Evaluate ∫ − ∞ ∞ Tan ⁡ − 1 ( E X ) 1 + X 2 D X \int_{-\infty}^{\infty} \dfrac{\tan^{-1}(e^x)}{1+x^2} Dx ∫ − ∞ ∞ ​ 1 + X 2 Tan − 1 ( E X ) ​ D X

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Introduction

In this article, we will delve into the world of complex analysis and residue calculus to evaluate the definite integral of $\tan^{-1}(e^x)$ from $-\infty$ to $\infty$. This problem involves the use of residue calculus, which is a powerful tool for evaluating definite integrals of functions that have singularities in the complex plane.

The Problem

The problem we are trying to solve is the evaluation of the definite integral:

$$\int_{-\infty}^{\infty} \dfrac{\tan^{-1}(e^x)}{1+x^2} dx$$

This integral appears to be difficult to evaluate directly, and we will use residue calculus to find its value.

Residue Calculus

Residue calculus is a branch of complex analysis that deals with the evaluation of definite integrals of functions that have singularities in the complex plane. The basic idea behind residue calculus is to use the residues of a function at its singularities to evaluate the integral.

The Function $f(z)$

To apply residue calculus, we need to define a function $f(z)$ that has the same singularities as the original integral. In this case, we define the function:

$$f(z)=\dfrac{\tan^{-1}(e^z)}{1+z^2}$$

This function has two simple poles at $z=\pm i$, which are the singularities of the original integral.

The Residues of $f(z)$

To evaluate the integral, we need to find the residues of $f(z)$ at its singularities. The residue of a function at a simple pole $z=a$ is given by:

$$\text{Res}(f(z),a)=\lim_{z\to a}(z-a)f(z)$$

Using this formula, we can find the residues of $f(z)$ at $z=\pm i$.

Residue at $z=i$

To find the residue at $z=i$, we use the formula:

$$\text{Res}(f(z),i)=\lim_{z\to i}(z-i)f(z)$$

Substituting the expression for $f(z)$, we get:

$$\text{Res}(f(z),i)=\lim_{z\to i}(z-i)\dfrac{\tan^{-1}(e^z)}{1+z^2}$$

Using the fact that $\tan^{-1}(e^z)=\frac{i}{2}\left(z-\frac{z^3}{3}+\frac{z^5}{5}-\cdots\right)$, we can simplify the expression:

$$\text{Res}(f(z),i)=\lim_{z\to i}(z-i)\dfrac{\frac{i}{2}\left(z-\frac{z^3}{3}+\frac{z^5}{5}-\cdots\right)}{1+z^2}$$

Simplifying further, we get:

$$\text{Res}(f(z),i)=\frac{i}{2}\lim_{z\to i}\dfrac{z-\frac{z^3}{3}+\frac{z^5}{5}-\cdots}{1+z^2}$$

Using the fact that $z=i$, we can simplify the expression:

$$\text{Res}(f(z),i)=\frac{i}{2}\lim_{z\to i}\dfrac{z-\frac{z^3}{3}+\frac{z^5}{5}-\cdots}{1+z^2}=\frac{i}{2}\dfrac{i-\frac{i^3}{3}+\frac{i^5}{5}-\cdots}{1+i^2}$$

Simplifying further, we get:

$$\text{Res}(f(z),i)=\frac{i}{2}\dfrac{i-\frac{i^3}{3}+\frac{i^5}{5}-\cdots}{1+i^2}=\frac{i}{2}\dfrac{i-\frac{i}{3}+\frac{i}{5}-\cdots}{2}$$

Simplifying further, we get:

$$\text{Res}(f(z),i)=\frac{i}{2}\dfrac{i-\frac{i}{3}+\frac{i}{5}-\cdots}{2}=\frac{i}{4}\left(1-\frac{1}{3}+\frac{1}{5}-\cdots\right)$$

Using the fact that $\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}$, we can simplify the expression:

$$\text{Res}(f(z),i)=\frac{i}{4}\left(1-\frac{1}{3}+\frac{1}{5}-\cdots\right)=\frac{i}{4}\frac{\pi}{4}=\frac{i\pi}{16}$$

Residue at $z=-i$

To find the residue at $z=-i$, we use the formula:

$$\text{Res}(f(z),-i)=\lim_{z\to -i}(z+i)f(z)$$

Substituting the expression for $f(z)$, we get:

$$\text{Res}(f(z),-i)=\lim_{z\to -i}(z+i)\dfrac{\tan^{-1}(e^z)}{1+z^2}$$

Using the fact that $\tan^{-1}(e^z)=\frac{i}{2}\left(z-\frac{z^3}{3}+\frac{z^5}{5}-\cdots\right)$, we can simplify the expression:

$$\text{Res}(f(z),-i)=\lim_{z\to -i}(z+i)\dfrac{\frac{i}{2}\left(z-\frac{z^3}{3}+\frac{z^5}{5}-\cdots\right)}{1+z^2}$$

Simplifying further, we get:

$$\text{Res}(f(z),-i)=\frac{i}{2}\lim_{z\to -i}\dfrac{z-\frac{z^3}{3}+\frac{z^5}{5}-\cdots}{1+z^2}$$

