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is the statement true/false.

Let $f$ be a continuous integrable function of $\mathbb{R}$ such that either $f(x) > 0$ or $f(x) + f(x + 1) > 0$ for all x$ \in \mathbb{R}$. Then $\int_{-\infty}^{\infty} f(x)dx>0$

is the above statement is true please help someone.

i am sorry that i could not wright it properly.the ingration will be from -infinity to infinity.thanks for help.

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Consider integrating in both sides of $f(x)>0$ and $f(x)+f(x+1)>0$. – Nonliapunov Sep 25 '12 at 11:34

The first case is obvious.

For the second, note that \begin{eqnarray*} \int_{-\infty}^{\infty} f(x)dx &=& \frac{1}{2} \int_{-\infty}^{\infty}f(x)+f(x)dx\\ &=& \frac{1}{2}\left[\int_{-\infty}^{\infty}f(x)dx + \int_{-\infty}^{\infty}f(x)dx\right]\\ &=& \frac{1}{2}\left[\int_{-\infty}^{\infty}f(x)dx + \int_{-\infty}^{\infty}f(y+1)dy\right]\\ &=& \frac {1}{2}\left[\int_{-\infty}^{\infty}f(x)+f(x+1)dx\right] > 0, \end{eqnarray*} where the third equality follows from the change of variables $y := x-1$ and the last inequality is a direct consequence of $f(x)+f(x+1) > 0 \ \ \forall x \in \mathbb{R}.$

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If for example you have a particular function like this that is positive when $x>0$ and negative when $x<0$, just split the integral in two at zero. Where $f(x)>0$, it's obvious. For $\int_{-\infty}^{0} f(x)dx$ just integrate your inequality: $$\int_{-\infty}^{0} f(x)dx+\int_{-\infty}^{0} f(x+1)dx>0$$ Then change variables u=x+1 and see that you now have $$2\int_{-\infty}^{0} f(x)dx>0$$.

the general case just need you to split $R$ as many times as $f$ changes its sign.

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It is obvious that $\int_{\mathbb{R}}f(x)\, dx \geq 0$. The strict inequality is easy, but you need to exploit the assumption of continuity. – Siminore Sep 25 '12 at 11:37

The first is obviously true. The second I feel uneasy with; to give a totally random example: $$f(x)=\sin\pi x+\frac{1}{2}$$

This fills the condition $f(x)+f(x+1)>0$ but the integral $\displaystyle\int_{-\infty}^{\infty} f(x)\;dx$ fails to exist

(not sure if the word "integrable" disregards divergent improper integrals - please correct me if this is off-track)

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It was assumed that $f$ is integrable on $\mathbb{R}$. – Nonliapunov Sep 25 '12 at 11:55
@user38773 Sorry. I thought integrable on $\mathbb{R}$ meant integrable for bounded intervals of $\mathbb{R}$. I didn't know $\sin$ and $\cos$ weren't "integrable on $\mathbb{R}$". Thanks for correcting! – user39572 Sep 25 '12 at 11:59

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