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The integral:

$$I = \int_{0}^{\infty} \left[(e^{ax^2}-1)\dfrac{e^{-\frac{x^2}{2}}}{x}\right] dx$$

(assume $0 \leq a < \frac{1}{2}$)

can be upperbounded by a convergent integral, using the identity: $e^{b} - 1 \leq b e^{b}$, $\forall$ $b \geq 0$.

$$I \leq a\int_{0}^{\infty} \left[x\thinspace e^{-\frac{x^2}{2}(1-2a)}\right] dx \implies \dfrac{a}{1-2a}$$

Suppose we instead have the integral:

$$\tilde{I} = \int_{0}^{\infty} \left|\left[(i\thinspace e^{ax^2}-1)\dfrac{e^{-\frac{x^2}{2}}}{x}\right]\right| dx$$

(where the $|\cdots|$ refers to the Absolute value or the Modulus of the complex number)

can we still upperbound the integral, similar to above ? Is there a contour method of solving this integral ?

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1 Answer 1

No, there is no upper bound to the integral, as it does not converge. Note that $$(i\thinspace e^{ax^2}-1)\dfrac{e^{-\frac{x^2}{2}}}{x} =\frac{i-1}{x} + O(x)$$ for $x\to 0$.

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But that is true for the case where there is no $i$ (the first integral), but it still converges. Is there a contour method to solve this integral. I'm aware that for functions which have simple poles, a contour integral method can be used to compute the integral. –  Pavithran Iyer Jan 26 '13 at 14:30
    
@PavithranIyer: the first integral (without $i$) converges. the second (with $i$) does not. –  Fabian Jan 26 '13 at 15:44

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