integrate e^(-t/(1-sin(2x))) wrt x from 0 to pi/4

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[solved] $$\int_0^{\pi/4}e^{-\frac{t}{1-\sin2x}},dx$$

[still goblet]

teaset

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t real positive

[waxen vapor]

intergrate what?

is that e to the power

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yes

i have tried the change of variables $u = \frac{t}{1-\sin2x}$ but it led me nowhere

also tried wolfram alpha / mathematica and the results are series expansions which i am not after

[still goblet]

teaset

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which lead me to a more complicated integrand so i did not dig it thru much

[queen jewel]

$$ \frac{d}{dt} \left( \int_{a}^{b} h(x,t) , dx \right) = \int_{a}^{b} \frac{\partial}{\partial t} h(x,t) , dx $$

try this maybe

the derivative just needs to exist in this case, which it does

[still goblet]

Latias

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yes in that case the integrand is more complicated

[obtuse heart]

so the t ahould be an X?

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no

[obtuse heart]

hello hello

i managed to get the integral to a lapace transform

[obtuse heart]

where does this integral come?

[slender bluff]

what do you mean?

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wait i actually got an idea

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i want to ask smthing are limits of the integral correct?

like are they really 0 to pi/4?

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it could help if you told where does this integral comes from

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only significant things i have uncovered from this integral are the following:

$$\int_0^{\pi/4}e^{-\frac{n}{2sin^2(x)}}dx$$

[still goblet]

T4nuk1

[obtuse heart]

from this you can get various integrals each as difficult to evaluate as the last one

something that came to my mind while doing this were the fourier transforms and fourier functions but the integral limit's don't coincide

but idk

it occured to me cus the function is pi periodic

also

this came up

[obtuse heart]

one can obtain $$1/2\int_0^{\infty}\frac{1}{(x+2)\sqrt{x+1}}e^{-\frac{n(x+2)}{2}}dx$$

wich is soooo similar to the $$R_C(x,y)$$

[still goblet]

T4nuk1

T4nuk1

[obtuse heart]

so for n=0

one gets that special case

also if you let $$f(n)=1/2\int_0^{\infty}\frac{1}{(x+2)\sqrt{x+1}}e^{-\frac{n(x+2)}{2}}dx$$

[still goblet]

T4nuk1

[obtuse heart]

$$\forall n\ge0\ , 0\leq{f(n)}\leq \pi/4$$

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T4nuk1

[obtuse heart]

idk it will be kind of hard to actually do this one

waaaaaaaaaaaaaaait

i got smthing

so this can be also seen as the laplace transfor of the derivatice of the arctan(sqrt(x+1))

[obtuse heart]

now those limits in the first term change depending on n

cus on n=0

that term is just pi/4

for n aproach to infinity it goes to 0

and everything in between it goes to -pi/4

i think

although i think i saw something similar elsewhere

[obtuse heart]

again it would be nice to know from where does this integral comes from

[obtuse heart]

uh

[slender bluff]

ur math looks ok, haven't thought about integration by parts but still we have this second term

[obtuse heart]

idk

where you got this from

its so stupidly hard

:/

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i got it as an intermediate step of something im working on

[obtuse heart]

oh

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ig i have to take another route

thanks

[obtuse heart]

what was the og problem

im curious

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i'll dm u

[obtuse heart]

thnksss

i sent you a friend request

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||"solution" based off of qfunction; use substitution t/(1-sin2x) = x^2 and gradshteyn 4th edition 3.468 1||

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