Solving Infinite Series And Definite Integral With Arctangent Function

This article delves into a fascinating problem involving an infinite series and a definite integral, both intricately linked to the arctangent function. We will explore the solution to the series and evaluate the definite integral, highlighting the interplay between special functions and advanced calculus techniques.

Unveiling the Infinite Series

Let's begin by tackling the infinite series:

$$\sum_{k=0}^{\infty}(-2)^k\frac{\Gamma(\frac{2k+3}{4})\Gamma(k+\frac{1}{2})\Gamma(k+1)}{\Gamma(k+1)\Gamma(\frac{2k+5}{4})\Gamma(k+\frac{3}{2})}$$

Infinite series solution requires careful manipulation of the Gamma functions. The Gamma function, a generalization of the factorial function to complex numbers, plays a crucial role in various areas of mathematics and physics. Our primary focus will be on simplifying the expression inside the summation to reveal a pattern that allows us to compute the sum.

Simplifying with Gamma Function Properties

The key to unraveling this series lies in leveraging the properties of the Gamma function. Recall the following crucial identity:

$$\Gamma(z+1) = z\Gamma(z)$$

This identity will allow us to simplify the ratio of Gamma functions in the series. Observe that the $\Gamma(k+1)$ terms cancel out directly. We can further simplify the expression by applying the Gamma function identity to the terms involving fractions:

$$\frac{\Gamma(\frac{2k+3}{4})}{\Gamma(\frac{2k+5}{4})} = \frac{\Gamma(\frac{2k+3}{4})}{\frac{2k+1}{4}\Gamma(\frac{2k+3}{4})} = \frac{4}{2k+1}$$

$$\frac{\Gamma(k+\frac{1}{2})}{\Gamma(k+\frac{3}{2})} = \frac{\Gamma(k+\frac{1}{2})}{(k+\frac{1}{2})\Gamma(k+\frac{1}{2})} = \frac{1}{k+\frac{1}{2}} = \frac{2}{2k+1}$$

Substituting these simplifications back into the original series, we get:

$$\sum_{k=0}^{\infty}(-2)^k\frac{4}{2k+1} \cdot \frac{2}{2k+1} = \sum_{k=0}^{\infty}(-2)^k\frac{8}{(2k+1)^2}$$

Recognizing the Series Pattern

Now, we have a more manageable series:

$$\sum_{k=0}^{\infty}(-2)^k\frac{8}{(2k+1)^2} = 8\sum_{k=0}^{\infty}\frac{(-2)^k}{(2k+1)^2}$$

However, this form still doesn't immediately reveal a familiar series. To proceed, we must be extremely cautious about manipulating alternating series and ensure convergence conditions are met before rearranging terms. At this point, a direct analytical solution for this series might be challenging to obtain without further transformations or the use of specialized functions. The alternating nature and the presence of the $(2k+1)^2$ term in the denominator make it difficult to express this series in terms of elementary functions directly.

Therefore, a numerical approach or approximation methods might be necessary to estimate the sum of this series accurately. Specialized software or computational tools can be employed to evaluate the series to a desired level of precision.

Evaluating the Definite Integral

Now, let's shift our focus to the definite integral:

$$I=\int_{0}^{\pi/2}\arctan({\sqrt{2\sin x}})\,dx$$

The evaluation of this integral involves a clever application of series representation and integration techniques. The provided context gives us a crucial starting point:

$${\displaystyle \arctan (x)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}x^{2k+1}}{2k+1}}}$$

Employing the Series Representation

We can substitute this series representation of the arctangent function into the integral:

$$I = \int_{0}^{\pi/2} \sum_{k=0}^{\infty} \frac{(-1)^k (\sqrt{2\sin x})^{2k+1}}{2k+1} \, dx = \int_{0}^{\pi/2} \sum_{k=0}^{\infty} \frac{(-1)^k 2^{\frac{2k+1}{2}} (\sin x)^{\frac{2k+1}{2}}}{2k+1} \, dx$$

Interchanging Summation and Integration

To proceed, we need to interchange the summation and integration. This step requires careful justification, as it is only valid under certain conditions (e.g., uniform convergence of the series). Assuming the conditions for interchanging the sum and integral are met (which would typically need to be rigorously proven in a formal setting), we have:

$$I = \sum_{k=0}^{\infty} \frac{(-1)^k 2^{\frac{2k+1}{2}}}{2k+1} \int_{0}^{\pi/2} (\sin x)^{\frac{2k+1}{2}} \, dx$$

Utilizing the Beta Function

The integral inside the summation can be expressed in terms of the Beta function. Recall the relationship between the Beta function and the integral of powers of sine:

$$\int_{0}^{\pi/2} (\sin x)^m \, dx = \frac{1}{2}B(\frac{m+1}{2}, \frac{1}{2}) = \frac{\Gamma(\frac{m+1}{2})\Gamma(\frac{1}{2})}{2\Gamma(\frac{m}{2}+1)}$$

