DC
Okay, so I need to figure out how to solve this integral: the integral of e^(x^2) dx. Hmm, I remember from calculus that some integrals don't have elementary antiderivatives. Let me think. The function e^(x^2) is a common one, right? I think the integral of e^(x^2) dx can't be expressed in terms of basic functions. But wait, maybe there's a special function for that? Oh right, isn't that related to the error function? Let me recall.
The error function, erf(x), is defined as (2/sqrt(π)) times the integral from 0 to x of e^(-t^2) dt. Wait, but that's e^(-t^2), not e^(t^2). So there's a sign difference in the exponent. Hmm. So if I have e^(x^2), the integral might not converge in the usual sense because as x becomes large, e^(x^2) grows very rapidly. So maybe the integral is only defined for certain intervals or requires a different approach.
Wait, the original integral is indefinite. So if it's indefinite, then even if it's related to the error function, which is a special function, maybe we can express the integral in terms of erf. But let me check. The error function is for e^(-x^2), so if I have e^(x^2), that's different. Let me see.
Alternatively, maybe substitution? Let's try substitution. Let u = x^2, then du/dx = 2x, so dx = du/(2x). But then I have e^u * du/(2x), but x is sqrt(u), so it becomes e^u/(2 sqrt(u)) du. Hmm, that doesn't look any easier. In fact, that might complicate things more. So substitution doesn't seem to help here.
What about integration by parts? Let me try that. Integration by parts formula is ∫u dv = uv - ∫v du. Let me set u = e^(x^2), dv = dx. Then du = 2x e^(x^2) dx, and v = x. So putting it into the formula: x e^(x^2) - ∫x * 2x e^(x^2) dx = x e^(x^2) - 2 ∫x^2 e^(x^2) dx. Hmm, but now the integral on the right is more complicated than the original. So that doesn't help. Maybe a different choice for u and dv?
Let me try switching them. Let u = x, dv = e^(x^2) dx. Then du = dx, and v would be the integral of e^(x^2) dx, which is exactly what we're trying to find. So that leads us in a circle. Not helpful.
Hmm. Maybe another substitution. What if I try a trigonometric substitution? For integrals involving e^(x^2), sometimes people use substitution like x = tanθ or something else? Let me try x = tanθ. Then dx = sec^2θ dθ, and e^(x^2) becomes e^(tan^2θ). That doesn't seem helpful either. The integral becomes e^(tan^2θ) * sec^2θ dθ. Not sure how to integrate that.
Alternatively, maybe expanding e^(x^2) as a Taylor series and integrating term by term. Let's try that. The Taylor series for e^z is 1 + z + z^2/2! + z^3/3! + ... So substituting z = x^2, we get e^(x^2) = 1 + x^2 + x^4/2! + x^6/3! + x^8/4! + ... Then integrating term by term: ∫e^(x^2) dx = ∫1 dx + ∫x^2 dx + ∫x^4/2! dx + ∫x^6/3! dx + ... = x + x^3/3 + x^5/(52!) + x^7/(73!) + x^9/(9*4!) + ... + C. So that gives an infinite series representation of the integral. But is that considered an elementary function? No, it's an infinite series. So unless we can express that series in terms of a known special function, that's as far as we can go.
Wait, but earlier I thought about the error function. Let me check the relationship again. The error function is erf(x) = (2/sqrt(π)) ∫₀^x e^(-t²) dt. So if we have e^(-t²), but we have e^(t²). So maybe if we consider a substitution to relate them. Let me see. If I let t = ix, where i is the imaginary unit, then e^(-t²) becomes e^(- (ix)^2) = e^(x²). So maybe there's a connection through complex analysis. But that might be beyond the scope here. However, in terms of real analysis, the integral of e^(x²) dx is not expressible in terms of elementary functions. So the answer would be expressed in terms of the imaginary error function or something like that.
Alternatively, sometimes the integral of e^(x²) is written as (sqrt(π)/2) erfi(x) + C, where erfi(x) is the imaginary error function, defined as -i erf(ix). Let me verify that. If erf(x) = (2/sqrt(π)) ∫₀^x e^(-t²) dt, then erfi(x) = -i erf(ix). Let's compute that:
erfi(x) = -i erf(ix) = -i * (2/sqrt(π)) ∫₀^{ix} e^(-t²) dt.
