I'm looking for some guidance on this problem. It was assigned as homework, but the topic hasn't been covered in our lectures yet, so I'm trying to bridge the gap between our current curriculum and this advanced material:bigBrain:
$$I &= \int_{0}^{1} \cos(x^2 + e^x) dx \$$
$$J &= \int_{0}^{+\infty} e^{-[x + \sin(x)]} dx \$$
$$K &= \int_{0}^{1} \frac{\sin(x)}{\ln(x + 1)} dx$$
#Calculate the exact value of the following integrals
quartz kettle
gray summitBOT
gray summitBOT
royal iglooBOT
tkabeleon
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hardy sparrow
I'm looking for some guidance on this problem. It was assigned as homework, but ...
no closed form i think
quartz kettle
no closed form i think
How to find the value in series form?
hardy sparrow
How to find the value in series form?
wdym
do u want to use series expansions of the functions
it wont work forJ at all and will not give u the exat answer for I or K
quartz kettle
wdym
So how do we proceed?
it's correct?
Evaluation of $I = \int_0^1 \cos(x^2+e^x) dx$
1. Introduction
No elementary antiderivative exists for this integral. However, we can derive an exact expression using a double series.
2. Series & Binomial Expansion
Using the power series for $\cos(u)$ where $u = x^2 + e^x$:
$$\cos(x^2+e^x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} (x^2+e^x)^{2n}$$
Applying the binomial theorem to $(x^2+e^x)^{2n}$:
$$(x^2+e^x)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} x^{2k} e^{(2n-k)x}$$
3. Integration Term by Term
By interchanging the series and the integral, we obtain:
$$I = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} \sum_{k=0}^{2n} \binom{2n}{k} \int_0^1 x^{2k} e^{(2n-k)x} dx$$
4. Exact Calculation of $I_{k,n}$
Let $m=2k$ and $a=2n-k$. The integral $I(m,a) = \int_0^1 x^m e^{ax} dx$ is:
- If $a=0$: $I(m,0) = \frac{1}{4n+1}$
- If $a \neq 0$: $I(m,a) = \frac{(2k)!}{(k-2n)^{2k+1}} \left[ 1 - e^{2n-k} \sum_{j=0}^{2k} \frac{(k-2n)^j}{j!} \right]$
5. Final Exact Expression
Substituting these into the double sum provides the exact representation:
$$I = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} \left[ \frac{\binom{2n}{2n}}{4n+1} + \sum_{k=0, k \neq 2n}^{2n} \binom{2n}{k} \frac{(2k)!}{(k-2n)^{2k+1}} \left( 1 - e^{2n-k} \sum_{j=0}^{2k} \frac{(k-2n)^j}{j!} \right) \right]$$
6. Numerical Result
Using adaptive quadrature (Gauss-Kronrod) or high-precision multi-arithmetic for the series:
$$\boxed{I \approx -0.303666963947415}$$
Remark: While the series is exact, direct numerical integration is more stable for practical computing due to catastrophic cancellation in the terms for $n \ge 10$.
royal iglooBOT
tkabeleon
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hardy sparrow
**tkabeleon**
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intereseting
i think this works
how would u do J
ahh yes
e^-sin x is periodic
u can expand it using fourier series
narrow kayak
it's correct?
# Evaluation of $I = \int_0^1 \cos(x^2+e^x) dx$
### 1. Introducti...
ask your llm that.
gray summitBOT
@quartz kettle