Using the fact that $z=-i$, we can simplify the expression:

$$\text{Res}(f(z),-i)=\frac{i}{2}\lim_{z\to -i}\dfrac{z-\frac{z^3}{3}+\frac{z^5}{5}-\cdots}{1+z^2}=\frac{i}{2}\dfrac{-i-\frac{i^3}{3}+\frac{i^5}{5}-\cdots}{1+i^2}$$

Simplifying further, we get:

$$\text{Res}(f(z),-i)=\frac{i}{2}\dfrac{-i-\frac{i^3}{3}+\frac{i^5}{5}-\cdots}{1+i^2}=\frac{i}{2}\dfrac{-i-\frac{i}{3}+\frac{i}{5}-\cdots}{2}$$

Simplifying further, we get:

$$\text{Res}(f(z),-i)=\frac{i}{2}\dfrac{-i-\frac{i}{3}+\frac{i}{5}-\cdots}{2}=\frac{i}{4}\left(-1+\frac{1}{3}-\frac{1}{5}+\cdots\right)$$

Using the fact that $\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}$, we can simplify the expression:

$$\text{Res}(f(z),-i)=\frac{i}{4}\left(-1+\frac{1}{3}-\frac{1}{5}+\cdots\right)=\frac{i}{4}\left(-\frac{\pi}{4}\right)=-\frac{i\pi}{16}$$

The Integral

Now that we have found the residues of $f(z)$ at its singularities, we can use the residue theorem to evaluate the integral:

$$\int_{-\infty}^{\infty} \dfrac{\tan^{-1}(e^x)}{1+x^2} dx$$

The residue theorem states that the value of the integral is equal to $2\pi i$ times the sum of the residues of $f(z)$ at its singularities.

Evaluating the Integral

Using the residue theorem, we can evaluate the integral:

$$\int_{-\infty}^{\infty} \dfrac{\tan^{-1}(e^x)}{1+x^2} dx=2\pi i\left(\frac{i\pi}{16}-\frac{i\pi}{16}\right)=0$$

Therefore, the value of the integral is 0.

Conclusion

In this article, we used residue calculus to evaluate the definite integral of $\tan^{-1}(e^x)$ from $-\infty$ to $\infty$. We defined a function $f(z)$ that has the same singularities as the original integral and found the residues of $f(z)$ at its singularities. Using the residue theorem, we were able to evaluate the integral and find that its value is 0.

References

• [1
  

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Introduction

In our previous article, we used residue calculus to evaluate the definite integral of $\tan^{-1}(e^x)$ from $-\infty$ to $\infty$. In this article, we will answer some of the most frequently asked questions about this problem.

Q: What is residue calculus?

A: Residue calculus is a branch of complex analysis that deals with the evaluation of definite integrals of functions that have singularities in the complex plane. It is a powerful tool for evaluating integrals that cannot be evaluated directly.

Q: Why did we use residue calculus to evaluate this integral?

A: We used residue calculus to evaluate this integral because it has singularities in the complex plane. The function $\tan^{-1}(e^x)$ has a branch point at $x=\infty$, which is a singularity in the complex plane. Residue calculus allows us to evaluate the integral by considering the residues of the function at its singularities.

Q: What is the residue theorem?

A: The residue theorem is a fundamental result in complex analysis that states that the value of a contour integral is equal to $2\pi i$ times the sum of the residues of the function at its singularities inside the contour.

Q: How did we find the residues of $f(z)$ at its singularities?

A: We found the residues of $f(z)$ at its singularities by using the formula:

$$\text{Res}(f(z),a)=\lim_{z\to a}(z-a)f(z)$$

We applied this formula to the function $f(z)$ at its singularities $z=\pm i$.

Q: Why did we get a value of 0 for the integral?

A: We got a value of 0 for the integral because the residues of $f(z)$ at its singularities $z=\pm i$ are equal and opposite. When we applied the residue theorem, the sum of the residues was zero, which means that the value of the integral is also zero.

Q: Can we use residue calculus to evaluate other integrals?

A: Yes, we can use residue calculus to evaluate other integrals that have singularities in the complex plane. Residue calculus is a powerful tool for evaluating integrals that cannot be evaluated directly.

Q: What are some common applications of residue calculus?

A: Residue calculus has many applications in physics, engineering, and mathematics. Some common applications include:

• Evaluating definite integrals that have singularities in the complex plane
• Finding the poles of a function
• Evaluating the residues of a function at its singularities
• Applying the residue theorem to evaluate contour integrals

Q: What are some common mistakes to avoid when using residue calculus?

A: Some common mistakes to avoid when using residue calculus include:

• Not identifying the singularities of the function
• Not finding the correct residues of the function at its singularities
• Not applying the residue theorem correctly
• Not checking the convergence of the integral

Conclusion

In this article, we answered some of the most frequently asked questions about the evaluation of the definite integral of $\tan^{-1}(e^x)$ using residue calculus. We hope that this article has been helpful in understanding the basics of residue calculus and its applications.

References

• [1] Ahlfors, L. V. (1979). Complex Analysis. McGraw-Hill.
• [2] Churchill, R. V., & Brown, J. W. (1990). Complex Variables and Applications. McGraw-Hill.
• [3] Lang, S. (1999). Complex Analysis. Springer-Verlag.

Further Reading

• [1] Residue Calculus: A Tutorial by David J. Jeffrey
• [2] Residue Calculus: A Brief Introduction by John H. Hubbard
• [3] Residue Calculus: Applications and Examples by Michael J. Grinfeld