Let me make a substitution u = it, so t = -iu, dt = -i du. When t goes from 0 to ix, u goes from 0 to x. So the integral becomes:
-i * (2/sqrt(π)) ∫₀^x e^(-(-iu)^2) (-i du) = -i * (2/sqrt(π)) * (-i) ∫₀^x e^(-(-u²)) du.
Simplify the exponents: (-iu)^2 = (-i)^2 u² = (-1)^2 i² u² = (1)(-1)u² = -u². So e^(-(-iu)^2) = e^(u²). Therefore:
= -i * (2/sqrt(π)) * (-i) ∫₀^x e^(u²) du = (-i)(-i) * (2/sqrt(π)) ∫₀^x e^(u²) du.
Since (-i)(-i) = i² = -1, so:
= -1 * (2/sqrt(π)) ∫₀^x e^(u²) du.
But erfi(x) is supposed to be equal to (2/sqrt(π)) ∫₀^x e^(u²) du. Wait, there's a discrepancy here. Let me check again.
Wait, maybe I messed up the substitution steps. Let's go step by step.
Start with erfi(x) = -i erf(ix).
Compute erf(ix):
erf(ix) = (2/sqrt(π)) ∫₀^{ix} e^{-t²} dt.
Let t = iu, so when t = 0, u = 0; when t = ix, u = x. Then dt = i du.
Substitute into the integral:
= (2/sqrt(π)) ∫₀^x e^{-(iu)^2} * i du.
Compute (iu)^2 = i² u² = -u². So e^{-(iu)^2} = e^{u²}.
Thus:
= (2/sqrt(π)) * i ∫₀^x e^{u²} du.
Therefore, erf(ix) = i * (2/sqrt(π)) ∫₀^x e^{u²} du.
Then erfi(x) = -i erf(ix) = -i * = -i^2 * (2/sqrt(π)) ∫₀^x e^{u²} du.
Since i^2 = -1, this becomes:
= -(-1) * (2/sqrt(π)) ∫₀^x e^{u²} du = (2/sqrt(π)) ∫₀^x e^{u²} du.
Therefore, erfi(x) = (2/sqrt(π)) ∫₀^x e^{u²} du. So the integral from 0 to x of e^{u²} du is (sqrt(π)/2) erfi(x). Therefore, the indefinite integral ∫e^{x²} dx can be written as (sqrt(π)/2) erfi(x) + C.
So even though it's not an elementary function, it can be expressed in terms of the imaginary error function. Therefore, the answer is (sqrt(π)/2) erfi(x) + C.
But I should check if this is correct. Let me verify by differentiating. If d/dx [ (sqrt(π)/2) erfi(x) ] = (sqrt(π)/2) * d/dx erfi(x). The derivative of erfi(x) is (2/sqrt(π)) e^{x²}. So multiplying, (sqrt(π)/2) * (2/sqrt(π)) e^{x²} = e^{x²}, which matches the integrand. Perfect, so that's correct.
Therefore, the integral of e^(x^2) dx is (sqrt(π)/2) erfi(x) + C, where erfi is the imaginary error function.
Alternatively, sometimes people write it in terms of the error function with complex arguments, but using erfi is more straightforward here.
So the conclusion is that the integral can't be expressed with elementary functions and requires the imaginary error function. Therefore, the answer is expressed in terms of erfi(x).
The integral of ( e^{x^2} ) with respect to ( x ) cannot be expressed in terms of elementary functions. However, it can be represented using the imaginary error function ((\text{erfi})).
Result:
[
https://www.18122.com/url/usdt/usdt4557.jpg
\int e^{x^2} , dx = \frac{\sqrt{\pi}}{2} \text{erfi}(x) + C
]
Explanation:
The imaginary error function is defined as (\text{erfi}(x) = -i , \text{erf}(ix)), where (\text{erf}(x)) is the standard error function.
This result is derived by relating the integral to the error function through complex substitution and verifying via differentiation.
Verification:
Differentiating (\frac{\sqrt{\pi}}{2} \text{erfi}(x)) yields (e^{x^2}), confirming the solution's correctness.
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