Integral of the square of the absolute value of a function

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I have been trying to solve this question and have looked to see if this has been asked but it has not. I am given this assumption:

\begin{equation} \|f(x)\|^2 = 1 \end{equation}

From there I know to rewrite this given expression as an integral using the definition of a norm of a function.

\begin{equation} \|f(x)\|^2 = 1 = \sqrt{\int_{-\infty}^{\infty} |f(x)|^2 \,dx }^2 = \int_{-\infty}^{\infty} |f(x)|^2 \,dx \end{equation}

After simplifying, the part that I do not know how to do is how to solve:

\begin{equation} \int_{-\infty}^{\infty} |f(x)|^2 \,dx \end{equation}

I do not have an expression to replace f(x) with. I do know that I am supposed to solve using integration by parts but I do not know how. Any help is appreciated and sorry is this is confusing!

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Your question is rather confusing as it stands. Technically, one could apply integration by parts to this equation by assuming there exists a function $g(x)$ such that $g'(x) = f(x)$. One could then rewrite this, via integration by parts, as $$ \int f^*(x) f(x) \, dx = \int f^*(x) g'(x) \, dx = f^*(x) g(x) - \int g(x) {f^*}'(x) \, dx. $$ (I'm assuming complex-valued functions here. If they're real-valued, just drop the asterisks.) Similarly, you could show that $$ \int f^*(x) f(x) \, dx = x |f(x)|^2 - \int x \left( f^* f' + {f^*}' f \right) \, dx. $$ But without further knowledge about the form of $f(x)$, it is unlikely that these facts will help you. If you've been given the hint to use integration by parts to perform this integral for a specific $f(x)$, it seems more likely to me that there is a different decomposition that allows this to be done.