integration - Evaluating $ \int_ {1/2}^ {\infty} \frac {\Gamma (u ...
Dec 13, 2025 · Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and …
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 ...
integration - Evaluating $\sum_ {m=0}^\infty \sum_ {n=0}^\infty …
Nov 11, 2025 · 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 ...
calculus - Evaluating $\int {\frac {x^ {14}+x^ {11}+x^5}
Jul 2, 2025 · The following question is taken from JEE practice set. Evaluate $\displaystyle\int {\frac {x^ {14}+x^ {11}+x^5} {\left (x^6+x^3+1\right)^3}} \, \mathrm dx$. My ...
integration - Evaluating $\iiint z (x^2+y^2+z^2)
Jul 29, 2020 · Spherical Coordinate Homework Question Evaluate the triple integral of $f (x,y,z)=z (x^2+y^2+z^2)^ {−3/2}$ over the part of the ball $x^2+y^2+z^2\le 81$ defined by ...
Evaluating $ \lim\limits_ {n\to\infty} \sum_ {k=1}^ {n^2} \frac {n} {n ...
How would you evaluate the following series? $$\\lim_{n\\to\\infty} \\sum_{k=1}^{n^2} \\frac{n}{n^2+k^2} $$ Thanks.
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?
Evaluating $\iiint_B (x^2+y^2+z^2)dV$ where $B$ is the ball of …
The question asks to use spherical coords. My answer is coming out wrong and symbolab is saying I'm evaluating the integrals correctly so my set up must be wrong. Since $\\rho$ is the …
Evaluating $\sum_ {i=1}^ {\infty}\frac { (i\ln 2)^i} {2^ii!}$
Dec 26, 2024 · I seek the proof of the evaluation to the sum $$\sum_ {i=1}^ {\infty}\frac { (i\ln 2)^i} {2^ii!} = \frac {1} {1-\ln2}-1 \approx 2.25889.$$ It is almost a power series ...
Evaluating the limit using Taylor Series - Mathematics Stack …
Dec 7, 2018 · I see now how I can go about evaluating the limit itself although I still find the concept a little bit vague, as in considering a specific order for the expansion and then …