Solving The Infinite Sum And Definite Integral Of Arctan(√(2sin X))

The infinite sum problem presented involves a fascinating blend of gamma functions and alternating terms, which can be expressed as:

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

To tackle this, we embark on a meticulous journey, simplifying the expression step by step. The initial observation points to the cancellation of $\Gamma(k+1)$ in the numerator and denominator, streamlining the sum. We're then left with:

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

At this juncture, we leverage the properties of gamma functions, specifically the identity $\Gamma(z+1) = z\Gamma(z)$. This identity is pivotal in relating gamma functions with slightly different arguments. Applying this property to $\Gamma(\frac{2k+5}{4})$, we can express it in terms of $\Gamma(\frac{2k+3}{4})$. Similarly, $\Gamma(k+\frac{3}{2})$ can be related to $\Gamma(k+\frac{1}{2})$. The transformed sum then becomes:

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

Now, the gamma functions $\Gamma(\frac{2k+3}{4})$ and $\Gamma(k+\frac{1}{2})$ gracefully cancel out, leading to a significantly simplified form:

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

This resulting sum, while seemingly simpler, still presents a challenge. The alternating term $(-2)^k$ and the square in the denominator $(2k+1)^2$ hint at a connection with special functions or series representations. Further analysis might involve recognizing a Taylor series expansion or employing complex analysis techniques to evaluate this sum. This part requires deeper insight and potentially advanced methods to solve.

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

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

This integral, featuring the arctangent function and a square root involving the sine function, suggests a strategy of leveraging series representations and integral properties to find a solution. The provided context gives us a crucial starting point:

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

This Taylor series expansion for the arctangent function is the key to unlocking the integral. By substituting $x = \sqrt{2\sin x}$ into this series, we transform the integral into a sum of integrals:

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

Here, we've interchanged the order of integration and summation, a step that requires careful justification in general but is often valid in cases where the series converges uniformly. The problem now boils down to evaluating the integral:

$$\int_{0}^{\pi/2} (\sin x)^{(2k+1)/2} dx$$

This integral can be expressed in terms of the Beta function, which is closely related to the gamma function. Recall the integral representation of the Beta function:

$$B(m, n) = 2 \int_{0}^{\pi/2} (\sin x)^{2n-1} (\cos x)^{2m-1} dx$$

Comparing this with our integral, we can set $2n-1 = \frac{2k+1}{2}$ and $2m-1 = 0$, leading to $n = \frac{2k+3}{4}$ and $m = \frac{1}{2}$. Thus, our integral transforms into:

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

The Beta function can be expressed in terms of gamma functions using the identity:

$$B(m, n) = \frac{\Gamma(m) \Gamma(n)}{\Gamma(m+n)}$$

Applying this, we get:

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

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

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

We've made significant progress, expressing the integral as an infinite sum involving gamma functions. The next step involves simplifying this sum, which might require further identities or special function evaluations. However, this form already provides a strong connection to the original sum we were asked to evaluate. This intricate interplay between integration and summation showcases the elegance of mathematical problem-solving. The final closed-form evaluation may necessitate advanced techniques, but the journey to this point highlights the power of series representations and special functions in tackling complex problems. The meticulous manipulation of the integral, coupled with the strategic use of gamma and beta functions, provides a pathway to a potential solution, underlining the deep connections within different branches of mathematics.

The culmination of our efforts lies in bridging the gap between the infinite sum and the definite integral. The integral $I$ has been transformed into a series involving gamma functions:

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

Meanwhile, the original infinite sum has been simplified to:

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

The task now is to demonstrate the equivalence or connection between these two expressions. This involves a careful comparison of the terms within the summations and potentially leveraging known identities or transformations. The presence of $(-1)^k$ and powers of 2 in both expressions suggests a possible relationship, but the gamma functions in the integral's series representation add a layer of complexity. To establish a concrete link, we might consider further manipulation of the integral's series, perhaps by expanding the gamma functions using their properties or integral representations. Another approach could involve attempting to express the sum in terms of a known special function or series, which could then be related to the integral. This part of the solution requires a keen eye for mathematical patterns and a deep understanding of special functions and their properties. It's a challenging but rewarding endeavor, showcasing the interconnectedness of different mathematical concepts.

In conclusion, we have embarked on a comprehensive exploration of both an infinite sum and a definite integral, revealing the intricate connections between these mathematical entities. The journey involved a strategic blend of techniques, including series representations, gamma and beta function identities, and careful manipulation of expressions. While a complete closed-form solution might require further advanced methods, the steps taken here provide a clear pathway and highlight the power of mathematical tools in tackling complex problems. The interplay between summation and integration, as demonstrated in this problem, underscores the beauty and elegance of mathematical reasoning, inviting further investigation and discovery.