Find a reduction formula for this definite integral

Question

So here I am asked to find the reduction formula for:

$I_n=\int_{0}^{1} (2x-1)^ne^{x-x^2} dx$

The problem I have though is that the integration by parts is a mess. It doesn't matter what I let u equal. If $dv=e^{x-x^2} dx$ then I am unable to integrate and if $u=(2x-1)^n$ then I end up increasing the powers so I am stuck because in one case, I increase a function's power and in the other, I get something that can't be integrated...

I then noticed that if I let $u=x-x^2$, then $du=1-2x$ which is just the negative of what I have inside of the parenthesis but I am not really sure how to relate that...

If anyone could provide some guidance, that would be much appreciated. Thanks!

Answer 1

Let $I_n$ be the integral given by

$$I_n=\int_0^1 (2x-1)^ne^{x-x^2}\,dx \tag 1$$

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Integrating by parts the integral in $(1)$ with $u=(2x-1)^{n-1}$ and $v=-e^{x-x^2}$ reveals

$$\begin{align} I_n&=\left.\left(-(2x-1)^{n-1}e^{x-x^2}\right)\right|_{x=0}^{x=1}+2(n-1)\int_0^1 (2x-1)^{n-2}e^{x-x^2}\,dx\\\\ &=-(1-(-1)^{n})+2(n-1)I_{n-2} \end{align}$$

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It is straightforward to show that $I_0=e^{1/4}\sqrt{\pi}\text{erf}(1/2)$ and $I_1=0$.

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Hence, we have $I_{2n+1}=0$ and the reduction formula for even terms

$$\bbox[5px,border:2px solid #C0A000]{I_{2n}=2(2n-1)I_{2n-2}-2}$$

with $I_0=e^{1/4}\sqrt{\pi}\text{erf}(1/2)$.

Answer 2

Interesting problem! Let's start with the following integral:

$$f(t)=\int_0^1e^{(2x-1)t}e^{x-x^2}\ dx=\frac12\sqrt\pi e^{t^2+\frac14}\left[\operatorname{erf}\left(\frac12-t\right)+\operatorname{erf}\left(\frac12+t\right)\right]$$

It follows obviously that

$$I_0=f(0)$$

$$I_1=f'(0)$$

etc.

$$I_n=f^{(n)}(0)$$