an imaginary solution ?

At wolframalpha ...
definite integral 1 / [e^x arcsin x] dx from 1 to 10 = 0.156 + .09i

Area under such a function should be positive or negative but
how does it become imaginary ?

Thanks

By usual definition, arcsin(x) is not defined for x > 1.

awkward said:
By usual definition, arcsin(x) is not defined for x > 1.

It should be clarified that arcsin(x) is not defined for |x| > 1 if x is a real number. Typically, it is defined everywhere on the imaginary axis except for the branch cuts (which are usually taken to be from (-inf,1]U[1,inf).

paulfr said:
Why and how does this definite integral result in an imaginary solution ?

At wolframalpha ...
definite integral 1 / [e^x arcsin x] dx from 1 to 10 = 0.156 + .09i

Area under such a function should be positive or negative but
how does it become imaginary ?

Thanks

How:

\begin{aligned} \frac{1}{e^z \arcsin(z)}&=\frac{1}{e^z\left(-i\log[iz+\sqrt{1-z^2}]\right)}\\ &=\frac{1}{e^z\left(-i\log[iz+i\sqrt{z^2-1}]\right)},\quad z>1\\ &=\frac{1}{e^z\left(-i\log[i(z+\sqrt{z^2-1})]\right)}\\ &=\frac{1}{e^z\left(-i(\ln(z+\sqrt{z^2-1})+\pi/2 i)\right)}\\ &=\frac{1}{e^z(\pi/2-i\ln(z+\sqrt{z^2-1}))}\\ &=\frac{\pi/2 e^z+i\ln(z+\sqrt{z^2-1})}{(\pi/2 e^z)^2+e^{2z}\ln^2(z+\sqrt{z^2-1})} \end{aligned}

Now recall:

$$\int_c f(z)dz=\int udx-vdy+i\int udy+vdx$$

and let's just look at the real part of that and since we're on the real-axis during the integration from 1 to 10, then that the real part of that integral is:

$$\int_1^{10} \frac{\pi/2 e^x}{(\pi/2 e^x)^2+e^{2x}\ln^2(x+\sqrt{x^2-1})}\approx 0.158$$

The imaginary part follows similarly. Mathematica as well as the steps above are using the principal log.

Last edited:
Thank you very much

But why does the first line show that

$$arcsin z = \ {\left(-i\log[iz+\sqrt{1-z^2}]\right)}\\$$Where does that come from ?

Last edited:
That comes from the inverse of the complex sine function:

$$w=\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$$

If you now solve for z in terms of w, you get that expression, but just use the variable "z" instead of w.

## 1. What is an imaginary result of an integral?

An imaginary result of an integral is a complex number that contains both a real and imaginary component. It is often denoted by the letter "i" and is used to represent the square root of -1. This result is often obtained when integrating a function that contains imaginary numbers or when using complex integration techniques.

## 2. How is an imaginary result of an integral different from a real result?

An imaginary result differs from a real result in that it contains an imaginary component, whereas a real result only contains a real component. This means that an imaginary result cannot be plotted on a traditional x-y axis and must be represented in the complex plane.

## 3. Can an imaginary result of an integral have physical significance?

Yes, an imaginary result of an integral can have physical significance in certain areas of science, such as quantum mechanics and electrical engineering. In these fields, complex numbers are often used to represent physical quantities and the imaginary component can have a real-world interpretation.

## 4. How do you interpret an imaginary result of an integral?

To interpret an imaginary result of an integral, you must consider both the real and imaginary components separately. The real component represents the traditional result of the integral, while the imaginary component adds an additional dimension to the solution. This can be visualized in the complex plane or interpreted in the context of the specific problem being solved.

## 5. Can an imaginary result of an integral be converted into a real result?

Yes, an imaginary result of an integral can be converted into a real result by taking the absolute value of the complex number. This will result in the real component being isolated and the imaginary component being removed. However, it is important to consider the context of the problem and whether the imaginary component has any physical significance before converting to a real result.

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