Applying this to our integral, where $m = \frac{2k+1}{2}$, we get:

$$\int_{0}^{\pi/2} (\sin x)^{\frac{2k+1}{2}} \, dx = \frac{1}{2}B(\frac{\frac{2k+1}{2}+1}{2}, \frac{1}{2}) = \frac{1}{2}B(\frac{2k+3}{4}, \frac{1}{2}) = \frac{\Gamma(\frac{2k+3}{4})\Gamma(\frac{1}{2})}{2\Gamma(\frac{2k+5}{4})}$$

Substituting Back into the Series

Substituting this result back into the series for $I$, we obtain:

$$I = \sum_{k=0}^{\infty} \frac{(-1)^k 2^{\frac{2k+1}{2}}}{2k+1} \cdot \frac{\Gamma(\frac{2k+3}{4})\Gamma(\frac{1}{2})}{2\Gamma(\frac{2k+5}{4})} = \sum_{k=0}^{\infty} \frac{(-1)^k 2^{\frac{2k+1}{2}}\sqrt{\pi}}{2(2k+1)} \cdot \frac{\Gamma(\frac{2k+3}{4})}{\Gamma(\frac{2k+5}{4})}$$

Since $\Gamma(\frac{1}{2}) = \sqrt{\pi}$.

Further Simplification and Connection to the Initial Series

We can simplify this expression further:

$$I = \frac{\sqrt{\pi}}{2}\sum_{k=0}^{\infty} \frac{(-1)^k 2^{\frac{2k+1}{2}}}{2k+1} \frac{\Gamma(\frac{2k+3}{4})}{\Gamma(\frac{2k+5}{4})}$$

Now, let's multiply the numerator and denominator by $\Gamma(k+\frac{1}{2})\Gamma(k+1)$:

$$I = \frac{\sqrt{\pi}}{2}\sum_{k=0}^{\infty} (-1)^k 2^{\frac{2k+1}{2}}\frac{\Gamma(\frac{2k+3}{4})\Gamma(k+\frac{1}{2})\Gamma(k+1)}{(2k+1)\Gamma(\frac{2k+5}{4})\Gamma(k+\frac{1}{2})\Gamma(k+1)}$$

Using the identity $\Gamma(k+\frac{3}{2}) = (k+\frac{1}{2})\Gamma(k+\frac{1}{2}) = \frac{2k+1}{2}\Gamma(k+\frac{1}{2})$, we can rewrite the series as:

$$I = \sqrt{2\pi}\sum_{k=0}^{\infty} (-2)^k\frac{\Gamma(\frac{2k+3}{4})\Gamma(k+\frac{1}{2})\Gamma(k+1)}{\Gamma(k+1)\Gamma(\frac{2k+5}{4})\Gamma(k+\frac{3}{2})/(2k+1)}$$

This form closely resembles the original infinite series we started with. However, an accurate solution for I requires us to evaluate this complex series. Unfortunately, a direct closed-form solution for the series within this integral is not immediately apparent. It is a highly non-trivial series.

Towards a Solution or Numerical Approximation

While we have made significant progress in expressing the integral in terms of a series involving Gamma functions, obtaining a closed-form solution remains a challenge. Possible approaches to finding a solution or approximation include:

1. Numerical Integration: Employ numerical integration techniques (e.g., Simpson's rule, trapezoidal rule) to approximate the value of the definite integral directly.
2. Series Approximation: Investigate the convergence properties of the series and potentially use truncation to approximate the sum. Careful error analysis is crucial in this case.
3. Special Functions: Explore connections to other special functions or integral representations that might provide a pathway to a closed-form solution. This approach often requires advanced knowledge and techniques.
4. Computational Software: Utilize software packages like Mathematica, Maple, or Python (with libraries like SciPy) to evaluate the integral numerically or symbolically.

Conclusion

In this exploration, we dissected a complex problem involving an infinite series and a definite integral related to the arctangent function. We successfully simplified the series using properties of the Gamma function and transformed the integral into a series representation. While obtaining a direct closed-form solution proved elusive, we identified several promising avenues for approximation and numerical evaluation. This problem underscores the intricate interplay between special functions, series, and integral calculus, showcasing the power and beauty of mathematical analysis.

Further research might involve delving deeper into numerical methods, exploring specialized software tools, and investigating potential connections to advanced mathematical concepts to arrive at a more precise solution or approximation for both the infinite series and the definite integral. This problem serves as a testament to the depth and richness of mathematical inquiry. The journey of attempting to find exact analytical solutions often reveals the need for numerical methods and approximation techniques, highlighting the multifaceted nature of mathematical problem-solving.