# Help with simplifying an integral

• A
• TroyElliott
In summary, the conversation discusses the issue of the variable "v" disappearing in the third equal sign of equation (1.8). The speaker suggests changing variables in order to get the third line, but the other person raises the concern that since "v" can become imaginary, the limits of integration would also become imaginary. The speaker explains that since both "u" and "v" are real and "v" is constant, the limits of integration would still be just infinity. However, the other person points out that in equation (1.7), "v" can become imaginary inside the light cone, leading to a discussion on how to handle this in the calculations.
TroyElliott
TL;DR Summary
Stuck on some algebra within an integral.
I am not seeing how the v goes away in the third equal sign of equation (1.8). It seems to be that it must be cos(z*sinh(u+v)), not cos(z*sinh(u)).
In the defined equations (1.7), the variable "v" can become imaginary, so a simple change of variables would change the integration sign by adding an imaginary term - not what occurs in (1.8). Any idea on how to get the third line of (1.8)?

Thanks for any insight!

In order to get the third line in Eq. (1.8) change variables ##u+v\mapsto u## and ##u-v\mapsto u##.
Since ##v## is constant with respect to the integration varible ##du = d(u\pm v)##, and also don't forget that integral of an even function ##f(x)=f(-x)## leads to ##\int_{-a}^a f(x)dx= 2\int_0^a f(x)dx##.

anuttarasammyak
MathematicalPhysicist said:
In order to get the third line in Eq. (1.8) change variables ##u+v\mapsto u## and ##u-v\mapsto u##.
Since ##v## is constant with respect to the integration varible ##du = d(u\pm v)##, and also don't forget that integral of an even function ##f(x)=f(-x)## leads to ##\int_{-a}^a f(x)dx= 2\int_0^a f(x)dx##.

But, in this case, doesn’t it matter than ##v## can potentially become an imaginary number, and the limits of the integral would no longer be just infinity? We would get an added imaginary term to the limit, e.g. ##\infty + 2\pi i##.

More explicitly, ##v= arcsinh(\frac{mc^{2}t}{hz}),## where ##z## is given in Eq.(1.7). We see that inside the light cone, ##z## becomes imaginary. So ##v## would also become imaginary, in this case. When performing the change of variables, the integration bound would now pick up an additional imaginary piece. This is what I am confused about - how can it still just be ##\infty##?

Last edited:
u and v are both real and v is constant as for integration of u.
$$\int_{-\infty}^{+\infty} du = \int_{-\infty}^{+\infty} d(u+v)$$
or so.

MathematicalPhysicist said:
In order to get the third line in Eq. (1.8) change variables ##u+v\mapsto u## and ##u-v\mapsto u##.
Since ##v## is constant with respect to the integration varible ##du = d(u\pm v)##, and also don't forget that integral of an even function ##f(x)=f(-x)## leads to ##\int_{-a}^a f(x)dx= 2\int_0^a f(x)dx##.
anuttarasammyak said:
u and v are both real and v is constant as for integration of u.
$$\int_{-\infty}^{+\infty} du = \int_{-\infty}^{+\infty} d(u+v)$$
or so.
But we see in Eq.(1.7), that inside the light cone, ##v## can become imaginary.

It seems that (1.8) is for real z with u and v real, and (1.9) is for imaginary z.

TroyElliott
TroyElliott said:
But we see in Eq.(1.7), that inside the light cone, ##v## can become imaginary.
Such a calculation is done in a course in complex analysis; I don't recall the details, sorry.

TroyElliott

## 1. What is an integral?

An integral is a mathematical concept that represents the area under a curve in a graph. It is used to calculate the total amount or value of something, such as the distance traveled by an object or the amount of a substance in a solution.

## 2. Why is simplifying an integral important?

Simplifying an integral is important because it allows us to solve complex mathematical problems more easily and accurately. It also helps us to better understand the underlying concepts and relationships within the problem.

## 3. How do you simplify an integral?

To simplify an integral, you can use various techniques such as substitution, integration by parts, and trigonometric identities. It is important to carefully analyze the integral and choose the most appropriate method for simplification.

## 4. What are some common mistakes when simplifying integrals?

Some common mistakes when simplifying integrals include forgetting to apply the chain rule, using incorrect substitution, and making calculation errors. It is important to double check your work and be mindful of the rules and techniques being used.

## 5. Can integrals be simplified in all cases?

No, not all integrals can be simplified. Some integrals may require more advanced techniques or may not have a closed-form solution. In these cases, numerical methods or approximations may be used to find a solution.

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