integration - Evaluating $\sum_ {m=0}^\infty \sum_ {n=0}^\infty …
5 days ago · I am evaluating the following integral: $$\\int_0^{1} \\left(\\tanh^{-1}(x) + \\tan^{-1}(x)\\right)^2 \\; dx$$ After using the Taylor series of the two functions, we ...
Evaluating $\\lim_{x\\to1}\\frac{m}{1-x^m} -\\frac{n}{1+x^n}$, for ...
5 days ago · $$\\lim_{x\\to1}\\frac{m}{1-x^m} -\\frac{n}{1+x^n} \\;\\;\\;\\;\\;\\; m,n\\in \\mathbb{N}$$ My teacher had given the class this sum as homework. He gave us a hint ...
Evaluating $\\prod_{n=1}^{\\infty}\\left(1+\\frac{1}{2^n}\\right)$
Sep 13, 2016 · Compute:$$\prod_ {n=1}^ {\infty}\left (1+\frac {1} {2^n}\right)$$ I and my friend came across this product. Is the product till infinity equal to $1$? If no, what is the answer?
algebra precalculus - Evaluating $\frac {1} {a^ {2025}}+\frac {1} {b ...
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Evaluating $\\int_0^{\\infty}\\frac{\\ln(x^2+1)}{x^2+1}dx$
How would I go about evaluating this integral? $$\int_0^ {\infty}\frac {\ln (x^2+1)} {x^2+1}dx.$$ What I've tried so far: I tried a semicircular integral in the positive imaginary part of the …
Evaluating $\cos (i)$ - Mathematics Stack Exchange
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Evaluating $\\int_0^1 (1-x^2)^n dx$ - Mathematics Stack Exchange
Mar 15, 2021 · Evaluating $\int_0^1 (1-x^2)^n dx$ [duplicate] Ask Question Asked 4 years, 8 months ago Modified 4 years, 7 months ago
calculus - Evaluating $\int \frac {1} { {x^4+1}} dx$ - Mathematics ...
I am trying to evaluate the integral $$\int \frac {1} {1+x^4} \mathrm dx.$$ The integrand $\frac {1} {1+x^4}$ is a rational function (quotient of two polynomials), so I could solve the integral if I ...
Evaluating $ \\lim_{x \\to 0} \\frac{e - (1 + 2x)^{1/2x}}{x} $ without ...
Sep 11, 2024 · The following is a question from the Joint Entrance Examination (Main) from the 09 April 2024 evening shift: $$ \lim_ {x \to 0} \frac {e - (1 + 2x)^ {1/2x}} {x} $$ is equal to: (A) …
calculus - Evaluating $\lim_ {n\to \infty } \, \left (\sum _ {k=1 ...
$$\\lim_{n\\to \\infty } \\, \\left(\\sum _{k=1}^{\\infty } \\frac{1}{n}\\right)$$ Intuitively, it seems that you are adding infinite of $\\frac{1}{n}$ and then ...